2021 AMC 10B Fall Problem 15

Question

In square

ABCD,

points

P

and

Q

lie on

\overline{AD}

and

\overline{AB},

respectively. Segments

\overline{BP}

and

\overline{CQ}

intersect at right angles at

R,

with

BR = 6

and

PR = 7.

What is the area of the square?

85

93

100

117

125

Answer: D

Concepts:square (geometry)congruence (geometry)similarity Pythagorean Theorem

Difficulty rating: 1950

Small hint:

BP=13

by the

6,7,13

triangle

Big hint:

Use the altitude theorem in right triangle

BQC

Solution:

Since

BR=6

and

PR=7,

we have

BP=13.

The right-angle and square-angle chasing gives

\triangle PAB\cong\triangle QBC,

so

CQ=BP=13.

Because

BP\perp CQ,

point

R

is the foot of the altitude from

B

to the hypotenuse

CQ

of right triangle

BQC.

Thus

QR\cdot RC=BR^2=36

and

QR+RC=CQ=13.

So

QR

and

RC

are

4

and

9.

Because

P

lies on

\overline{AD},

we have

AP\le AB.

Congruence gives

BQ=AP,

while the right-triangle projection formulas give

BQ^2=CQ\cdot QR

and

BC^2=CQ\cdot RC.

Hence

QR\le RC,

so

QR=4

and

RC=9.

Therefore,

\begin{aligned} BC^2&=BR^2+RC^2\\ &=6^2+9^2\\ &=117. \end{aligned}

The area of the square is

117.

Thus, the answer is D .