Blended digital and print resources specifically created for the new AQA GCSE Mathematics specification, available from early 2015.
Brighter Thinking
Brighter thinking for the new curriculum: • Written by an experienced author team of teachers, partners and advisers. • Rich digital content to engage and motivate learners. • Differentiated resources to support all abilities. • Progression and development at the heart of all our resources.
For more information or to speak to your local sales consultant, please contact us:
www.cambridge.org/ukschools ukschools@cambridge.org 01223 325 588 CUPUKschools
MATHEMATICS Higher Problem-solving Book Sample
Written from draft specification
3 It depends how you look at it
3 It depends how you look at it Sometimes it’s useful to make a change to a diagram or scenario so that it is easier to see what’s going on. Here is an example.
The diagram shows a red square with a purple circle drawn through its vertices and a blue square drawn so it touches the purple circle on all sides. What is the relationship between the area of the red square and the area of the blue square? Explain how you know. Before you turn the page, have a go at the question. What ideas have you got? Here are some things you can try. Get a rough idea of what the answer will be; you know that the blue square is bigger than the red square, but it is certainly less than, say, four times the area. This is not the way to work out the final answer, not least because you can’t provide an explanation, but it will help you decide if your answer seems reasonable later. You don’t know any measurements, so do you need to provide some? Will it help to call the side length of the red square x? Or the radius of the circle could be r. Maybe this will be useful and maybe it won’t. Try putting on some extra lines. Will they help?
Tip If you can’t do it please don’t worry. This is a mathematical problem that is supposed to be difficult. Remember, if you can just ‘see’ the answer then it wouldn’t be a mathematical ‘problem’.
© Cambridge University Press 2014 Written from draft specification
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GCSE Mathematics Higher Problem-solving Book
Did you manage to answer the problem? If not, here’s one way that you could have solved it by changing the way you look at the diagram.
If you rotate the red square nothing vital changes. The area of the red square is still the same and the purple and blue shapes are unaffected. Does this help you to see the relationship between the red square and the blue square?
To make this very clear you can remove the circle.
Then you can divide the shape up like this.
Tip The extra lines are helpful after you’ve rotated the red square.
Now you can explain that the area of the red square is half the area of the blue square. Here is an example of a full answer.
If you rotate the red square so the vertices of the red square meet the middle of the sides of the blue square you do not change its area. The two extra lines I have added join the vertices of the red square, cutting it into quarters that are right-angled triangles. These lines also join the middle of the sides of the blue square, cuttting it into quarters that are squares. The total area of the red right-angled triangles is half the area of the blue square. So the area of the red square is half the area of the blue square. Tip You may have answered this question by working out some measurements in terms of x or r instead of using this method. This is also a good way to solve the problem.
2
© Cambridge University Press 2014 Written from draft specification
Tip You had a rough idea that the area of the blue square is between one and four times the area of the red square, and it is. Estimating the answer can be a useful check that your final answer is sensible.
3 It depends how you look at it
1
A nna knows that she can create a heart shape from a triangle and two semi-circles. She makes a Valentine’s card by cutting a heart shape out of a piece of card, as shown.
13 cm
What is the area of card that has been removed? Give your answer correct to 3 significant figures.
2
The mean of a set of five numbers is 12.
The numbers are in the ratio 1 : 1 : 2 : 3 : 5.
Find the largest number in this set. 3
9 cm
130 m
Sadiq was planning to fence an L-shaped field. He drew a plan and started to measure the sides. He didn’t finish measuring, but he knows that the perimeter of the field is 2.28 km.
a
Calculate the value of a. Give your answer in kilometres.
Tip
130 m
Remember that there are 1000 m in 1 km.
4
650 m
Eric is reorganising his hamsters in his pet shop. If he puts three in each cage, he has one hamster left with no cage to go into. If he puts four in each cage, one cage is left empty. How many hamsters and how many cages does Eric have?
The exploded segment(s) from the colour wheel tells you what strand(s) of maths each question covers:
Algebra
Probability
Ratio, proportion and rates of change
Statistics
Number
Geometry and measures
© Cambridge University Press 2014 Written from draft specification
The star rating indicates the level of difficulty.
3
GCSE Mathematics Higher Problem-solving Book
5
a Explain how Stephan can use these two triangles to show that √8 = 2√2.
2
b Draw a pair of triangles to demonstrate that:
i
√18 = 3√2
ii
√12 = 2√3
1
B A 1
2
Tip Why is it useful that the two triangles are right-angled? The two triangles are mathematically similar. What does this mean? How can it help you?
6
A
B
Mia is experimenting with shapes she can draw inside circles. First, she inscribes a square inside a circle of radius 1 cm (A). a
Calculate the area of the square.
Next, she inscribes an equilateral triangle in the same circle (B).
C
b Show that, to the nearest whole number, the ratio of the area of the triangle to the area of the square in part a is 2 : 3.
