Answer(s)
∆ AQM and ∆ BQM
In
AM =MB ……. Radii QM = QM ………. Common
¿ AMQ = ¿ QMB ∆ AQM ≡
∴
∆ BQM ….. S, S, A
This means that corresponding angles and corresponding are also equal. So, AQ = BQ
AQ BQ
=
AM BM
ratio of congruent sides in these two triangles
You need to express sides in terms of values that are eligible to merge with x values as given It is unavailable, so convert sides to figures.
AQ BQ
=
AM BM
AQ AQ
=
AM BM
1=
since AQ = BQ (crash them to one value **).
AM BM
Substitute values of AM and BM as they appear from the given drawing 1=
5+ X MN + NB
(Note that BM = MN + MB). ……………………… (1)
The above expression is called first equation ………………….…..(1) The next step is to express MN + NB in terms of x. Note that: CN = QN = BN …….. they are all radii to smaller semi-circle. And NB is BN whichever way you express this side. CN = MN + X So MN + x = QN = NB This means NB = MN +X …………………………………………………. (2) The above expression is called second equation ………………… (2) Substitute MN + x into the first equation ………………………. (1) In other words substitute the second equation into the first equation Hence substitute (2) into (1) (see heighted sections from previous sections).
1=
5+ x MN + NB
(Note that BM = MN + MB). ……………………… (1)
This means NB = MN +X …………………………………………………. (2) How often do you see simultaneous equations? * 1=
5+ X MN + NB
NB = MN +X
……………………………………………….. (1)
…………………………………………………. (2)
5+ X MN + MN + X
Therefore: 1 =
5+ X 2 MN + X
1=
The next step is to make MN the subject of a formula. (2MN+X) 1 =
5+ X 2 MN + X
2MN +X = 5+X Solve for MN:
2MN +X-X = 5+X –X
2MN + 0 = 5 + 0 2MN = 5
2 MN 2
=
5 2
Therefore, MN =
5 2
(2MN +X)
Now you need to recall what is …radius, diameter and meaning of word perpendicular You also need to recall what Pythagoras theorem is. Your next focus, as ever it should have been …is in Right Angled -triangle QMN You need to express QN in terms of x, so that you use these values to complete for hypotenuse (the longest side opposite to the right angle of any triangle). This went goes as follows: Equate radii as long as they are equal MN + x is a radius, but MN = So QN =
5 2
5 + x = NB = QN 2 5 +x 2
(you can complete the fraction, at your at your own convenient method/option).
Since, MN + x = QN = NB…
So QN = QN =
5 +x 2
…radii (they are all the radii of smaller semi-circle).
Last thing (and perhaps the first to others) is to express QM in terms of x So, Hint: M is center of larger semi –circle So , AM = PM But AM = 5+X ,
………………. Radii and PM = 3 + QM
Therefore 5+X = 3+ QM Make QM the subject of a formula, so as to express QM in terms of X. 5+X = 3+ QM 5-3+X = 3-3+ QM 2+X = 0 +QM 2+X = Q Therefore QM = 2+ x This is same like to say QM = X+ 2 (whichever way you need to express this).
