Measurement *
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Measurement * Copyright 2008 Some rights reserved. * For Review
This material is licensed under the Creative Commons Attribution-Share Alike license (http://creativecommons.org/licenses/by-sa/3.0/). Print date:2009-04-16 12:05 CK12 psn: 77e64602d563c25d9b52509cb31b6faf
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Contents 1. Geometry and Measurement........................................................................................................... 4
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1. Geometry and Measurement
AUTHOR: Sarah Brockett SOURCE: Math Grade 6 – 1st Evolution LICENSE: CCSA
Lesson 1: Area of Parallelograms Learning Objectives
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Recognize the relationship between the area of parallelograms and rectangles.
•
Find the area of parallelograms.
•
Find unknown dimensions of parallelograms.
•
Solve real-world problems involving area of parallelograms.
Introduction In this lesson we will explore the area of parallelograms. A parallelogram is a figure that has two pairs of parallel sides. The figures below are parallelograms.
[art 10.1.Ia: slanted parallelogram, rectangle] You can see that opposite sides are parallel in a parallelogram. It doesn’t matter what the angles are, as long as there are two pairs of parallel sides. Area, as we know, is the amount of space a figure covers. In this lesson we will learn how to find the area of parallelograms using a formula. Then we will practice finding the area of parallelograms in real-world situations.
Recognizing the Relationship between the Area of Parallelograms and Rectangles As we saw in the example above, rectangles are parallelograms because they have two pairs of parallel sides. They are special parallelograms because they have four right angles. Therefore we can use what we know about finding the area of rectangles to write a formula for finding the area of parallelograms.
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Look closely at the parallelogram below. Is there a way we can transform it into a rectangle?
What if we cut off one end of the parallelogram to make a triangle? If we move it to the other side, we make a rectangle.
[art 10.1.1a: two parallelograms refigured] Notice that the total area remains the same when we rearrange the parallelogram to make a rectangle. Now let’s think about finding the area of a rectangle. What is the formula we use? A=lxw In the formula, l represents the length of the rectangle and w represents the width. Now let’s look at the parallelogram. Which sides does it share with the rectangle? When we transformed the parallelogram, the length of the long side did not change. Therefore we can use this side in the formula to find the area of a parallelogram as well. We call it the base of the parallelogram. What about the height? The width of the rectangle is always perpendicular to the length. In a parallelogram, however, this is not always true. The angle at which the sides meet can vary. We need something constant in order to write a formula. Notice that the dashed line we drew to “cut off” a triangle from one end of the parallelogram is equivalent to the width of the rectangle we created. We can use this measurement to find the area. We call it the height of the parallelogram, rather than the width. This gives us the formula for the area of a parallelogram: A=bxh As you can see in the parallelograms below, it doesn’t matter how slanted the parallelograms are. We can always use their base and height.
[art 10.1.1b various parallelograms]
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Now let’s practice using the formula to find the area of some parallelograms
Finding the Area of Parallelograms As we’ve learned, we can find the area of a parallelogram when we know the length of its base and height. Let’s practice using the formula for area. Example 1 Find the area of the parallelogram below.
[art 10.1.2a: parall. with dashed height=2 in., base =3.5 in.] We can see that the base is 3.5 inches and the height is 2 inches. We simply put these numbers into the appropriate places in the formula. A = bh A = 3.5(2) 2
A = 7 in.
Remember that we always measure area in units squared. The area of this parallelogram is 7 square inches. Let’s try another. Example 2 What is the area of the parallelogram below?
[art 10.1.2b: parall. with dashed height=4.5 cm, base =6.5cm] Again, we are given the base and the height, so we put these numbers into the formula and solve.
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A = bh A = 6.5(4.5) 2
A = 29.25 cm
The area of this parallelogram is 29.25 square centimeters. Great job! Now let’s use the formula for area in other ways.
Finding Unknown Dimensions of Parallelograms Sometimes a problem will give us the area and ask us to find one of the dimensions of the parallelogram—either its base or its height. All we have to do is put the numbers we know into the formula and solve for the missing piece. Let’s try an example. Example 3 2
A parallelogram has an area of 75 m . The base of the parallelogram is 5 m. What is its height? In this problem, we know the area and the height of the parallelogram. We put these numbers into the formula and solve for the height, h. A 75
=bh =5h
75 ÷ = h 5 15 m = h By solving for h, we have found that the height of the parallelogram is 15 m. Let’s check our calculation to be sure. We can check by putting the base and height into the formula and solving for area: A = bh A = 5(15) 2
A = 75 m
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We know the area is 75 m , so our calculation is correct. Nice work! Let’s try another. Example 4 The area of a parallelogram is 143 square yards and its height is 11 yards. Find its base. This time we know the area of the parallelogram and its height. We can put these into the formula and solve for the base, b. A 143
= bh = 11b
143 ÷ = b 11 13 yd = b We have found that the parallelogram has a base of 13 yards. Again, let’s use the formula to check our work. 7
A = bh A = 13(11) 2
A = 143 yd
Our calculation is correct! Whenever we are given two pieces of information about a parallelogram, we can use the formula for area to find the third measurement.
Solving Real-World Problems Involving Area of Parallelograms We have seen that we can apply the formula for finding the area of parallelograms to different kinds of situations. Sometimes we need to solve for the area, but other times we may need to find the height or the base. We can also use this formula when we are given real measurements. Let’s try a few problems involving parallelograms in the real world. Example 5 Mr. Chen is building a deck in the shape of a parallelogram. One side will be 14 feet long. Mr. Chen measured a line perpendicular to the 14-foot side and it was 12 feet. What will be the area of the deck? Let’s begin by figuring out what the problem is asking us to find. We need to find the area of the deck, so we will use the area formula to solve for A. In order to use the formula, we need to know the base and height of the deck. We know that one side of the deck will be 14 feet. Let’s call this the base. We also know that Mr. Chen made a perpendicular line in order to measure the height. The height given in the problem is 12 feet. Let’s put this information into the formula and solve for area. A = bh A = 14(12) 2
A = 168 ft
The area of Mr. Chen’s deck will be 168 square feet. Let’s move on. Example 6 The front wall of the Seaside Hotel forms a parallelogram. The owner wants to paint the front orange. The wall has a base of 37 yards and a height of 22 yards. If one can of paint can cover 50 square yards, how many cans of paint must the owner buy? This problem is a little different, so read carefully. What is it asking us to find? We need to find the number of cans of paint necessary to cover the whole wall. In order to figure this out, though, we first need to find the area of the wall. What information have we been given? We know that the base is 37 yards and the height is 22 yards, so let’s put this information into the formula: A = bh A = 37(22) 2
A = 814 yd
The area of the Seaside Hotel’s front wall is 814 square yards. But we’re not done yet! Remember, we need to find a number of cans of paint. What information have we been given about the paint? We know that one can of paint covers 50 square yards. To find the number of cans of paint necessary to cover the whole wall, 8
we need to divide the area by 50. 814 ÷ 50 = 16.28. The owner therefore must buy 17 cans of paint to cover the area of the wall. Let’s try one more. Example 7 Claire made a flag for her scout troop in the shape of a parallelogram. She used 89.25 square feet of fabric. If the height of the flag is 8.5 feet, what is its base? First, let’s make sure we understand the question in the problem. We need to find the base of the parallelogram. This time we are given the area and the height. As we’ve seen, we simply put these numbers into the formula and solve for the base, b. A 89.25 89.25 8.5 10.5 ft
= bh = b(8.5) ÷ =b =b
The base of the flag is 10.5 feet. As long as we have any two pieces of the area formula, we can solve for the third. We have now had a good look at finding the area of parallelograms. Specifically, we have learned:
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that the formula for the area of a parallelogram is derived from the formula for the area of a rectangle
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to use the formula to find the area of a parallelogram
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to use the formula to find unknown dimensions in a parallelogram
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to solve real-world problems using the formula
We have seen that parallelograms and rectangles are related. We can move the triangles at the ends of a parallelogram to make a rectangle, and then apply a formula for area. That formula is A = bh. With this formula, we can solve for area. We simply fill in the numbers for the base and height. We can also solve for either the base or height if we are given the area. As long as we have two of the three pieces of information that go into the area formula, we can use it to find the missing piece. This is also true for problems involving real-world situations. The area formula does not change. We need to make sure we understand what the problem is asking us to find. When we find the facts given in the problem, we put them into the formula to solve for the missing fact.
Homework Find the area of the parallelograms below. [art 10.1.HPa: 1-6: 4in/6in, 7m/13m, 2.5ft/11ft, .5yd/9 yd, 120cm/50cm, 4.3 in./12in]
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1. 2
24 in.
2. 2
91 m
3. 2
27.5 ft
4. 2
4.5 yd
10
5. 2
6,000 cm
6. 2
51.6 in
7. A parallelogram has an area of 112 square centimeters. If its height is 8 cm, what is its base? 14 cm 8. What is the height of a parallelogram whose base is 31 inches and area is 527 square inches? 17 in. 9. Jaime is stitching a parallelogram patch into a quilt. The base is 14.3 cm and the height is 12.2 cm. What 2
is the area of the patch? 174.46 cm
10. John is planting grass in a patch of lawn that is shaped like a parallelogram. The height of the parallelo2
gram is 12 feet. The other border is 25 feet. How many square feet of grass will John plant? 300 ft
11. Ravi is decorating his room by painting shapes on his walls. He paints a red parallelogram with an area of 20 square inches and a height of 4 inches. He paints a blue parallelogram with an area of 30 square inches and a height of 3 inches. Which parallelogram has the longer base? By how many inches is it longer? blue; 5 inches 12. Mrs. Johnson bought a mirror in the shape of a parallelogram. The area of the mirror is 9,690 square centimeters. If its base is 114 centimeters, what is its height? 85 cm
Lesson 2: Area of Triangles Learning objectives
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Recognize the relationship between the area of triangles and rectangles.
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Find the area of triangles.
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Find unknown dimensions of triangles.
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Use triangles to find the area of larger figures.
Introduction In this lesson we will explore the area of triangles. Triangles have three sides and three angles. They can be acute, obtuse, or right. They can have three equal sides (equilateral), two equal sides (isosceles), or no equal sides. Area, as we know, is the amount of space a figure covers. In this lesson we will learn how to find the area of triangles using a formula. Being able to find the area of a triangle can often come in handy in geometry. As we’ll see in this lesson, we can divide figures into triangles in order to calculate their area. Let’s see how this works by first understanding the formula for the area of triangles.
Recognizing the Relationship Between the Area of Triangles and Rectangles Area, as we have said, is the amount of space a figure covers. To find area, we multiply the dimensions, or sides, of the figure. The simplest example of this is finding the area of a rectangle. How many squares does the rectangle below cover?
[art: 3 (width) x 4 (length) rectangle on grid paper. side marked w and length marked l] We can count the squares to see that the rectangle covers 12 squares. We can also think of the squares as rows and columns. There are 4 columns of squares and 3 rows. If we multiply these, we also get 12. Therefore, we can say that the formula for a rectangle is the number of columns, or the length (l), times the number of rows, or the width (w): A = l(w) Now we can use this formula and the basic idea of area of rectangles to make a formula for finding the area of triangles. Look again at the rectangle. Is there a way we can draw a line and divide the rectangle into triangles?
