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Bunuel
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Bunuel

In rectangle ABCD, a semicircle CPS is drawn as shown above. The semicircle’s diameter is shared with and extends beyond side BC, it intersects side AB at Q, and it is tangent to side AD at S. If CQ = 10, what is the area of rectangle ABCD?


A. 50
B. 60
C. 75
D. 100
E. 120


Note the shape is not fixed, hence we can adjust the shape freely as we wish and we can use special examples to find an easy way out.

For example, QC = PC = BC = 10 and it is clear that BC = 10 with AB = 5. Another example is QC = SC = 10, in that case 10 is the diagonal of square ABCD.

The answer is A, 50, but I'll show how to prove the answer is 50 in any scenario for the math enthusiasts!

Connect PQ, triangle PQC must be a right triangle from Thales's theorem.

Triangle PQC and triangle QBC are both right triangles that share an angle, thus they are similar.

Then we can use: \(\frac{PC}{QC} = \frac{QC}{BC}\)

\(PC * BC = QC^2 = 100\).

Now note the area we are finding is \(DC*BC\), and DC = r is half of PC = 2r. Thus \(DC*BC = \frac{PC * BC }{2} = 100/2 = 50\)

Q.E.D
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Bunuel

In rectangle ABCD, a semicircle CPS is drawn as shown above. The semicircle’s diameter is shared with and extends beyond side BC, it intersects side AB at Q, and it is tangent to side AD at S. If CQ = 10, what is the area of rectangle ABCD?


A. 50
B. 60
C. 75
D. 100
E. 120


Note the shape is not fixed, hence we can adjust the shape freely as we wish and we can use special examples to find an easy way out.

For example, QC = PC = BC = 10 and it is clear that BC = 10 with AB = 5. Another example is QC = SC = 10, in that case 10 is the diagonal of square ABCD.

The answer is A, 50, but I'll show how to prove the answer is 50 in any scenario for the math enthusiasts!

Connect PQ, triangle PQC must be a right triangle from Thales's theorem.

Triangle PQC and triangle QBC are both right triangles that share an angle, thus they are similar.

Then we can use: \(\frac{PC}{QC} = \frac{QC}{BC}\)

\(PC * BC = QC^2 = 100\).

Now note the area we are finding is \(DC*BC\), and DC = r is half of PC = 2r. Thus \(DC*BC = \frac{PC * BC }{2} = 100/2 = 50\)

Q.E.D

This is brilliant. Thank you for solving this both ways.
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Bunuel

In rectangle ABCD, a semicircle CPS is drawn as shown above. The semicircle’s diameter is shared with and extends beyond side BC, it intersects side AB at Q, and it is tangent to side AD at S. If CQ = 10, what is the area of rectangle ABCD?


A. 50
B. 60
C. 75
D. 100
E. 120


Note the shape is not fixed, hence we can adjust the shape freely as we wish and we can use special examples to find an easy way out.

For example, QC = PC = BC = 10 and it is clear that BC = 10 with AB = 5. Another example is QC = SC = 10, in that case 10 is the diagonal of square ABCD.

The answer is A, 50, but I'll show how to prove the answer is 50 in any scenario for the math enthusiasts!

Connect PQ, triangle PQC must be a right triangle from Thales's theorem.

Triangle PQC and triangle QBC are both right triangles that share an angle, thus they are similar.

Then we can use: \(\frac{PC}{QC} = \frac{QC}{BC}\)

\(PC * BC = QC^2 = 100\).

Now note the area we are finding is \(DC*BC\), and DC = r is half of PC = 2r. Thus \(DC*BC = \frac{PC * BC }{2} = 100/2 = 50\)

Q.E.D
———
Hi

Could you help me understand why you would say that the shape is not fixed?

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kajaldaryani46 Sure!

In order to get an idea, we can start from drawing the half-circle PSC, then draw the rectangle part ABCD touching the top of the circle and you may notice we can "move" the segment AB around between point P and S. Pick any AB. then label the intersection of AB and arc PSC as Q, connect QC and label it 10. Now imagine moving AB around, we will get different QC's while obeying all the conditions. We will get different figures that are not meant to be compared side by side, and we have to scale the dimensions of each figure to have the same QC lengths but each figure follows the conditions. (I have an attachment for two different graphs that are scaled properly, you can see I shrank the 2nd figure to have equal QC's, the rectangle is longer but less wide).

When doing complicated geometry questions, checking whether the shape of a figure is fixed is generally quite useful. If the shape of the figure is fixed, that means all lengths follow some fixed proportion among each other. Then finding any length of the shape is enough to find the rest of the lengths, in theory. For this question, we can use the fact that is it not fixed to our advantage and find special cases to calculate the area.
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Circle+Square+Sample.png [ 41.46 KiB | Viewed 2700 times ]

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Bunuel

In rectangle ABCD, a semicircle CPS is drawn as shown above. The semicircle’s diameter is shared with and extends beyond side BC, it intersects side AB at Q, and it is tangent to side AD at S. If CQ = 10, what is the area of rectangle ABCD?


A. 50
B. 60
C. 75
D. 100
E. 120

Are You Up For the Challenge: 700 Level Questions


Attachment:
The attachment Untitled.png is no longer available

Join P and Q to get PQ. Now angle PQC is on the diameter PC, so \(\angle PQC=90\)

In triangles PQC and QBC ,
\(\angle QPC=\angle BQC\), \(\angle QCP \) is common.
\(\angle PQC=\angle QBC=90\)
Thus both triangles are similar.

\(\frac{QC}{PC}=\frac{BC}{QC}\)
\(PC*BC=10^2=100\)

Now We can say that one side CD is equal to the radius and so diameter PC=2*radius=2*CD

Area=BC*CD=\(\frac{BC*PC}{2}\)=100/2=50


A
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Bunuel

In rectangle ABCD, a semicircle CPS is drawn as shown above. The semicircle’s diameter is shared with and extends beyond side BC, it intersects side AB at Q, and it is tangent to side AD at S. If CQ = 10, what is the area of rectangle ABCD?


A. 50
B. 60
C. 75
D. 100
E. 120

Are You Up For the Challenge: 700 Level Questions


Attachment:
Untitled.png

Join P and Q to get PQ. Now angle PQC is on the diameter PC, so \(\angle PQC=90\)

In triangles PQC and QBC ,
\(\angle QPC=\angle BQC\), \(\angle QCP \) is common.
\(\angle PQC=\angle QBC=90\)
Thus both triangles are similar.

\(\frac{QC}{PC}=\frac{BC}{QC}\)
\(PC*BC=10^2=100\)

Now We can say that one side CD is equal to the radius and so diameter PC=2*radius=2*CD

Area=BC*CD=\(\frac{BC*PC}{2}\)=100/2=50


A

Thanks a lot. This is the best solution posted so far for this question.
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