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# A circle is inscribed in an isosceles trapezoid with bases 8 and 18

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Re: A circle is inscribed in an isosceles trapezoid with bases 8 and 18 [#permalink]
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vedha0 wrote:
PU = PT = a. but this need not be equal to QV or QU. QV=QU=b. and then RV=RW=c and ST=SW=d

However, since its isoceles trapezoid, PS = QR. so we eventually get PS = QR = 13 on solving using given information. 5-12-13 triplet implies dia is 12 and thus radius is 6. hence, area is 36*pi . so the answer is still C. How is the answer A? Bunuel can you please give OE?

GMATinsight wrote:
Bunuel wrote:

A circle is inscribed in an isosceles trapezoid with bases 8 and 18 as shown above. What is the area of the circle?

A. $$25\pi$$
B. $$30\pi$$
C. $$36\pi$$
D. $$49\pi$$
E. $$64\pi$$

Attachment:
2023-06-28_11-18-47.png

Please check teh explanation in attachment

_________________________________________
The OA is A, not C. Edited. Thank you for notifying!
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Re: A circle is inscribed in an isosceles trapezoid with bases 8 and 18 [#permalink]

A circle is inscribed in an isosceles trapezoid with bases 8 and 18 as shown above. What is the area of the circle?

A. $$25\pi$$
B. $$30\pi$$
C. $$36\pi$$
D. $$49\pi$$
E. $$64\pi$$
Area of circle = $$r^2\pi$$
Therefore the answer choice must be a multiple of square of a number(most likely an integer as we see the choice available).
Here itself we can reject B as 30 is not a square of an integer.

Since this is an isosceles trapezoid, the sides non parallel sides are equal. The figure is such that circles fits in such a manner that all sides touch circle uniformly i.e. point of touch divides parallel sides in half and non parallel ones in equal ratios. the smaller part of those non parallel sides is of the length 4(8/2) and bigger one is of the length 9(18/2). Now that we know the length of the non parallel sides is 13(4 + 9) we can get the distance between the tow parallel sides using Pythagoras theorem, which gives us 12. Thus, radius is 6.

Now if we see the answer choices we have only one choice that is a multiple of 6 or 6ˆ2.

Note: the non parallel sides would touch circle such that the point of touch divides them in the ratio of length of parallel sides i.e. 8/18 = 4/9. Actually, here it is 4 and 9.

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Re: A circle is inscribed in an isosceles trapezoid with bases 8 and 18 [#permalink]
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Tangents drawn to a circle are equal in length.

Therefore,
We take one tangent of 18/2 = 9 and other of 8/2 = 4
Total length of hypotenuse = 9+4 = 13
Base = 5

Height = 12 (Pythagoras)