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[art: same rectangle with diagonal line and right angle marks in two triangles] Whenever we draw a diagonal through a rectangle, we create two congruent triangles. Together, they make a rectangle. Therefore the area of each triangle must be half the area of the whole rectangle. The area formula for triangles is A = (½) bh In the formula, b stands for “base.” The base of a triangle is the same as the length of a rectangle. When dealing with triangles, we say “height” instead of “width.” In the formula,h stands for “height.”
[art 10.2.1a: rectangle with l and w marked and area formula below; rectangle divided into triangle with b and h and formula below] There is one other thing to remember when finding the area of triangles. The height of a triangle is always perpendicular to the base. The height is not necessarily a side of the triangle; this happens only in right triangles, because the two sides joined by a right angle are perpendicular. Let’s look at the height of some other triangles.
[art 10.2.1b: various triangles with dashed height line, including one that extends past acute slanted triangle] We can still use the formula to find the area of any of these triangles. We just have to be careful that we use the height in the formula. As long as we are given the base and the height of a triangle, we can use this formula to find the area, no matter what shape or size the triangle is. Let’s see how this works.
Finding the Area of Triangles As we’ve learned, we can find the area of a triangle when we know the length of its base and its height. Let’s practice using the formula for area. Example 1 Find the area of the triangle below.
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[art 10.2.2a: right tri with height= 6 cm., base =14 cm] We can see that the base is 14 centimeters and the height is 6 centimeters. We simply put these numbers into the appropriate places in the formula. A = (½)bh A = (½)14(6) A = (½)(84) ²
A = 42 cm
Remember that we always measure area in units squared. The area of this triangle is 42 square centimeters. Let’s try another. Example 2 What is the area of the triangle below?
[art 10.2.2b: triangle with dashed height down center of isosc= 5 cm, base =7cm] Notice that the height is shown by the dashed line. It is perpendicular to the base. We put it and the base into the formula and solve. A = (½)bh A = (½) 5(7) A = (½) (35) ²
A = 17.5 cm
The area of this parallelogram is 17.5 square centimeters. Great job! Let’s try one more. 14
Example 3 Find the area of the triangle below.
[art 10.2.2c: triangle with dashed height down side of scalene= 2.5 yd, base =5 yd] A = (½)bh A = (½) 2.5(5) A = (½) (12.5) 2
A = 6.25 yd
This triangle has an area of 6.25 square yards. Now that we are familiar with the area formula, let’s try using it in other ways.
Finding Unknown Dimensions of Triangles Sometimes a problem will give us the area and ask us to find one of the dimensions of the triangle—either its base or its height. All we have to do is put the numbers we know into the formula and solve for the third. Let’s try an example. Example 4 2
A triangle has an area of 28 m . The base of the triangle is 8 m. What is its height? In this problem, we know the area and the height of the triangle. We put these numbers into the formula and solve for the height, h. A 28
=(½) bh =(½)8h
2 8 / =8h (½) 28(2/1) =8h 56 =8h 7
=h
Remember, when you divide both sides by a fraction, you need to multiply by its reciprocal. To divide by (½), then, we multiply by 2. Keep this in mind when you use the area formula. By solving for h, we have found that the height of the triangle is 7 m. Let’s check our calculation to be sure. We can check by putting the base and height into the formula and solving for area: A = (½)bh A = (½) 8(7) 15
A = (½) (56) ²
A = 28 m
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We know from the problem that the area is 28 m , so our calculation is correct. Nice work! Let’s try another. Example 5 The area of a triangle is 96 square yards and its height is 12 yards. Find its base. This time we know the area of the triangle and its height. We can put these into the formula and solve for the base, b. A= 96 =
(½) bh
(½) 12b 96 / (½) 12b = 96 (2/1) 12b = 192 = 12b 16 = b We have found that the triangle has a base of 16 yards. Again, let’s use the formula to check our work. A = (½) bh A = (½) 12(16) A = (½) (192) ²
A = 96 yd
Our calculation is correct! Whenever we are given two pieces of information about a triangle, we can use the formula for area to find the third measurement.
Using Triangles to Find the Area of Larger Figures We mentioned in the Introduction that we can use triangles to find the area of larger figures. Remember that we could divide a rectangle into two triangles? If we know the area of the triangles, we can add their areas together to find the area of the rectangle. We can do this for all kinds of figures. If we can divide the figure into triangles, we can find the area of each triangle and add the areas together. Also, we have seen that we can apply the formula for finding the area of triangles to different kinds of situations. Sometimes we need to solve for the area, but other times we may need to find the height or the base. We can use information given about the figure whenever that information corresponds to the height, base, or area of one of the triangles contained with in it. Let’s try an example to get a better idea of how this works. Example 6 Find the area of the figure below.
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[art: 10.2.4a: trapezoid with measurements and dashed height line] We need to find the area of the whole figure. To do so, we can divide into shapes whose area formulas we know. We can see that one triangle has been drawn already. Can we draw another to divide the figure again?
[art: 10.2.4b: trapezoid with measurements and two dashed height lines] Now we have divided the figure into a rectangle and two triangles. If we can find the area of each of these, we can add them together to find the area of the whole figure. Let’s give it a try. First, let’s calculate the area of the rectangle in the center. We know the formula for the area of rectangles is A = lw. Do we know the length and width of the rectangle? The width is represented by the dashed line. It is 6 inches. The length of the rectangle is the same as the top edge of the figure: 9 inches. Let’s put these numbers into the formula and solve for area: A = lw A = 9(6) 2
A = 54 in.
Now let’s find the area of one of the rectangles. We know that the height is 6 inches. What is the base? Look carefully at the figure. We know that its bottom edge is 15 inches. We also know that the length of the rectangle in the middle is 9 inches.
[art: reprint figure but highlight bottom edge of rectangle and add 9 in. label] That means there are 15 – 9 = 6 inches left of the bottom edge. If we divide this equally, we find that each triangle has a base of 3 inches.
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[art: reprint figure, label 3 in., 9 in., 3 in. along the bottom for each internal shape] Now we have the height and base of each triangle (they have the same height and base), so we can calculate the area: A = (½) bh A = (½) 3(6) A = (½) (18) ²
A = 9 in.
Great! Now we have the area of each shape within the figure. All we have to do is add these together to find the area of the whole figure. triangle 1 2
9 in.
rectangle triangle 2 whole figure +54 in.2 +9 in.2 =72 in.2
The area of the whole figure is 72 square inches. Let’s try another. Example 7 What is the area of the figure below?
[art 10.2.4c: rectangle/triangle] This time we need to divide the figure into smaller shapes ourselves. Can you draw a line to make a triangle on the left side?
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[reprint shape with dashed line dividing it into a triangle and rectangle] Now we have a triangle and a rectangle. Let’s see if we have all the measurements we need to use the area formulas for these two shapes. We know that the opposite sides of a rectangle are congruent, so we can fill in the measurements for the other two sides. This gives us the height of the triangle. We still need to find its base, however. Look at the bottom edge of the triangle. We know its entire length is 8 centimeters, and of that 2 centimeters are taken up by the rectangle. That leaves 8 – 2 = 6 centimeters for the base of the triangle.
[art 10.2.4d: shape with all measurements] Now let’s calculate the area of each smaller shape. Rectangle A = lw A = 2(4) 2
A = 8 cm
Triangle A = (½) bh A = (½) 6(4) A = (½) (24) 2
A = 12 cm
Great! We have found the area of each smaller shape. Let’s add them together to find the area of the whole figure.
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rectangle 2
8 cm
triangle whole figure +12 cm2 =20 cm2
The area of the whole figure is 20 square centimeters. Nice job! We have now had a good look at finding the area of triangles. Specifically, we have learned:
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that the formula for the area of a triangle is derived from the formula for the area of a rectangle
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to use the formula to find the area of a triangle
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to use the formula to find unknown dimensions in a triangle
•
to use triangles to find the area of larger figures
We have seen that triangles and rectangles are related. We can divide rectangles into two triangles by drawing a diagonal line. This gives us the formula for the area of a triangle: A = (½) bh. With this formula, we can solve for area. We simply fill in the numbers for the base and height. We can also solve for either the base or height if we are given the area. As long as we have two of the three pieces of information that go into the area formula, we can use it to find the missing piece. We can also use the area formula for triangles to find the area of larger figures. We do this by dividing the figure into smaller shapes such as triangles and rectangles. If we can find the area of each smaller shape, we can add their areas together to find the area of the whole figure. This is an important concept in geometry, so remember to look for triangles within other figures!
Homework Problems Find the area of the triangles below.
[see art scrap 10.2.HPa: 1-6: 3in/8in, 9m/10m, 1.5ft/13ft, .5yd/6 yd, 110cm/60cm, 2.7 in./12in]
1. 20
2
12 in.
2. 2
45 m
3. 2
9.75 ft
4. 2
15 yd
5. 2
3,300 cm
21
6. 2
16.2 in
7. Duane drew a triangle that had a base of 7 inches and a height of 3 inches. What is the area of the triangle Duane drew? 2
10.5 in.
8. A triangle has an area of 144 square centimeters. If its height is 9 cm, what is its base? 32 cm 9. What is the height of a triangle whose base is 26 inches and area is 234 square inches? 18 in. 10. Tracy is planting flowers in a triangular shaped garden. The height of the triangle is 22 feet. The other border is 37 feet. How many square feet of flowers will Tracy plant? 2
407 ft
Find the area of each figure below.
11. [10.2.HPb: shape with internal lines given] 2
9 ft
12. 2
162 cm 22
Lesson 3: Circumference of Circles Learning Objectives
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Recognize that pi is the ratio of circumference to diameter.
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Find the circumference of circles.
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Find unknown dimensions of circles.
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Solve real-world problems involving circumference of circles.
Introduction In this lesson we will begin to explore circles. A circle is a figure whose edge is made of points that are all the same distance from the center. Imagine a drop of water hitting a lake. A ripple spreads equally in every direction from the point where the drop hit the surface of the lake. Circles are a measurement of that distance the ripple travels from the center. Let’s look at some circles.
[art 10.3.Ia: circles with radii in several places. 1st 2 cm, last 4 ft] You can see that the distance from the center of a circle to its edge is always the same, no matter where we try to measure it. We call this distance from the center to a point on the circle the radius. In first circle above, the radius is 2 centimeters. In the last circle, it is 4 feet. The diameter of a circle is the distance across it through the center. The diameter divides a circle into two equal halves. It is also twice as long as the radius. Can you see why?
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[art 10.3.Ib: circle with diameter marked. underneath write d = 2r] Circumference is the distance around a circle. It varies in size depending on how long the radius is, or far the edge of the circle is from the center. Can you see why the circumference of the first circle is much larger than the circumference of the second?
[art 10.3.Ic: large and small circle with radius marked and arrow around to show circumference] Now we know the basics of circles: the radius, the diameter, and the circumference. Let’s look more closely at how they are related.
Recognizing Pi as the Ratio of the Circumference to the Diameter Let’s recall the example of the ripple spreading out from a drop hitting a lake. As the ripple expands, the measurements of the circle change proportionally. In other words, as the diameter of the circle grows, the circumference of the circle grows at the same rate.