Finally, she inscribes a regular pentagon in a circle (C). c Show that the area of the pentagon is approximately 6 the area of 5 the square in part a.
7
Use the diagram of an equilateral triangle to find the exact value of tan 60°, sin 60° and cos 60°.
What do you know about the properties of an equilateral triangle? Why is the triangle divided in half? How might this help you to answer the question?
1
Show that sin 60° = cos 30° and sin 30° = cos 60° and state their exact values.
4
Tip
© Cambridge University Press 2014 Written from draft specification
3 It depends how you look at it
WORKED SOLUTIONS
3 It depends how you look at it 1
Diameter of each semi-circle = 9 cm ÷ 2 = 4.5 cm Radius of each semi-circle = 4.5 ÷ 2 = 2.25 cm
2
The heart shape can be split into two semi-circles and a triangle.
Area of the two semi-circles together = π × 2.252 = 15.9043… cm2
First find the area of the semi-circles.
Height of triangle = 13 cm – 2.25 cm = 10.75 cm
Then find the area of the triangle.
Area = 0.5 × 9 cm × 10.75 cm = 48.375 cm2
Area of triangle = 0.5 × base × height.
15.9043… cm2 + 48.375 cm2 = 64.2793... cm2
Total area of heart shape.
64.2793... cm2 = 64.3 cm2 to 3 s.f.
Answer rounded to 3 significant figures.
The total of the numbers is the mean multiplied by how many numbers there are.
First find the total, using the property of the mean (mean = total of values ÷ number of values).
Total of the numbers = 5 × 12 = 60
3
Sum of parts in the ratio = 1 + 1 + 2 + 3 + 5 = 12
Then find how many parts there are.
60 ÷ 12 = 5
This is the value of each part.
5 × 5 = 25
The largest number is five parts.
Missing horizontal length = 650 − 130 = 520
Find the missing horizontal length.
Missing vertical length = a − 130
Find the missing vertical length.
Total perimeter = 650 + a + 130 + (a − 130) + 520 + 130
Write the perimeter of the field in terms of a.
= 1300 + 2a
Simplified perimeter in terms of a.
So 2280 = 1300 + 2a
Equation for perimeter, in metres.
2a = 2280 − 1300 = 980
Rearrange to find the value of 2a.
a = 980 ÷ 2 = 490
Use inverse operations to solve the equation for a.
Convert to kilometres.
= 0.49 km
650 m + 490 m + 130 m + 490 m – 130 m + 520 m + 130 m = 2280 m = 2.28 km
Check the answer by substituting 0.49 km (or 490 m) for a.
© Cambridge University Press 2014 Written from draft specification
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GCSE Mathematics Higher Problem-solving Book
4
Let the number of hamsters be H and the number of cages be C.
Define the variables.
H = 3C + 1
Model scenario 1.
H = 4(C − 1)
Model scenario 2.
3C + 1 = 4(C − 1)
Find the number of cages by putting these equations equal to each other.
3C + 1 = 4C – 4 3C + 5 = 4C 5=C
Solve to find C.
H = 3C + 1 = 3 × 5 + 1 = 16
Put C back into equation 1 to find the value of H.
H = 4(C − 1) = 4(5 – 1) = 4 × 4 = 16
Check with equation 2.
So Eric has 16 hamsters and 5 cages. 5
a The general form of Pythagoras’ theorem is: 2 2 2 a +b =c
The triangles are right-angled so apply Pythagoras’ theorem to find the length of the hypotenuse.
In triangle A, a = 1, b = 1: 2 2 2 1 +1 =c 2 2 = c c = √2
First take triangle A.
In triangle B, a = 2, b = 2: 2 2 2 2 +2 =c 2 8 = c c = √8
Next take triangle B.
Length of hypotenuse on triangle B = 2 × length of hypotenuse on triangle A.
6
√8 = 2 × √2 = 2√2
The two triangles are mathematically similar. Triangle B is an enlargement of triangle A by a scale factor of 2. This gives two values for the hypotenuse of the larger triangle. These must be equal.
© Cambridge University Press 2014 Written from draft specification
3 It depends how you look at it
b i 1
In part a, you saw a hypotenuse of length 2√2 was created by enlarging triangle A by a scale factor of 2.
√2
A 1
It follows that you can create a triangle, C, with a hypotenuse of length 3√2 by enlarging triangle A by a scale factor of 3. 3√2
3 C
3
2
2
2
a + b = c This time, a = 3, b = 3: 2 2 2 3 + 3 = c 18 = c2 c = √18
The two values found for the hypotenuse of triangle C must be equal.
c = √18 = 3√2
ii √3
Consider the sides of triangle C. The length of the hypotenuse can be found using Pythagoras’ theorem.
This time the length is not a multiple of √2 so you cannot solve it by thinking about triangle A.
2
You need a right-angled triangle that includes a length of √3.
