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

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 .

2021 AMC 10B Fall Problem 15

Answer: D

Difficulty rating: 1950

Solution:

Problem 15 in Other Years