[art 10.3.1a: two circles. first d = 2, under write c = 6.28. second d = 4, c = 12.56] Notice that the second circle is exactly twice as large as the first, because its diameter is twice as long. Now compare their circumferences. The circumference of the second circle is also exactly twice as big. However the diameter of the circle changes, the circumference of the circle must change exactly the same way. We express this proportional relationship as a ratio. A ratio simply means that two numbers are related to each other. Circles are special in geometry because this ratio of the circumference and the diameter always stays the same. We can see this when we divide the circumference of a circle by its diameter. No matter how big or small the circle is, we will always get the same number. Let’s try it out on the circles above. circumference/diameter =6.28/2 =3.14 circumference/diameter =12.56/4 =3.14
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Even though we have two different circles, the result is the same! Therefore the circumference and the diameter always exist in equal proportion, or a ratio, with each other. This relationship is always the same. Whenever we divide the circumference by the diameter, we will always get 3.14. We call this number pi, and we represent it with the symbol π . Pi is actually a decimal that is infinitely long—it has no end. We usually round it to 3.14 to make calculations easier. Using the equations above, we can write a general formula that shows the relationship between pi, circumference, and diameter. C/d =π If we rearrange this formula, we can also use it to find the circumference of a circle when we are given the diameter: C =π d We can use this formula to find the circumference of any circle. Remember, the number for π is always the same: 3.14. We simply multiply it by the diameter to get the circumference. Let’s try it.
Finding the Circumference of Circles As we’ve learned, we can find the circumference of a circle when we know its diameter. We simply put the diameter into the formula and multiply by pi, 3.14. Example 1 Find the circumference of the circle below.
[art 10.3.2a: circle with d = 6 in.] We can see that the diameter of the circle is 6 inches. Let’s put this number into the formula. C= π d C = π (6) C = 18.84 in. The circumference of a circle that has a diameter of 6 inches is 18.84 inches. In other words, if we could unroll the circle into a flat line, it would be 18.84 inches long. Let’s try another.
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Example 2 What is the circumference of the circle below?
[art 10.3.2b: circle with d = 3.2 m] Again, we know the diameter, so we put it into the formula and solve. C= π d C = π (3.2) C = 10.048 m We can round this number up and say that a circle with a diameter of 3.2 meters has a circumference of approximately 10.05 meters. Good job! Let’s try one more. Example 3 What is the circumference of the circle below?
[art 10.3.2c: circle with r = 7 cm] Ah! This is a bit tricky. Notice that the diagram gives us the measure of the radius, not the measure of the diameter. Can we still solve for circumference? Sure we can, as long as we remember that the radius is exactly half the length of the diameter. We simply have to multiply the radius by 2. C= π d 26
C = π (7 x 2) C = π (14) C = 43.96 cm We can also write the formula for circumference like this: C=2π r It means the same thing as our other formula. This one just makes it easier to use the radius when it is given instead of the diameter. Let’s try it to be sure. C=2π r C = 2 π (7) C = 14 π C = 43.96 cm We can use either of these formulas to find the circumference of a circle. Now let’s look at some other ways we can use the formula.
Finding Unknown Dimensions of Circles Sometimes a problem will give us the circumference of a circle and ask us to find either its diameter or its radius. We can still use the formula for circumference; we just need to rearrange it a bit. All we have to do is put the numbers we know into the formula and solve for the missing piece. Let’s try an example. Example 4 A circle has a circumference of 15.7 m. What is its diameter? In this problem, we are given the circumference and we need to find the diameter. We put these numbers into the formula and solve for d. C 15.7
=π d = π d
15.7 ÷ π = d 5 =d By solving for d, we have found that the diameter of the circle is 5 meters. Let’s check our calculation to be sure. We can check by putting the diameter into the formula and solving for the circumference: C= π d C = π (5) C = 15.7 m We know the circumference is 15.7 meters, so our calculation is correct. Nice work! Let’s try another. Example 5 The circumference of a circle is 204.1 yards. Find its radius. Again, we have been given the circumference. Read carefully! This time we need to find the radius, not the diameter. We can use the formula for radii and solve for r.
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C 204.1
=2π r
204.1
= 6.28r ÷ =r
204.1 6.28 32.5 yd
=2π r
=r
We have found that the circle has a radius of 32.5 yards. This time let’s try checking our work by using the other formula to find the diameter. Remember the diameter is twice the length of the radius. C 204.1
= π d = π d
204.1 ÷ π = d 65 yd =d We have found that the diameter of the circle is 65 yards. The radius must be half this length, or 65 ÷ 2 = 32.5 yards. Our calculation is correct! Whenever we are given the circumference, we can use the formula to solve for the diameter or the radius. The number for pi always stays the same, so we only need one piece of information about a circle to find the other measurement.
Solving Real-World Problems Involving Circumference of Circles We have seen that we can apply the formula for finding the circumference of circles to different kinds of situations. Sometimes we need to solve for circumference, but other times we may need to find the diameter or the radius. We can also use this formula when we are given real measurements. Let’s try a few problems involving circles in the real world. Example 6 Carlos and his family ordered a 16-inch pizza. What is the circumference of the pizza? [art: pizza with diameter drawn over and length shown] Let’s begin by figuring out what the problem is asking us to find. We need to find the circumference of the pizza, so we will use the formula to solve for C. In order to use the formula, we need to know either the diameter or the radius of the pizza. The problem tells us that the diameter of the pizza is 16 inches. Let’s put this information into the formula and solve for the circumference. C= π d C = π (16) C = 50.24 in. The circumference of the pizza is 50.24 inches. Good job! Let’s move on. Example 7 Shawna wants to paste some ribbon around a circular mirror to make a border. If the mirror is 40 centimeters across, how many centimeters of ribbon will she need? What is this problem asking us to find? We need to find how much ribbon is necessary to go around the mirror. Therefore we need to find the circumference of the mirror. We know that it is 40 centimeters across. In other words, the diameter of the mirror is 40 centimeters. Let’s put this information into the formula and solve for the circumference.
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C= π d C = π (40) C = 125.6 cm The circumference of the mirror is 125.6 centimeters, so Shawna will need 125.6 centimeters of ribbon to put around it. That wasn’t bad! Let’s try one more. Example 8 The circumference of Jesse’s bicycle wheel is 78.5 inches. What is the length of each spoke in the wheel? [art: bicycle wheel with spokes set at odds so none are across from each other so that no diameter is shown] First, let’s make sure we understand the question in the problem. We need to find the length of the spokes in the wheel. Look at the picture. Each spoke is a radius of the wheel, so we need to find the radius of a circle whose circumference is 78.5 inches. This time we use the formula to solve for the radius, r. C
=2π r
78.5
=2π r
78.5
= 6.28r 78.5 ÷ =r 6.28 12.5 in. =r The radius of the circle is 12.5 inches, so each spoke in the wheel is 12.5 inches long. We can check our answer by finding the circumference: C=2π r C = 2 π (12.5) C = 25 π C = 78.5 in. This is the circumference given in the problem, so we know our calculation is correct. Great work! In this lesson, we have investigated the properties of circles. Specifically, we have learned:
•
that the circumference and the diameter exist in relation to one another, and this ratio is represented by pi
•
to use the formula to find the circumference of a circle
•
to use the formula to find unknown dimensions in a circle
•
to solve real-world problems using the formula
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We have seen that the circumference of a circle and the diameter are related. We use pi ( π ) to represent this relationship. Pi is an infinite decimal, so we usually use the number 3.14. We can use pi to write a formula for the circumference of a circle. That formula is C = π d. This is equivalent to C = 2 π r, which we can also use to find the circumference. This is because the diameter is always twice the length of the radius: d = 2r. With these formulas, we can solve for the circumference. We simply fill in the number for the diameter or the radius. We can also solve for the diameter or radius if we are given the circumference. As long as we have one piece of information about the circle that goes into the formula, we can use it to find the missing piece. This is also true for problems involving real-world situations. The circumference formula does not change. We need to make sure we understand what the problem is asking us to find. When we find the facts given in the problem, we put them into the formula to solve for the missing fact.
Homework Find the circumferences of the circles below. [art 10.3.HPa: 1-4: d = 8 cm, r = 6.5 in, d = 42 yd, r = 17 ft]
1. 25.12 cm
2. 40.82 in.
3. 131.88 yd
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4. 106.76 ft 5. What is the circumference of a circle whose diameter is 22.5 meters? 70.65 m 6. A circle has a radius of 60 centimeters. What is its circumference? 376.8 cm 7. What is the circumference of a circle whose radius is 7.5 feet? 47.1 ft 8. What is the diameter of a circle whose circumference is 172.7 inches? 55 in. 9. A circle has a circumference of 628 centimeters. What is the radius of the circle? 100 cm 10. Trisha is stitching a circular pillow. The diameter of the pillow is 14 inches. What is the circumference of the pillow? 43.96 cm 11. Farmer Graff is building three circular pens for his pigs. Each pen has a diameter of 26 feet. How much fence will he need to buy to surround all three pens? 244.92 ft 12. Kelly baked a cake in a 9-inch cake pan. What is the circumference of the cake? 28.26 in. 13. The circumference of a circular rug is 125.6 feet. What is the radius of the rug? 20 ft 14. Terrence swam around the edge of a circular pool and found that it took him 66 strokes to swim one complete time around the pool. About how many strokes will it take him to swim across the pool? 21 strokes
Lesson 4: Area of Circles Learning Objectives
•
Use a formula to find the area of circles.
•
Find the radius or diameter of circles given area.
•
Find the area of complex figures involving parts of circles.
Introduction In this lesson we will learn to find the area of circles. Area is the amount of flat space a figure or object covers. Circles are unique shapes. A circle is the set points that are equidistant from a center point. 31
[art: circle with center point labeled “center” and outline labeled “set of points”] The radius of a circle is the distance from the center to any point on the circle. The diameter is the distance across the circle through the center. The diameter is always twice as long as the radius. Can you see why?
[art: circle with radius marked and labeled “radius (r)”. circle with diameter marked in same place as radius on other circle, labeled “diameter (d)”] We also use a special number when dealing with circle calculations. We call that number pi. Pi is a decimal that is infinitely long (3.14159265...), but in our calculations we round it to 3.14. We use the symbol π to represent this number. Pi is the ratio of the circumference, or distance around a circle, to the diameter. In other words, these two measurements are related. If we change the diameter, the circumference changes proportionally. Now let’s look at how we can use these measurements, π and the radius, to find the area of a circle.
Finding the Area of a Circle We can use a formula to find the area of any circle. That formula is 2
A = π r or A = π x r x r We already know that the symbol π represents the number 3.14, so all we need to know to find the area of a circle is its radius. We simply put this number into the formula in place of r and solve for the area, A. Let’s try it out. Example 1 What is the area of the circle below? 32
[art: circle with radius labeled r = 2 cm] We know that the radius of the circle is 2 centimeters. We put this number into the formula and solve for A. 2
A= π r
2
A = π (2) A=4π
2
A = 12.56 cm
Remember that squaring a number is the same as multiplying it by itself. The area of a circle with a radius of 2 centimeters is 12.56 square centimeters. We always show area in square units. Let’s try another. Example 2 Some students have formed a circle to play dodge ball. The radius of the circle is 16 feet. What is the area of their dodge ball circle?
[art: circle with radius labeled r = 16 ft] The dodge ball court forms a circle, so we can use the formula to find its area. We know that the radius of the circle is 16 feet, so let’s put this into the formula and solve for area, A. 2
A= π r
2
A = π (16) A = 256 π
2
A = 803.84 ft
Notice that a circle with a larger radius of 16 feet has a much larger area: 803.84 square feet. Great job! Let’s try one more.