D 1
One possible triangle is shown. Enlarging triangle D by a scale factor of 2 will produce a triangle including a length of 2 √3. 4
2√3 E
2
© Cambridge University Press 2014 Written from draft specification
7
GCSE Mathematics Higher Problem-solving Book
The general form of Pythagoras’ theorem is: 2 2 2 a + b = c This time, b = 2, c = 4, so: 2 2 2 a + 2 = 4 2 2 2 a = 4 − 2 2 a = 12 a = √12
As before, the two values for the vertical side length must be equal.
√12 = 2√3
6
Consider triangle E in isolation and use Pythagoras’ theorem to calculate the length of the vertical side as if you didn’t already know that it was 2√3.
a
Diagonal of square = diameter of circle = 2 cm.
2 cm
Use Pythagoras’ theorem to calculate the length of each side of the square. a
2 cm
a
2
2
2
a +a =2 2a2 = 4 a2 = 2 a = √2 (= 1.414… cm)
8
Area of square = a2 = 2 cm2
Area of the square.
© Cambridge University Press 2014 Written from draft specification
3 It depends how you look at it
b 1 cm 1 cm
h
The equilateral triangle can be split into three isosceles triangles. 1 cm 30° a
Draw a diagram and label all the known angles.
30°
Let the height of each isosceles triangle be h, let
Use trigonometry to calculate the height of each isosceles triangle.
a be half the base, b, of each isosceles triangle. h = 1 × sin 30° h = 0.5 cm
a = 1 × cos 30° a = 0.866… cm 2a = 1.732… cm
Use trigonometry to calculate the base of each isosceles triangle.
Area of each isosceles triangle.
Area = 1bh 2 Area = 0.5 × 1.732… cm × 0.5 = 0.433… cm2
3 × 0.433… cm2 = 1.299… cm2
Area of equilateral triangle.
1.299… : 2
Ratio of area of equilateral triangle to area of square.
1 : 1.539…
Write ratio in the form 1 : n by dividing both sides by 1.299…
2 : 3.079…
Multiply ratio by 2 to match answer required.
2:3
Round to nearest whole number as required.
c
The pentagon can be split into five identical isosceles triangles. Draw a diagram and label all the known angles; the sum of the interior angles of a regular pentagon is 540°.
1 cm 1 cm 1 cm 1 cm
54° a
h
1 cm 54°
Let the height of each isosceles triangle be h, let
a be half the base, b, of each isosceles triangle. h = 1 × sin 54° = 0.809… cm a = 1 × cos 54° = 0.587…. cm 2a = 2 × cos 54° = 1.175… cm 1
Area = bh 2 2 = sin 54° × cos 54° = 0.475… cm
5 × 0.475… cm2 = 2.377… cm2
Use trigonometry to calculate the height and base of the isosceles triangles.
Area of each isosceles triangle.
Area of pentagon.
© Cambridge University Press 2014 Written from draft specification
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GCSE Mathematics Higher Problem-solving Book
2.377… ÷ 2 = 1.188… = 1.2 to 2 s.f.
Area of pentagon divided by area of square. Write answer as a fraction.
1.2 = 12 = 6 10 5 6
The area of the pentagon is times the area of 5 the square in part a. 7
You know the angles because the triangle is equilateral.
30° 1
Conclusion.
x
60° 1 2
The general form of Pythagoras’ theorem is: a2 + b 2 = c 2 x2 + 12 = 12 2 2 x = 12 − (1)2 = 1 − 1 = 3 2 4 4 √ 3 x= 2 tan θ = opposite adjacent √3 tan 60° = 2 = √3 1 2 sin θ = opposite hypotenuse √3 2 √3 sin 60° = = 2 1 adjacent hypotenuse 1 2 1 cos 60° = = 1 2 cos θ =
1 2 1 sin 30° = = 1 2 √3 2 √3 cos 30° = = 2 1 √ sin 60° = cos 30° = 23 sin 30° = cos 60° = 1 2
10
Use Pythagoras’ theorem to calculate the length of the vertical line, x, dividing the triangle in half.
From the diagram calculate the value of tan 60°.
From the diagram calculate the value of sin 60°.
From the diagram calculate the value of cos 60°.
Using the angle at the top of the diagram instead, you can calculate the value of sin 30° and cos 30°.
Conclusion.
© Cambridge University Press 2014 Written from draft specification
Blended digital and print resources specifically created for the new AQA GCSE Mathematics specification, available from early 2015.
Brighter Thinking
Brighter thinking for the new curriculum: • Written by an experienced author team of teachers, partners and advisers. • Rich digital content to engage and motivate learners. • Differentiated resources to support all abilities. • Progression and development at the heart of all our resources.
For more information or to speak to your local sales consultant, please contact us:
www.cambridge.org/ukschools ukschools@cambridge.org 01223 325 588 CUPUKschools
MATHEMATICS GCSE for AQA Teacher’s Resource Sample
Written from draft specification