Student often feel nervous to express things whichever way as they felt it is against the memorandum, and that is why some sections are emphatic. Provided you have a right angled triangle QMN, because AB is perpendicular to PM, you may now use the figures you have been determining to calculate the value of radii. In order to do that you may simple use the Theorem of Pythagoras to determine hypotenuse So in
¿ angled ∆ QMN:
QN2 = MN2 + QM2 QN =
5 +x 2
Therefore: (
…………………. Theorem of Pythagoras
5 2
, and MN =
5 + x )2 = ( 2
5 2
, and QM = 2+X
)2 + (2+X)2
Once you do that , you solve for X If you did it correctly , you will get the value of X = 4
But let do it together:
(
5 2
+ x )2 = (
(
5 2
+x) (
5 2
5 2
)2 + (2+X)2
+x) =(
5 2
)(
5 2
) + (2+X) (2+X)
This is same as : 5 2
5 2
(
25 4
+ x ) +X ( 5 2
+
X+
5 2
5 2
+x) =
X + X2 =
25 4
25 4
+ [ 2(2+X)+X(2+X) ] + [4+2X + 2X + X2)
][ Look for common factors and group them (following signs and rules) Do not break BOMDAS rules, so if you are not aware of that .. be careful . 25 4
+
5 2
X+
So,
25 4
+
5 2
5 2
X + X2 = 5 2
X+
25 4
X + X2 =
+ [4+2X + 2X + X2) 25 4
+ (4 + 4X + X2)
If you have nothing into situation , then you impose 1 e.g. : + ( b ) is same as +1 ( b ) and also same as +1(+1xb) and sometimes +1(+1xb x1x1 ) no matter where it goes . 25 4
So, 25 4
+
+
5 2
X+
5 2
X + X2 =
25 4
+ (4 + 4X + X2)
5 2
X+
5 2
X + X2 =
25 4
+1 (4 + 4X + X2)
25 4
+
5 2
X+
5 2
X + X2 =
25 4
+1 (4) + (1)4X + (1)X2
25 4
+
5 2
X+
5 2
X + X2 =
25 4
+4 + 4X + X2
Notice that:
5 2
X+
One can let
5 2
X = K , so K + K = 2K
Where K =
5 2
Therefore:
25 4 25 4
5 2
X are also like terms **
5 2
X , 2K = 2( +
5 2
X+
5 2
+ 5X + X2 =
X) = 5x**** X + X2 = 25 4
25 4
+4 + 4X + X2
+4 + 4X + X2
Now eliminate denominators, by multiplying by 4 to all sides Please notice that you are moving either from Left to Right and vice versa. In this case, we are moving from Left to Right Hand Side. 25 4
4
(
+ 5X + X2 =
25 ) 4
25 4
+4 + 4X + X2
+ (4)5X + (4)X2 =
25 + 20X + 4X2
=
(4 )
25 4
+ (4)4 + (4)4X + (4) X2
25 + 16 + 16 X + 4X2
Your last step is to group like terms once again and solve for X You can choose to move from Left to Right Hand Side or Move from Right Hand Side to Left Hand Side. In this case, we have chosen to move from Right to Left 25 + 20X + 4X2
=
25 + 16 + 16 X + 4X2
25 + 20X + 4X2
=
41 + 16 X + 4X2 -
+25-25 + 20X -20X + 4X2 – 4X2 = 41-25 +16X -20X + 4X2 -4X2
0 +
0
+
0
= 16
+ (-4X) + 0
0
= 16
- 4X
You may now arrange to solve for X and then determine the actual size of a radius as follows: 0
= 16
- 4X
-16 = -4X −16 −4
=
−4 X −4
4=X Therefore X = 4 Now substitute X into all values where radii values are found to determine the actual size of all radii. That goes as follows:
If you chose to use CN = X + MN You will remember that MN = So radius = X + MN = 4 +
5 2
5 2
=
13 2
You can also check the correctness of this status Since QN =
5 2
+x
The radii of smaller semi-circle =
13 2
The radii of larger semi-circle = AM or PM or BM For AM , Radius = 5 +X Therefore AM = 5 +4 = 9 The radius of larger circle = 9 units You can also double check the validity of this answer Since AM = PM Recall 5+X = 3+ QM If you express QM in terms of X : 5+X = 3+ QM 5-3+X = +3-3+ QM 2 + X = 0 + QM 2+ X = QM But X = 4 So QM = 6 units Mind You PM = PQ + QM Given PQ = 3 So PM = 3+ QM PM = 3+ 6 = 9
‌ since QM =9 from calculations.
The End: Extra thought : Do you remember that Tangent is perpendicular to radius So at any time you introduce a tangent along long B , triangle BQC becomes a right angled triangle
Figure out how you would use existing values to prove that Angle BQC is right angle. Hint: triangle MQC and triangle MQB are right angled triangles. Do you remember that Radii of smaller circles are equal in this case? Evaluate this: NB = x +
5 2
by knowing that QN = NB
But Left alone NB + MN is also equal to size of radius of larger circle So NB + MN = 5 + X NB +
5 2
=5+X
NB = (5 +X) -
5 2
But X =4 So, NB = (5 +4) -
5 2
=9-
5 2
=
13 2