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Example 3 What is the area of a circle with a diameter of 25 centimeters? Read the problem carefully! We need to find the area, but what information is given in the problem? This time we know the diameter, not the radius. How can we find the radius so that we can use the area formula? We know that the diameter of a circle is always twice the radius. If the diameter is 25 centimeters, then the radius must be 25 ÷ 2 = 12.5 centimeters. Now we can put this number into the formula. 2
A= π r
2
A = π (12.5)
A = 156.25 π 2
A = 490.63 cm
The area of a circle with a diameter of 25 centimeters (and a radius of 12.5 centimeters) is 490.63 square centimeters. Nice work! Now let’s look at other ways we can use the area formula.
Finding the Radius or Diameter of Circles We have seen that when we are given the radius or the diameter of a circle, we can find its area. We can also use the formula to find the radius or diameter if we know the area. Let’s see how this works. Example 4 The area of a circle is 78.5 square inches. What is its radius? This time we know the area and we need to find the radius. We can put the number for area into the formula and use it to solve for the radius, r. A
= π r
78.5
= π r
2 2
78.5 ÷ π = r2 2 25 =r √25
=r
5 in.
=r
To solve this problem, we need to isolate the variable r. First we divide both sides by π , or 3.14. Then, to remove the exponent, we take the square root of both sides. A square root is a number that, when multiplied by itself, gives the number shown. We know that 5 is the square root of 25 because 5 x 5 = 25. The radius of a circle with an area of 78.5 square inches is 5 inches. We can also use the formula to check if our calculation is correct. If we put in a radius of 5 inches, we should get an area of 78.5 square inches. Let’s try it. 2
A= π r
2
A = π (5) A = 25 π
2
A = 78.5 in.
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It worked, so we know our calculation of 5 inches was correct. Nice job! Let’s try another. Example 5 2
What is the diameter of a circle whose area is 153.86 cm ? What is this problem asking us to find? We need to find the diameter (not the radius!). What information is given in the problem? We know the area. Therefore we can use the formula to solve for the radius, r. Once we know the radius, we can find the diameter. Let’s give it a try. A
= π r
153.86
= π r
2 2
153.86 ÷ π = r2 2 49 =r √49
=r
7 cm
=r
The radius of a circle with an area of 153.86 square centimeters is 7 centimeters. Remember, this problem asked us to find the diameter, so we’re not done yet. How can we find the diameter? The diameter is always twice the length of the radius, so the diameter of this circle is 7 x 2 = 14 centimeters. As we have seen, we can use the area formula whenever we are given information about a circle. If we know the diameter or radius, we can solve for the area, A. If we are given the area, we can solve for the radius, r. If we know the radius, we can also find the diameter. Now that we are familiar with different ways to use the formula, let’s try some more complex calculations.
Finding the Area of Complex Figures We can also calculate the area of a portion of a circle. As long as we know the radius of the circle, we can find its area. Then we can divide that area into smaller pieces or subtract a portion to find the area of part of the circle. Let’s try this out. Example 6 What is the area of the figure below?
[art 10.4.3a: semicircle with diameter of 22 in] This figure is a semicircle, or half of a circle. Remember that a diameter always divides a circle in half. Therefore the edge measuring 22 inches is the circle’s diameter. Can we use it to find the area of the whole circle?
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We sure can! The radius of the circle must be 22 ÷ 2 = 11 inches. Now let’s use the formula to solve for area: 2
A= π r
2
A = π (11) A = 121 π
2
A = 379.94 in.
We know that a whole circle with a radius of 11 inches (and a diameter of 22 inches) is 379.94 square inches. Therefore the semicircle figure has an area of 379.94 ÷ 189.97 square inches. As long as we can find the area of a whole circle, we can divide or subtract to find the area of a portion of a circle. We can also use what we have learned about finding the area of circles and portions of circles to find the area of other complex figures. When we are given complex figures, we can divide them into smaller shapes and use the different formulas for area to find the area of each component. Different geometric shapes have different formulas for area. Let’s review a few: area of a circle:
2
A= π r
area of a rectangle: A = lw area of a square: A = s2 area of a triangle: A = ½ bh Look for ways to divide figures into circles, rectangles, squares, or triangles. Take a look at some examples below.
[art 10.4.3b: shapes divided] In each case, we have found a way to divide the figure into rectangles or squares and portions of a circle. If we can find the area of each shape, we can add the areas together to find the area of the whole figure. Let’s see how this works. Example7 Find the area of the figure below.
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[art 10.4.3c: rect/semi 4 x 5 cm] First, we need to find a way to divide the figure into smaller shapes. Can we draw a line to make a rectangle? We can, like this:
[reprint art 10.4.3c with dashed line dividing rectangle and semicircle] Now we have a rectangle and a semicircle. If we can find the area of these two pieces, we can add them together to find the area of the whole figure. Let’s look at the rectangle first. As we have seen, the area formula for a rectangle is A = lw. Do we know the length and width of the rectangle? We do, so let’s solve for area, A. A = lw A = 5(4) 2
A = 20 cm
The area of the rectangle is 20 square centimeters. Now let’s look at the other shape. The other shape is a semicircle. If we can find the area of the whole circle, we can divide it by 2 to find the area of the half shown here. What information do we need to find the area of a circle? We need to know the radius of the circle. Can we find the radius from the information we have? Let’s look closely. The semicircle has one straight edge where it meets the rectangle. This edge is actually a diameter! Can you see why?
37
[art 10.4.3d: rectangle/circle superimposed] The diameter is the same as one side of the rectangle, so it is 4 centimeters long. The radius therefore must be 4 ÷ 2 = 2 centimeters. Now we can solve for the area of the whole circle: 2
A= π r
2
A = π (2) A=4 π
2
A = 12.56 cm
The area of a circle with a radius of 2 centimeters is 12.56 square centimeters. Remember, only half of the circle makes up part of this figure, so we need to find half the area: 12.56 ÷ 2 = 6.28 square centimeters. Now we know the area of each component of the figure.
[art 10.4.3e: rectangle/circle with areas written in] We add the areas of the components together to find the area of the whole figure: 2
2
2
6.28 cm + 20 cm = 26.28 cm
The figure has an area of 26.28 square centimeters. Nice work! We found the area by dividing the figure into smaller shapes. In this lesson, we have used a formula to find out information about circles. Specifically, we have learned:
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•
to use a formula to find the area of a circle
•
to use the formula to find the radius or diameter of a circle
•
to find the area of complex figures
We have seen that the area of a circle depends on the length of the circle’s radius. The longer the radius, 2
the greater the area the circle has. We can use the formula A = π r to find the area of a circle. We simply fill in the number for the radius. We can also use the formula to solve for the radius or diameter if we are given the area. We put in the number for area and solve for the radius, r. As long as we have one piece of information about the circle that goes into the formula, we can use it to find the missing piece. Finally, we can use the formula to help us find the area of parts of circles and of complex figures. The area of a semicircle, or half circle, for example, is half the area of the whole circle. To find the area of a complex figure, we divide it into smaller shapes. We can use formulas to find the areas of each shape, and then add them together to find the area of the whole complex figure. Homework Find the area of the circles below. [art 10.4.HPa: 1-4: r = 12 cm, d = 9 in, r = 21 yd, d = 14 ft]
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2
1. 452.16 cm 2
2. 63.59 in.
2
3. 1,384.74 yd 2
4. 153.86 ft
2
5. What is the area of a circle whose radius is 15.5 meters? 754.39 m
2
6. A circle has a diameter of 40 centimeters. What is its area? 1,256 cm 2
7. What is the area of a circle whose radius is 6 feet? 113.04 ft
8. What is the radius of a circle whose area is 314 square inches? 10 in. 9. A circle has an area of 530.66 square centimeters. What is the diameter of the circle? 26 cm 2
10. Chloe baked a pie in a 9-inch pie pan. What is the area of the pie Chloe baked? 63.59 in.
11. Ari is painting large polka dots on his wall. He painted 6 polka dots, each with a radius of 2 feet. What 2
is the total area that the polka dots cover? 75.36 ft
12. Josh cut a sheet into a circle to make part of a fort. The area of the circular sheet was 50.24 square feet. What was the radius of the circular sheet he cut? 4 feet
40
Find the area of the figures below. [art 10.4.HPb: complex figures: quarter circle, rectangle MINUS semicircle at one end] 2
13. 50.24 in.
2
14. 39.87 cm
Lesson 5: Classifying Solid Figures Learning Objectives
•
Identify faces, edges, and vertices of solid figures.
•
Classify solid figures as prisms, cylinders, pyramids, cones, or spheres.
•
Select real-world examples of given solid figures.
Introduction In this lesson we will examine solid figures. Solid figures are shapes that exist in three-dimensional space. Unlike plane shapes, which have only length and width, solid figures have length, width, and height. We will learn to identify each kind of solid figure. Let’s take a look at some solid figures.
[art 10.5.Ia: rectangle/ rect. prism with l, w, h marked. square/cube labeled circle/ sphere labeled.] First we will examine the features of solid figures. The number of faces, edges, and vertices a solid figure has tells us what kind of solid figure it is. We can use this information to classify solids. We see solid figures around us every day. Take a look around you. What solids do you see? How many faces or edges do they have? Recognizing and understanding these figures is an important key to doing geometry. Let’s get started! [art: solids in real world: cones, pyramids, spheres, boxes]
Identifying Faces, Edges, and Vertices As we’ve said, solid figures have many faces, edges, and vertices. We classify, or identify, them by the number of each that they have. A face is a flat side of a solid figure. Faces are in the form of plane shapes, such as triangles, rectangles, and squares. Take a look at the faces highlighted below.
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[art 10.5.1a: pyramid with triangle, rect. prism with rectangle, cube with square] Every solid figure has several faces. We can count the number of faces the figure has. How many faces does each figure above have? Let’s look at the first figure. Count the number of triangular faces. There are four. Are there any other faces? There is the base, which is in the shape of a square. That makes five faces. The next figure has a face on the bottom and the top. It has four faces around the sides. Therefore it has six faces in all. The last figure is similar. It also has a face on the bottom and the top, as well as four faces around the sides. It also has six faces. We can also count the edges in solid figures. An edge is the place where two faces meet. Edges are straight; they cannot be curved. How many edges does this figure have?
[art 10.5.1b: pyramid with one edge highlighted and labeled] Count all of the straight edges where two faces meet. This figure has 8 edges. Some figures do not have any edges because they do not have flat sides. The figures below have no edges.
[art: sphere, cone, cylinder] The place where two or more edges meet is called a vertex. A vertex is like a corner. We can also count the number of vertices in order to identify solid figures. This figure has six vertices. Can you find them all?
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[art: triangular prism (triangle base and top, rectangular sides] Now let’s practice finding faces, edges, and vertices. Example 1 How many faces, edges, and vertices does the figure below have?
[art: rectangular prism] Let’s count the faces first. Remember, each face is a flat plane shape. In this figure, the faces are all rectangles. There is one on the top and one on the bottom, plus four around the sides. This figure has six faces. Next let’s count the edges where each rectangular face meets another. There are four around the top rectangle where it meets each side, and four more around the bottom rectangle where it meets each side. And there are four more where each side meets another. This figure has twelve edges. Now let’s find the vertices. Remember, a vertex is like a corner. This figure has four corners, or vertices, on the top and four on the bottom. It has eight vertices in all. Nice work! Now let’s use what we know about faces, edges, and vertices to classify solid figures.
