2022 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 15 Easy

Quadrilateral ABCD with side lengths AB=7, BC=24, CD=20, DA=15 is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form  \frac{a\pi-b}{c} , where a, b, and c are positive integers such that a and c have no common prime factor. What is a+b+c?

  • A.

    260

  • B.

    855

  • C.

    1235

  • D.

    1565

  • E.

    1997

Answer:D

7^2+24^2=25^2=15^2+20^2

\angle{B} + \angle {D}= 180 ^{\circ}

If \angle{B}>90^{\circ}, AC>75

\because \angle{B}+ \angle{D}=180^{\circ}

\therefore \angle D<90^{\circ},AC<25, so there is a contradiction.

If \angle{B}<90^{\circ}, we also have a contradiction.

\therefore \angle B=\angle D = 90^{\circ},

\therefore AC is the diameter of the circle.

\therefore A_{ABCD}=A_{ABC}+A_{ACD}=\frac 12 \times 7 \times 24+\frac 12 \times 15 \times 20=234,

A_{circle}=\pi \cdot \left(\frac{25}{2} \right)^2=\frac{625\pi}{4}.

\therefore \frac{a\pi-b}{c}=\frac{625\pi}{4}-234=\frac{625\pi-936}{4}

\therefore a+b+c=625+936+4=1565.

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