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?

$85$

$93$

$100$

$117$

$125$

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 $9$ . From the diagram $RC=9$ , and therefore $BC^2=BR^2+RC^2=6^2+9^2=117.$

The area of the square is $117$ .

Thus, the answer is D .

2021 AMC 10B Fall Problem 15

Solution:

Problem 15 in Other Years