Classifying Solid Figures As we have said, we can use the number of faces, edges, and vertices to classify, or identify, solid figures. Let’s look at each kind of solid figure. A sphere is a solid figure that has no faces, edges, or vertices. This is because it is completely round; it has no flat sides or corners.
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[art: sphere] A cone has one face but no edges or vertices. Its face is in the shape of a circle. Because a circle is a flat, plane shape, it is a face. But because it is round around the outside, it does not form any edges or vertices.
[art: cone. highlight circle and label “face”] A cylinder has two circular faces but also no edges or vertices. Can you see why?
[art: cylinder. highlight both circles and label both “face”] A pyramid has one rectangular base and four triangular sides. Therefore it has five faces. It has edges and vertices where all of its faces meet, since none of its faces are curved. Thus it has eight edges and five vertices.
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[art: pyramid. highlight rectangular base and one side and label both “face”. Label one edge “edge” and point at top “vertex”] A prism is a solid figure that has two parallel bases and any number of sides. In other words, it can have any number of faces, but at least two of them must be parallel. The shape of the two parallel bases can be a triangle, square, rectangle, pentagon, hexagon, or any other kind of polygon. The number of sides the bases have determines the number of faces the figure has. The number of edges and vertices also depends on the shape of the base. Here are some different prisms.
[art 10.5.2a: triangular prism, rectangular, hexagonal. label some faces/bases, edges, and vertices] As you can see, there are many types of prisms. The shape of the bases tells us which kind of prism it is. Triangular prisms have the least possible number of faces, edges, and vertices in a prism because their bases have only three sides. Therefore we can say that a prism must have at least five faces, nine edges, and six vertices as a triangular prism does. Let’s sum up the properties we’ve learned about these solid figures. The table below tells how many faces, edges, and vertices each kind of solid has. [table] Figure
Figure Name Number of Faces Number of Edges Number of Vertices sphere
0
0
0
cone
1
0
0
45
cylinder
2
0
0
pyramid
5
8
5
prism
at least 5
at least 9
at least 6
[end table] Now let’s try classifying some figures. Use the table above to help you. Example 2 Count the number of faces, edges, and vertices in the figure below. Then identify the figure.
[art: cylinder on its side] First, we should look for any faces the figure has. This has two that are parallel bases. Can it be a prism? It cannot, because the bases are circles, not triangles or some other polygon. Which solid has circular bases? A cylinder does. Let’s check the number of edges and vertices to be sure. Are there any? No—because the circular bases are curved, there are no edges and therefore no vertices. Two faces, no edges, no vertices...this must be a cylinder. Let’s try another. Example 3 Count the number of faces, edges, and vertices in the figure below. Then identify the figure
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[art: rectangular prism] Again, let’s look at the faces first. This figure has six faces. Two of them are parallel and congruent. That makes these the bases. What shape are they? This is a prism with rectangular bases, so it is a rectangular prism. Now let’s count the edges and vertices. There are four edges around each base, and four where each of the side faces meet. There are twelve edges in all. There are four vertices on the top and four on the bottom. This rectangular prism has six faces, twelve edges, and eight vertices. Nice work!
Finding Solid Figures Around Us We have learned the names and properties of some important geometric figures in this lesson. Now let’s use what we have learned to find examples of these figures in the real world. Take a look around you...we see spheres, cones, cylinders, pyramids, and prisms around us every day! Let’s try to identify some of these objects. Example 4 Which of the following shows a pyramid? [art: answer choices A-D: a) soccer ball b) box of cereal c) metronome d) tall glass of water] We need to find a pyramid. So first let’s remember what we know about pyramids. A pyramid has only one rectangular base. Can we rule out any of the choices based on this information? Choice A has one base, but it is a circle, not a rectangle, so it cannot be a pyramid. Choices B and D both have two bases. Only choice C has one base. Let’s make sure it is a pyramid by counting its faces, edges, and vertices. It has five faces, 8 edges, and five vertices. Choice C is a pyramid. Good job! Let’s try another. Example 5 Which of the following shows a cone? [art: answer choices A-D: a) triangular tent b) canister of oatmeal c) baseball d) traffic cone] Again, let’s begin by recalling what we know about cones. A cone has one circular face, or base. Choice A definitely has no circles or curves in it, so let’s rule it out. Choice B actually has two circular faces, both the base and the top. It is a cylinder, not a cone. Does choice C have any faces? It does not; it is a sphere. That leaves choice D. It has one circular face, no edges, and no vertices. Therefore it is a cone. You can see why it is important to be able to recognize and classify these objects by their geometric properties. We see and use these objects every day! We have learned all about solid figures in this lesson. Now we can:
•
identify their faces, edges, and vertices
•
classify them by these properties
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•
recognize them in the world around us
We know that solid figures exist in three dimensions, not two like plane figures. This means they have faces, or flat sides, edges where the faces meet, and vertices where the edges meet. We classify, or name, solid figures by the number of their faces, edges, and vertices. For example, any figure with two circular faces, no edges and no vertices is a cylinder. We have seen that these geometric solids exist in the world around us. We can also identify these objects by their faces, edges, and vertices. Homework Count the number of faces, edges, and vertices in each figure. Then name the figure.
1. [art: cylinder] 2 faces, 0 edges, 0 vertices ; cylinder
2. [art: pyramid] 5 faces, 8 edges, 5 vertices; pyramid
3. [art: cone] 1 face, 0 edges, 0 vertices; cone 4. [art: basketball] 0 faces, 0 edges, 0 vertices; sphere 5. [art: cube shaped present] 6 faces, 12 edges, 8 vertices; prism 6. [art: thick cylindrical candle] 2 faces, 0 edges, 0 vertices; cylinder
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7. [art: triangular prism] 5 faces, 9 edges, 6 vertices; prism 8. [art: earth] 0 faces, 0 edges, 0 vertices; sphere 9. [art: conical party hat] 1 face, 0 edges, 0 vertices; cone 10. [art: rectangular fish aquarium] 6 faces, 12 edges, 8 vertices; prism 11. Which of the following objects is a sphere? [art choices A-D: A) cylindrical jar B) marble C) ice cream cone D) light bulb] B 12. Which of the following objects is a prism? [art choices A-D: A) box of crackers B) rectangular table C) roll of paper towels D) pearl necklace] A 13. A figure has one circular face, no edges, and no vertices. What kind of figure is it? cone 14. A figure has five faces, five vertices, and 8 edges. What kind of figure is it? pyramid
Lesson 6: Surface Area and Volume of Prisms Learning Objectives
•
Find the surface area of prisms using nets.
•
Find the surface area of prisms using formulas.
•
Find the volume of prisms using unit cubes.
•
Find the volume of prisms using formulas.
Introduction In this lesson we will learn to find the surface area and volume of prisms. A prism is a solid shape that exists in three-dimensional space. Prisms have faces that give it length, width, and height. Let’s look at some prisms.
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[art 10.6.Ia: rect. prism with l, w, h marked. cube labeled. triangular prism labeled with b not l] The surface area of a prism is the total of the areas of each face. Imagine you could wrap one of the figures above in wrapping paper, like a present. The amount of wrapping paper needed to cover the figure represents its surface area. To find the surface area, we must be able to calculate the area of each face and then add these areas together. [image:math6-10-06-nIa.jpg] [art: cube shaped wrapped present; net for a cube with the same wrapping paper print on it] The volume of a prism is the measure of how much three-dimensional space it takes up or holds. Imagine a fish aquarium. Its length, width, and height determine how much water the tank will hold. If we fill it with water, the amount of water tells the volume of the tank. We measure volume in cubic units, because we are multiplying three dimensions: length, width, and height. We will look at several ways to calculate these two measurements. First let’s learn how to find the surface area of a prism.
Finding Surface Area Using Nets As we’ve said, surface area is the total area for all the faces of a prism. That means we need to find the area of each face of the prism. One way to do this is to use a net. A net is a two-dimensional diagram of a three-dimensional prism. Imagine you could unfold a box so that it is completely flat. You would have something that looks like this:
[art 10.6.1a: rectangle net two shaded; rect. prism with top and bottom shaded] It is as if we have unfolded the box. The shaded rectangles show the top and bottom of the box, and the unshaded rectangles show each side. Can you see how to fold the net to make the box? With the net, we can see each face of the box more clearly. To find the surface area, we need to calculate the area for each rectangle in the net. How do we find the area of a rectangle? We use the formula A = lw. If we know the length and width of each rectangle, we can calculate its area. Let’s put in the measurements for all the rectangles. Remember, each face represents two dimensions of the box: length and width, length and height, or width and height. [art: same rectangle net with all side measurements labeled]
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Now that we have all the information we need, we can calculate the area of each face and then add their areas together. bottom face top face long side long side short side short side 7x2 +7 x 2 +7 x 3 +7 x 3 +3 x 2 +3 x 2 14 +14 +21 +21 +6 +6 = 82 in.² We found the area of each rectangular face and then added all of these areas together. The total surface area of the rectangular prism is 82 square inches. Using a net helped us to locate all of the faces and find the measurements of each side. Let’s try another example. Example 1 What is the surface area of the figure below?
[art: triangular prism with triangles top and bottom] The first thing we need to do is draw a net. Get ready to exercise your imagination! It may help to color the top and bottom faces to keep you on track. Begin by drawing the bottom face. It is a triangle. Each side of the face is connected to a side face. What shape is each side face? They are rectangles, so we draw rectangles along each side of the triangular base. Lastly, we draw the top face, which can be connected to any of the side faces.
[art 10.6.1b: steps: base; base and sides; base, sides, and top. all matching whole triangular prism above] Next let’s fill in the measurements for the sides of each face so that we can calculate their area. Be careful! This time two of the faces are triangles. Remember, we calculate the area of triangles with the formula A = 1/2 bh. We need to know the height of the triangles. [art: reprint full net from 10.6.1b and put all side measurements in, except triangles which have only base and height (dashed line down middle of triangle)] Now that we have the measurements of all the faces, let’s calculate the area of each. Remember to use the correct area formula.
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bottom face top face side side side 1/2(3 x 4) +1/2(3 x +4 x 9 +5 x 9 +5 x 9 4) 6 +6 +36 +45 +45 = 138 cm² We used the formula A = ½ bh to find the area of the top and bottom faces. We used the formula A = lw to find the area of the three side faces. When we add these together, we get a surface area of 138 square centimeters for this triangular prism. Nice work! We can always draw a net to help us organize information in order to find the surface area of a prism. A net helps us see and understand each face of the prism. We can also use a formula to find the area for rectangular and triangular prisms. In the next section we’ll see how.
