In rectangle ABCD, a semicircle CPS is drawn as shown above. The semic

Question

https://gmatclub.com/forum/download/file.php?id=58181 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 Untitled.png

IanStewart · Accepted Answer

We don't have any information about exactly where point Q is. So if this question has a unique answer (which must be true when there are five exact answer choices), the answer to the problem must be the same no matter where point Q is. So we can just imagine points Q, S and A are in the exact same location. Then ABCD is a square with a diagonal of length 10, and thus with sides of length 10/√2, and its area is 100/2 = 50.

Alternatively you could imagine Q is in the same location as P. Then ABCD is a rectangle with length 10, and since that's the diameter of the circle too, its width, which is the radius of the circle, would be 5. So in that case the area is again 50.

Alternatively you could imagine Q is in the same location as P. Then ABCD is a rectangle with length 10, and since that's the diameter of the circle too, its width, which is the radius of the circle, would be 5. So in that case the area is again 50.

aviddd · Answer

I have scratched my head over this for the past 10 minutes. I ended up with 3 equations with 4 different variables. I gave up.

Will wait for OA.

TestPrepUnlimited · Answer

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

aviddd · Answer

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

kajaldaryani46 · Answer

———
Hi

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

Posted from my mobile device

TestPrepUnlimited · Answer

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.

chetan2u · Answer

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

shivam10 · Answer

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

In rectangle ABCD, a semicircle CPS is drawn as shown above. The semic

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In rectangle ABCD, a semicircle CPS is drawn as shown above. The semic

Prep Toolkit

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