2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 15 Medium

In square A B C D, points P and Q lie on \overline{A D} and \overline{A B}, respectively. Segments \overline{B P} and \overline{C Q} intersect at right angles at R, with B R=6 and P R=7. What is the area of the square?(2021 AMC Fall 10B, Question #15)

  • A.

    85

  • B.

    93

  • C.

    100

  • D.

    117

  • E.

    125

Answer:D

Solution 1:

Note that \triangle A P B \cong \triangle B Q C. Then, it follows that \overline{P B} \cong \overline{Q C}. Thus, Q C=P B=P R+R B=7+6=13. Define x to be the length of side C R, then R Q=13-x. Because \overline{B R} is the altitude of the triangle, we can use the property that Q R \cdot R C=B R^{2}. Substituting the given lengths, we have (13-x) \cdot x=36 \text {. } Solving, gives x=4 and x=9. We eliminate the possibilty of x=4 because R C>Q R. Thus, the side length of the square, by Pythagorean Theorem, is \sqrt{9^{2}+6^{2}}=\sqrt{81+36}=\sqrt{117} \text {. } Thus, the area of the sqaure is (\sqrt{117})^{2}=117. Thus, the answer is (D) 117

Solution 2:

As above, note that \triangle B P A \cong \triangle C Q B, which means that Q C=13. In addition, note that B R is the altitude of a right triangle to its hypotenuse, so \triangle B Q R \sim \triangle C B R \sim \triangle C Q B. Let the side length of the square be x; using similarity side ratios of \triangle B Q R to \triangle C Q B, we get \frac{6}{x}=\frac{Q B}{13} \Longrightarrow Q B \cdot x=78 Note that Q B^{2}+x^{2}=13^{2}=169 by the Pythagorean theorem, so we can use the expansion (a+b)^{2}=a^{2}+2 a b+b^{2} to produce two equations and two variables; \begin{aligned} &(Q B+x)^{2}=Q B^{2}+2 Q B \cdot x+x^{2} \Longrightarrow(Q B+x)^{2}=169+2 \cdot 78 \Longrightarrow Q B+x=\sqrt{13(13)+13(12)}=\sqrt{13 \cdot 25}=5 \sqrt{13} \\ &(Q B-x)^{2}=Q B^{2}-2 Q B \cdot x+x^{2} \Longrightarrow(Q B-x)^{2}=169-2 \cdot 78 \Longrightarrow Q B-x=\sqrt{13(13)-13(12}=\sqrt{13 \cdot 1}=\sqrt{13} \end{aligned} We want x^{2}, so we want to find x. Subtracting the first equation from the second, we get 2 x=6 \sqrt{13} \Longrightarrow x=3 \sqrt{13} Then x^{2}=\left(3 \sqrt{13}^{2}\right)=9 \cdot 13=117=D