Finding Surface Area Using Formulas Nets let us see each face so that we can calculate their areas. However, we can also use formulas to represent the faces as we find their area. Let’s look again at our calculations the first prism we dealt with. bottom face top face long side long side short side short side 7(2) +7(2) +7 (3) +7(3) +3(2) +3(2) 2 14 +14 +21 +21 +6 +6 = 82 in. Notice that our calculations repeat in pairs. This is because every face in a rectangular prism is opposite a face that is congruent. In other words, the top and bottom faces have the same measurements, the two long sides have the same measurements, and the two short sides have the same measurements. Therefore we can create a formula for surface area that gives us a short cut. We simply double each calculation to represent a pair of sides. The formula looks like this: SA = 2lw + 2lh + 2hw In this rectangular prism, l = 7 inches, w = 2 inches, and h = 3 inches. We simply put these numbers into the formula and solve for surface area. Let’s try it. SA = 2lw + 2lh + 2hw SA = 2(7)(2) + 2 (7(3)) + 2(3(2)) SA = 2(14) + 2(21) + 2(6) SA = 28 + 42 + 12 2
SA = 82 in.
As we have already seen, the surface area of this prism is 82 square inches. This formula just saves us a little time by allowing us to calculate pairs of sides at a time. Triangular prisms have a different formula because they have two triangular faces opposite each other. Remember, the formula for the area of triangles is not the same as the area formula for rectangles, so we’ll have to proceed differently here to find a formula for surface area. [art: prism with b and h in triangle as in 10.6.1b] First, we know that we need to find the area of the two triangular faces. Each face will have an area of 1/2bh. Since we need the area for the pair, we double the formula to get 2(1/2bh). The numbers cancel each other out, and we’re left with bh. That part was easy!
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Next, we need to calculate the area of each side face. The length of each rectangle is the same as the height of the prism, so we’ll call this H. The width of each rectangle is actually the same as each side of the triangular base. Instead of multiplying the length and width for each rectangle, we can combine this information. We can multiply the perimeter of the triangular base, since it is the sum of each “width” of a rectangular side, by the height of the prism, H. Let’s take a closer look.
[art 10.6.2a: H and sides of triangle s1, 2, 3] If we put these pieces together, the area of the bases and the area of the side faces, we get this formula: SA = bh + (s1 + s2 + s3)H To use this formula, we fill in the base and height of the prism’s triangular base, the lengths of the base’s sides, and the height of the prism. Don’t confuse the height of the triangular base with the height of the prism! Let’s try out the formula. Example 2 What is the surface area of the figure below?
[art 10.6.2b: triangular pris, H=8, h=3, b=8, s2/3= 5] We have all of the measurements we need. Let’s put them into the formula and solve for surface area, SA. SA = bh + (s1 + s2 + s3)H SA = 8(3) + (5 + 5 + 8)8 SA = 24 + 18(8) SA = 24 + 144 2
SA = 168 cm
This triangular prism has a surface area of 168 square centimeters. That wasn’t so bad! We just have to be careful to put each measurement in the right place in the formula. Let’s try another. 53
Example 3 What is the surface area of the figure below?
[art: rectangular prism, l=11, w=4, h=5] Notice that all of the faces of this prism are rectangles. Therefore we can use our first formula for surface area. Let’s put the numbers into the formula and solve. SA = 2lw + 2lh + 2hw SA = 2(11(4)) + 2 (11(5)) + 2(5(4)) SA = 2(44) + 2(55) + 2(20) SA = 88 + 110 + 40 2
SA = 238 in.
This rectangular prism has a surface area of 238 square inches. Nice work! Now let’s find out how to calculate the volume of these prisms.
Finding the Volume of Prisms Using Unit Cubes Volume, as we have said, is the amount of space a three-dimensional solid takes up. One way to find the volume of a prism is to consider how many unit cubes it can contain. A unit cube is simply a cube measuring one inch, centimeter, foot, or whatever units of measurement we are using, on all sides. Here are some unit cubes. [art: 1 cm cube with sides labeled and 1 inch cube labeled] We use unit cubes as a way to measure the space inside a solid figure, or its volume. We simply count the number of unit cubes that “fit” into the prism. We begin by counting the number of cubes that cover the bottom of the prism, and then count each layer. Let’s see how this works.
[art 10.6.3a: empty 3 x 2 x 4(h) prism. same with one layer. same with two layers. same with all layers]
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As you can see, the first layer is 3 cubes long and 2 cubes wide. There are 6 cubes in the first layer. Length and width are only two dimensions, however. Remember that prisms have three dimensions. The third dimension is height. We add this dimension when we add layers. Each layer has 6 cubes. When we add the second layer, then, we have a total of 12 unit cubes. We add 6 more for the third layer to get a total of 18 unit cubes. The volume of this prism is 18 unit cubes. Let’s practice with another example. Example 4 What is the volume of the figure below in unit cubes? [art: rect prism 5 x 3 x 4] Let’s think about the bottom layer first. The measurements tell us that the length is 5 cubes long, so there are 5 cubes in each row of the first layer. How many rows are in this layer? It is 3 cubes wide, so there are 3 rows with 5 unit cubes in each row. Draw a picture of the first layer if you’re not sure. In all, there are 15 unit cubes in the first layer. Each layer will have 15 more unit cubes. How many layers are there in this prism? The prism has a height of 4, so there will be 4 layers of unit cubes. Each time we add a layer, we add 15 unit cubes. This gives us a total of 15 + 15 + 15 + 15 = 60 unit cubes This prism therefore has a volume of 60 unit cubes. Nice job! As you may have guessed, we can also use a formula to find the volume of prisms.
Finding the Volume of Prisms Using Formulas The formula for volume is a simplified version of the method we just learned. As we saw, we used length and width to find the number of unit cubes in the first layer of the prism. This is the same as finding the area of the prism’s base. Let’s see how this works.
[art 10.6.4a: 3x3x5 rectangle, 2x7x3 rectangle with first layer of cubes drawn in] For each of these prisms, we can find the number of unit cubes in the first layer by multiplying the length and the width of the prism. Recall that the formula for the area of a rectangle tells us to do exactly the same thing: A = lw. Once we find the area, we simply multiply it by the height to add the rest of the layers. Therefore the formula for the volume of a rectangular prism is V = Bh B represents the base area of the prism. In other words, we find the area of the face that is the base and then multiply this by the height of the prism. Let’s try it out. Example 5 Find the volume of the prism below. [art: rectangular prism 6(l) x 4(w) x 3(h) cm]
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First we find the area of the base. This is the rectangular side on the bottom. Remember, to find the area of a rectangle we multiply the length times the width. B = lw B = 6(4) 2
B = 24 cm
The base area is 24 square centimeters. Now we simply multiply this by the height, which represents the number of layers in the prism. V = Bh V = 24(3) 3
V = 72 cm
The volume of this rectangular prism is 72 cubic centimeters. Remember, we measure volume in cubic units because we are dealing with three dimensions. As you can see, the formula is a shortcut that allows us to put the layers of cubic units together quickly and easily. Now let’s try finding the volume of a different kind of prism. Example 6 What is the volume of the prism below?
[art 10.6.4b: triangular prism 2(h) x 6(b) x 7(h) in.] Let’s follow the same process we used in the previous example. The volume formula for any prism is V = Bh. First we need to find the base area. Take a look at the prism above. What shape is its base? It is a triangle. That means we need to find the area of the triangle. What is the formula we use to find the area of a triangle? To find the area of the base triangle, we use the formula B = ½ bh. Remember, we use the height and base measurements for the triangular face, not the height measurement for the whole prism. Look carefully at the image! B = ½ bh B = ½ (6) (2) B = ½ (12) 2
B = 6 in.
We now have the base area. We simply multiply it by the height of the prism, according to the volume formula: V = Bh
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V = 6(7) 3
V = 42 in.
That wasn’t bad at all! Whenever we calculate volume, we find the area of the base using the appropriate area formula, and then multiply this by the height of the prism. We have learned to calculate the surface area and volume of prisms in this lesson. Specifically, we learned to:
•
use nets to find the surface area of prisms
•
use formulas to find the surface area of prisms
•
use unit cubes to find the volume of prisms
•
use formulas to find the volume of prisms
We can use nets to help us see all of the faces of a prism. This is useful for calculating the surface area of the prism. We can find the area of each face and then add their areas together. We can also use formulas to find the surface area of prisms. The formula for the surface area of a rectangular prism is SA = 2lw + 2lh + 2hw. The formula for the surface area of a triangular prism is SA = bh + (s1 + s2 + s3)H. We also learned two ways to find the volume of prisms. We can use unit cubes to count how many fill the space that the prism takes up. We can also use a formula to find the volume of prisms. The formula is V = Bh, where B is the area of the prism’s base. We use the appropriate area formula depending on the shape of the base, and then multiply this by the height of the prism.
Homework Find the surface area of each prism. [art 10.6.HPa: see below]
1. 2
[art: rect net 3x5x7] 142 cm
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2. 2
[art: triag net 8(b) 6(h) 10(s) 5(H)] 188 in.
3. 2
[art: rect prism 2x9x3] 102 ft
4. 2
[art: rect prism 3x3x5] 78 m
5. 2
[art: triang prism 8(b) 3(h) 5(s) 2(H)] 60 in.
Find the volume of each prism. [art 10.6.HPb: see below]
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6. [art: rect prism with cubes filled in 3x7x3] 63 unit cubes
7. [art: rect prism with cubes filled in 2x2x6] 24 unit cubes 3
8. [art: triang prism 7(b) 4(h) 7(H)] 98 ft 3
9. [art: rect prism 4x10x5] 200 cm 3
10. [art: rect prism 6x2x8] 96 in.
11. What is the volume of Jesse’s fish tank? [art 10.6.HPc: fish tank with measurements: 40x20x35 in] 28,000 3
in.
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12. What is the volume of the box of crackers? [art 10.6.HPd: box of crackers with measurements: 16x8x30 3
cm] 3,840 cm
Lesson 7: Surface Area and Volume of Cylinders Learning objectives
•
Find the surface area of cylinders using nets.
•
Find the surface area of cylinders using a formula.
•
Find the volume of cylinders using a formula.
Introduction In this lesson we will learn to find the surface area and volume of cylinders. A cylinder is a solid shape that exists in three-dimensional space. A cylinder has two faces that are circles. We do not call the side of a cylinder a face because it is curved. We still have to include its area in the total surface area of the cylinder, however.
[art: cylinder with circular face labeled “face” and height labeled “height”] The surface area of a cylinder is the total of the area of each circular face and the side of the cylinder. Imagine a can of soup. The top, bottom, and label around the can would make up the surface area of the can. To find the surface area, we must be able to calculate the area of each face and the side and then add these areas together. The volume of a cylinder is the measure of how much three-dimensional space it takes up or holds. Imagine a thermos. Its size determines how much water the thermos will hold. If we fill it with water, the amount of water tells the volume of the thermos. We measure volume in cubic units, because we are multiplying three dimensions: length, width, and height. The width of the thermos is the same as the diameter of the circular faces.
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We will look at several ways to calculate these two measurements. First let’s learn how to find the surface area of a cylinder.
Finding Surface Area Using Nets As we’ve said, surface area is the total area for the faces and the side of a cylinder. That means we need to find the area of each face of the cylinder, and then the area of the side. One way to do this is to use a net. A net is a two-dimensional diagram of a three-dimensional cylinder. Imagine you could unfold a box so that it is completely flat. You would have something that looks like this:
[art 10.7.1a: cylinder nets with two circles shaded, couple different sizes] It is as if we have unrolled the cylinder. The shaded circles show the top and bottom faces of the cylinder, and the unshaded rectangle shows the side. Can you see how to fold the net back up to make the cylinder? With the net, we can see each face of the cylinder more clearly. To find the surface area, we need to calculate ²
the area for each circle in the net. How do we find the area of a circle? We use the formula A = πr . If we know the radius or diameter of each circle, we can calculate its area. Look closely again at the cylinders above. The two circular faces are congruent! That means they have the same radius or diameter. Let’s calculate the area for the bottom face and then we will also know the area of the top face. Use the READ-WRITE-WRITE-SUBSTITUTE-OPERATE-SOLVE strategy.
