Question

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 $\dfrac{a \pi - b}{c},$ where $a, b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a + b + c?$

$260$

$855$

$1235$

$1565$

$1997$

Solution:

Notice that $7^2 + 24^2$ and $15^2 + 20^2$ are both the same. This forces $AC = 25$ since otherwise $\angle B$ and $\angle D$ would both be acute or obtuse, violating the fact that their sum is $180^{\circ}.$

Also since $\angle B$ is right, we know that $AC$ is the diameter of the circle. The area of the circle is then $\dfrac{625}{4} \pi.$

To find the area of the quadrilateral, we can find the area of each of the triangles, which is $\dfrac{1}{2}(7 \cdot 24 + 20 \cdot 15) =$ $84 + 150 = 234.$

To find the area outside the quadrilateral, we subtract to get $\dfrac{625}{4} \pi - 234 = \dfrac{625 \pi - 936}{4}.$

Therefore, $a + b + c = 625 + 936 + 4$ $= 1565.$

Thus, D is the correct answer.

2022 AMC 10A Problem 15

Solution:

Problem 15 in Other Years