[art: same net with all measurements labeled: r=3, h=6] step1:We see the measurements 1: Read the problem or the diagram. step2:r = 3cm; h = 6cm; A = ? 2: Write given information and what is needed 2 3: Write the formula step3:A = π r 2
step4:=π (3cm)
4: Substitute given information into formula
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5: Operate
2
step5:=π 9cm
6: Solve! Multiply by
2
step6: A = 28.26 cm
2
The area of the bottom face is 28.26 cm therefore so is the top face We found the area of each circular face and then added these areas together. Now we need to find the area of the side. The net shows us that, when we “unroll” the cylinder, the side is actually a rectangle. Recall that the formula we use to find the area of a rectangle is A = lw. For cylinders, the width of the rectangle is the same as the height of the cylinder. In this case, the height of the cylinder is 6 centimeters. What about the length? Well, the length is the same as the perimeter of the circle, which we call its circumference. When we “roll” up the side, it fits exactly once around the circle. To find the area of the cylinder’s side, then, we multiply the circumference of the circle by the height of the cylinder. First we find the circumference of a circle with the formula C = 2 π r, and then we multiply C by the height (h). Let’s try it. step1:We see the measurements 1: Read the problem or the diagram. step2:r = 3cm; h = 6cm; A = ? 2: Write given information and what is needed 3: Write the formula step3:C = 2π r step4:= 2π (3cm)
4: Substitute given information into formula
step5: = π 6cm step6:= 18.84 cm
5: Operate 6: Solve! Multiply by
C = 18.84 cm; and h = 6 cm Write given information and what is needed Write the formula for the area of the side of the cylinder A = Ch = 6cm(18.84 cm)
Substitute given information into formula Solve!
2
= 113.04 cm
Now we know the area of both circular faces and the side. Let’s add them together to find the surface area of the cylinder. bottom top face face side surface area 2
2
2
2
28.26 cm + 28.26 cm + 113.04 cm =169.56 cm
The total surface area of the cylinder is 169.56 square centimeters. Using a net helped us to locate the faces and the find the measurements of the side. Let’s try another example. Example 1 What is the surface area of the figure below?
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[art: cylinder with r=5 and h=4 in] The first thing we need to do is draw a net. Get ready to exercise your imagination! It may help to color the top and bottom faces to keep you on track. Begin by drawing the bottom face. It is a circle with a radius of 5 inches. What shape is the side when we “unroll” the cylinder? It is a rectangle, so we draw a rectangle above the circular base. Lastly, we draw the top face, which is also a circle with a radius of 5 inches.
[art 10.7.1b: steps: base; base and side; base, side, and top. all matching whole triangular cylinder above] Next let’s fill in the measurements for the side and radius of each face so that we can calculate the area of each component. Now we can calculate their areas. Remember to use the correct area and circumference formulas for circles. bottom face top face A=πr
2
A=πr
side C = 2π r
2
= π(5)
A = π(5)
C = 2π(5)
= 25π
A = 25π
C = 10π
2
2
2
A = 78.5 in.
2
2
A = 78.5 in. C = 31.4 in.
x 4 = 125.6 in. 2
Now we add these areas together to find the surface area of the cylinder. 2
78.5 + 78.5 + 125.6 = 282.6 in. 2
We used the formula A = πr to find the area of the top and bottom faces. We used the formula C = 2πr to find the circumference of the circular base and multiplied this by the height to find the area of the side. When we add these together, we get a surface area of 282.6 square inches for this cylinder. Nice work!
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We can always draw a net to help us organize information in order to find the surface area of a cylinder. A net helps us see and understand each face of the cylinder. We can also use a formula to find the area of cylinders. In the next section we’ll see how.
Finding Surface Area Using a Formula Nets let us see each face so that we can calculate its area. However, we can also use formulas to represent the faces and the side as we find their area. You may have noticed in the previous section that the two circular faces always had the same area. This is because they have the same radius. We can therefore simplify matters by multiplying the area of a circular 2
base by 2, which would give us 2 π r . We can also combine the measurements for the side into a simpler equation. We need to find the circumference by using the formula 2π r, and then we multiply this by the height of the cylinder. Therefore we can just write 2π rh. When we combine the formula for the faces and for the side we get: 2
SA = 2 π r + 2 π rh This formula may look long and intimidating, but all we need to do is put in the values for the radius of the circular faces and the height of the cylinder and solve. And we’ve already done these calculations piece by piece anyway. Let’s use the formula to find the surface area of the cylinder back in Example 1.
[art: reprint 5/4 cylinder from example 1 last section] In this cylinder, r = 5 inches and h = 4 inches. We simply put these numbers into the formula and solve for surface area. Let’s try it. 2
SA = 2 π r + 2π rh 2
SA = 2 π (5) + 2π(5)(4) SA = 2 π 25 + 2 π 20 SA = 50π + 40 π SA = 90 π 2
SA = 282.6 in.
As we have already seen, the surface area of this cylinder is 282.6 square inches. This formula just saves us a little time. Let’s try another. Example 2 What is the surface area of the figure below?
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[art: cylinder h=8, r=4 cm] We have all of the measurements we need. Let’s put them into the formula and solve for surface area, SA. 2
SA = 2 π r + 2 π rh 2
SA = 2 π (4) + 2 π(4)(8) SA = 2 π 16 + 2 π 32 SA = 32π + 64 π SA = 96π 2
SA = 301.44 cm
This cylinder has a surface area of 301.44 square centimeters. That wasn’t so bad! We just have to be careful to put each measurement in the right place in the formula and take it one step at a time. Let’s try one more. Example 3 What is the surface area of the figure below?
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[art: cylinder on its side, r=2, h=11 ft] We have the radius and the height, so let’s put the numbers into the formula and solve. 2
SA = 2 π r + 2π rh 2
SA = 2 π (2) + 2π(2)(11) SA = 2 π 4 + 2 π22 SA = 8 π + 44π SA = 52π 2
SA = 163.28 ft
This cylinder has a surface area of 163.28 square feet. Nice work! Now let’s find out how to calculate the volume of these cylinders.
Finding the Volume of Cylinders Using a Formula Volume, as we have said, is the amount of space a three-dimensional solid takes up. We measure area in two dimensions. In a cylinder, the circular faces are flat, and the area tells us how much space they cover. To calculate volume, we add another dimension: height. In other words, we build up from the flat circle that is the bottom face. Imagine we could stack many flat circles until we have a tall cylinder. The volume formula for a cylinder helps us do just that. First, we have to find the area of the bottom face. We did this to find the surface area as well. 2
Remember that the formula we use to find the area of a circle is A = π r . Then, to “stack” the circles to form a cylinder, we simply multiply this by the height of the cylinder. This gives us the formula for the volume of a cylinder:
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2
V=πrh Again, all we need to know is the radius of the circular faces and the height of the cylinder. We simply put these numbers into the formula and solve for volume, V. Let’s give it a try. Example 4 Find the volume of the cylinder below.
[art: cylinder r=6, h=3 cm] We have been given all the information we need in order to solve for volume. Let’s put the numbers into the formula: 2
V=πrh 2
V = π (6) (3) V = π (36) (3) V = 108 π 3
V = 339.12 cm
The volume of this cylinder is 339.12 cubic centimeters. Remember, we measure volume in cubic units because we are dealing with three dimensions. That wasn’t bad! Let’s try it again. Example 5 What is the volume of the cylinder below?
[art: sideways cylinder r=1.5 yd, h=6.5 yd]
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Again, we have everything we need to use be able to use the formula, so let’s get right to it. 2
V=πrh 2
V = π (1.5) (6.5) V = π (2.25) (6.5) V = 14.63π 3
V = 45.94 yd
Nice work! This cylinder has a volume of 45.94 cubic yards. Keep in mind during your calculations that you should round to the nearest hundredth for decimals that are long. Also, remember to show your answer in cubic units. We have learned to calculate the surface area and volume of cylinders in this lesson. Specifically, we learned to:
•
use nets to find the surface area of cylinders
•
use a formula to find the surface area of cylinders
•
use a formula to find the volume of cylinders
We can use nets to help us see the faces and side of a cylinder. This is useful for calculating the surface area of the cylinder. We can find the area of each face and the side and then add their areas together. We can also use a formula to find the surface area of cylinders. The formula for the surface area of a cylinder 2
is SA = 2 π r + 2π rh. We also learned how to use a formula to find the volume of cylinders. The formula is V = π 2
r h. This formula represents the area of the bottom face as we “stack” it in the third dimension of height.
Homework Questions: Find the surface area of each cylinder.
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1.
2.
3.
4.
[art: cylinder net r=3, h=9cm]
[art: cylinder net r=7, h=4in]
[art: cylinder r=2, h=12ft]
[art: cylinder r=4, h=4m]
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5.
[art: cylinder r=5, h=10m]
Find the volume of each cylinder.
6.
[art: cylinder r=2, h=7m]
7.
[art: cylinder r=1.5, h=8in]
8. 70
[art: cylinder r=6, h=5ft]
9.
[art: cylinder r=3.5, h=13 cm]
10.
[art: cylinder r=5, h=3in]
11. What is the volume of Keisha’s thermos? [image:math6-1007-n18.jpg] [art: thermos with measurements: r=2.5, h=10 in] 12. What is the volume of the can of soup? [image:math6-1007-n19.jpg] [art: soup can with measurements: r=4, h=12cm] Answers: 2
1. 226.08 cm
2
2. 483.56 in. 2
3. 175.84 ft
2
4. 200.96 m 2
5. 471 in.
Find the volume of each cylinder. 3
6. 87.92 m
3
7. 56.52 in.
71
3
8. 565.2 ft
3
9. 500.05 cm
3
10. 235.5 in.
3
11. 196.25 in.
3
12. 602.88 cm
Lesson 8: Solving Strategies: Solve a Simpler Problem Learning Objectives
•
Read and understand given problem situations.
•
Develop and use the strategy: Solve a Simpler Problem.
•
Solve real-world problems using the best strategy.
Introduction In this lesson we will learn how to use problem solving strategies to solve problems involving geometric figures. Some problem solving strategies can help us classify shapes and objects. Others can help us find the area or unknown measurement in a figure. Still others can help us decide how to approach real world applications of geometry. For example, one useful strategy to use when working with geometric figures is to solve a simpler problem. Many figures and objects are complex and many-sided. If we can find a way to break them down into smaller shapes, we can combine information about each piece to find out more about the whole figure. No matter which strategy we choose to help us solve a problem, we must always be sure to understand the problem itself. We need to recognize what it is asking us to find and what information it is giving us to help us solve the problem.
Read and Understand Given Problem Situations Whenever we are given a problem to solve, the first thing we must do is recognize what the question is asking us to find. We can look for clues in the problem to help guide us to the question within it. Sometimes what we need to find is written in the form of a question, like this: What is the circumference of the circle? Which figure shows a cylinder? In other cases, it may be written as a command: Find the area of the triangle.
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Calculate the volume of John’s fish tank. Look for a question mark or a command to help you understand what the problem is asking you to do. Let’s see if we can find the question in the problem below. Example 1 Kendra sewed squares of cloth together to make a quilt. Each square had sides of 8 inches. She used 10 red squares and 14 blue squares. What is the total area of the red portion of the quilt? Read the problem through. It is full of information. How do we know what it wants us to find? Look for the question mark and read carefully. We need to find the area of the red squares in the blanket, not the area of the whole blanket. It should only take you a minute to find the question in the problem, but this is a crucial step before you proceed! If you’re not sure, read the problem again. It may help to circle the question once you find it so that you can make sure you have answered it when you are done. Once you know what you need to find, you should look any information given in the problem that will help you solve it. Let’s look at our quilt example again. What facts are we given? It may help to underline them as you find them. Example 2 Kendra sewed squares of cloth together to make a quilt. Each square had sides of 8 inches. She used 10 red squares and 14 blue squares. What is the total area of the red portion of the quilt? [art: in the text above, underline “sides of 8 inches”, “10 red squares and 14 blue squares ”. Circle “What is the total area of the red portion”] Keep in mind that the facts give information related to the question. The question asks us to find the area of part of the quilt. Notice that the facts are measurements or quantities. Look for numbers in a problem because they usually are facts you can use to answer the question. Now that we have the information we need to answer the question, we must figure out how to put it together. We know that each square has sides of 8 inches. We also know that there are 10 8-inch red squares and 14 8-inch blue squares. Remember, we need to find the total area of the red squares, not the area of the whole quilt. If we find the area of one square, we simply multiply this by the number of red squares. The squares are 8 inches on each 2
side. To find the area of a square, we use the formula A = s . 2
A = ss
2
A = 8s
2
A = 64 in.s
The area of each square in the quilt is 64 square inches. There are 10 red squares, so we multiply 64 x 10 to get a total of 640 square inches. Of the total area of the quilt, 640 square inches are red. Nice work! Let’s review the steps we need to follow in order to be able to read and understand problems. 1. Look for the question in the problem. 2. Look for the facts in the problem. 3. Use the facts to answer the question.
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We can use this method to solve any geometry problem. Now let’s look at a useful strategy in problem solving.
Solve a Simpler Problem One problem solving strategy that may be useful in geometry is to solve a simpler problem. In geometry, we often are asked to solve complex problems. For example, we may have to calculate the surface area of a prism or find the area of complex figures. If a problem is complicated, it often helps to break it down into steps and solve pieces one at a time. Let’s look at an example. Example 3 Find the area of the figure below. [image:math6-1008-n1.jpg] [art: trapezoid with top=6, bottom=12, and dashed height in center=4 cm] We need to find the area of this trapezoid. We could use the formula for the area of trapezoids, which is 1/2 x h(l1 + l2). However, what if you’re not familiar with this formula? Can we still find the area of the trapezoid? We can if we divide the trapezoid into shapes that are easier to manage. For example, you should be more familiar with the area formulas for rectangles and triangles. Can you find a way to divide the trapezoid into rectangles and triangles? We can divide the trapezoid into one rectangle and two triangles like this: [image:math6-1008-n2.jpg] [art: repeat art above but show two dashed lines from top corners to bottom line, making a rectangle] Now we can solve for the area of each piece and then add their areas together to find the total area for the trapezoid. Let’s find the area of the rectangle first. The area formula for rectangles is A = lw, where l is the length of the rectangle and w is its width. Here, the width of the rectangle is the same as the height of the trapezoid, 4 centimeters. The length of the rectangle is the same as the top side of the trapezoid, 6 centimeters. Now let’s solve for the area of this piece of the trapezoid. A = lw A = 6(4) 2
A = 24 cms
Great, we have found the area of one piece of the trapezoid. Now let’s look at the triangles. The area formula for triangles is A = ½ bh, where b is the base of the triangle and h is the height. We can see from our drawing that the height of each triangle is the same as the height of the trapezoid, 4 centimeters. What about the base of each triangle? Let’s look more closely at the bottom edge of the trapezoid. It is 12 centimeters long. However, we know that a portion of this side makes up the rectangle, which has a length of 6 centimeters. If we subtract the 6 centimeters of the rectangle from the 12 centimeters of the edge, there are 6 centimeters remaining. Dividing this by 2 tells us that the base of each triangle must be 3 centimeters (3 + 6 + 3 = 12). Now that we have the base and height of each triangle, let’s solve for the area. A = ½ bh A = ½ (3)(4) A = ½ (12)
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2
A = 6 cms
Nice work! Now we have the area of each triangle and the area of the rectangle. We can add their areas together to find the area of the trapezoid. triangle 2
6 cms
rectangle triangle trapezoid +24 cms2 +6 cms2 =36cms2
The area of the trapezoid is 36 square centimeters. By dividing the trapezoid into smaller shapes, we were able to solve simpler problems (the area of each shape) and then add the answers together. You can see how this technique might come in handy in geometry! Let’s try it in another context. Example 4 What is the surface area of the prism below? [image:math6-1008-n3.jpg] [art: rectangular prism with l=7, w=3, h=2 in] Again, we may be able to use a formula to find the surface area of this rectangular prism. If you’re not sure, you can apply the strategy of solving a simpler problem. Surface area is the total area of a geometric solid. If we can find the area of each of the solid’s faces, we can add these together to find the surface area of the prism. Let’s use the formula for area, A = lw, to find the area of each face. You might want to draw a net to help you keep track of the faces. face 1 face 2 face 3 face 4 face 5 face 6
total area
7 = 21 (3) 3(2) = 6 7(2) = 14 7(3) = 21 3(2) = 6 7(2) =+14
surface 82 in.s2
We found the area of each side and then added these together to find the surface area of the prism. We made a tricky problem much easier by taking it one piece at a time. Solving a simpler problem can make daunting problems a breeze!
Solve Real Problems Using the Solve a Simpler Problem Strategy Now that we’ve seen how the strategy of solving a simpler problem can make our work easier, let’s try applying the strategy to situations in the real world.
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We can use this strategy as part of our problem solving method. Let’s review this method again: 1. Understand the problem: find the question and the facts. 2. Choose the best strategy to approach the problem. 3. Solve the problem. We can choose during the second step whether to use the solve a simpler problem strategy. Now let’s use this method to solve a few real-world problems. Example 5 Karen is decorating her crafts box by covering it in fabric. The craft box is 12 inches long, 8 inches wide, and 6 inches high. She covers the top and the front face in gold fabric. What is the total surface area Karen covers in gold fabric? [image:math6-1008-n4.jpg] [art: rectangular prism: should look like a real box. Label dimensions: 12(l), 8(w), 6(h). “Front is 12 x 6 side] First, let’s make sure we understand the problem. What do we need to find? We need to find the surface area of two sides of the box. What facts are in the problem? We know the length, width, and height of the box. We also know which two faces Karen covers with fabric. Now that we understand the question and know the faces, we select the best strategy to answer the question. We need to find only part of the box’s surface area. Using the formula for surface area is probably not the easiest way. We can use our strategy of solving a simpler problem, however. How can we break the question down into simpler problems? We can find the area of each of the two faces, and then add their areas together. Let’s try it. area of top face area of front face A = lw
A = lh
A = 12(8)
A = 12(6) 2
A = 96 in.s
2
A = 72 in.s
Now we add their areas together to find the surface area of these two faces. 2
96 + 72 = 168 in.s
The surface area of the two faces Karen covered in gold fabric is 168 square inches. Nice work! Let’s try another problem. Example 6 Jason is carpeting his bedroom. Three walls of the room are straight, but the fourth wall is curved. If the straight walls measure 15 feet and 10 feet, many square feet of carpet will Jason need to buy? [image:math6-1008-n5.jpg] [art: three sides of 10 x 15 rectangle with fourth short side replaced by curve making a semicircle] First, let’s make sure we understand the question. In order to know how many square feet of carpet Jason must buy, we need to find the area of his room. We know the length and width of his room, and we know that one of its walls is curved.
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What strategy can we use to solve the problem? There is no formula for finding the area of a shape like this with one curved side. We can, however, apply our strategy of solving a simpler problem. If we can divide the room into smaller shapes, we can find the area of each shape and then add their areas together. What is the best way to divide the room? We can divide the room into a rectangle and semicircle, like this: [image:math6-1008-n6.jpg] [reprint art from above. add dashed line to complete rectangle and form semicircle] Now let’s find the area of each piece, one at a time. Let’s work on the rectangle first. A = lw A = 15(10) 2
A = 150 fts
How can we find the area of the semicircle? Well, we can find the area of the whole circle and then divide 2
it by 2. Remember, the area formula for a circle is A = π rs . What is the radius of the circle? Since this is a semicircle, the width of the rectangle is the same as the diameter of the circle. The diameter is therefore 10 feet. The radius is always half the length of the diameter, so the radius of the circle is 5 feet. Now let’s solve for area. 2
A = π rs
2
A = π (5)s A = 25 π
2
A = 78.5 fts
The area of the circle is 78.5 square feet. We need the area of half of the circle, so we divide by 2 to get 39.25 square feet. Now we know the area of the rectangle and the area of the semicircle. We can add their areas together to get the area of Jason’s room. 2
150 + 39.25 = 189.25 fts
The area of the room is 189.25 square feet, so Jason will need to buy 189.25 square feet of carpet. Great job! We have learned a lot about problem solving in this lesson. Let’s review the steps we have learned. First we need to understand the problem. We need to find the question in the problem and then look for the facts, or pieces of information given in the problem. Next, we need to choose a problem solving strategy to help us answer the question. We can choose from many different strategies, but the best one will suit the question we need to answer and the facts we have been given. Solving a simpler problem is a useful strategy in geometry when we are given complex or threedimensional figures.
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Once we have selected the best strategy, we can use it to solve the problem. We can use this procedure for geometry problems involving real world situations, too.
Homework Circle the question and underline the facts. Then solve the problem. 1. What is the circumference of a circle that has a diameter of 8 meters? 25.12 m 2. Diana drew a triangle with a height of 6 centimeters and an area of 45 square centimeters. What is the base of the triangle Diana drew? 15 cm Use the strategy solve a simpler problem to solve each problem.
3. 2
What is the area of the figure below? [art 10.8.HPa] 35 in.s
2
4. Find the surface area of a cube with sides of 7 centimeters. 294 cms
2
5. What is the area of the unshaded portion of the square below? [art 10.8.HPb] 20 fts 2
6. Find the area of the figure below. [art 10.8.HPc] 166 ms
7. Ravi’s fish tank is 20 inches long, 10 inches wide, and 10 inches tall. To clean the tank, he emptied out all the water. Now he has put some back in. The water in the tank is 3 inches deep. How many more cubic 3
inches of water must Ravi add to fill up the tank? 1,400 in.s
8. Mr. Lee stacked 4 soup cans on a shelf in his store. What is the total surface area of the stack of cans, not counting the surfaces where they meet?
2
[art 10.8.HPd] 276.32 in.s
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9. Chloe is painting one wall of her bedroom. The wall is 16 feet long and 12 feet high. There are two windows 2
in the wall, each of which is 2 feet long and 3 feet high. What is the total area that Chloe will paint? 180 fts
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