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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 15"

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==Problem==
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In square <math>ABCD</math>, points <math>P</math> and <math>Q</math> lie on <math>\overline{AD}</math> and <math>\overline{AB}</math>, respectively. Segments <math>\overline{BP}</math> and <math>\overline{CQ}</math> intersect at right angles at <math>R</math>, with <math>BR = 6</math> and <math>PR = 7</math>. What is the area of the square?
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<asy>
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size(170);
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defaultpen(linewidth(0.6)+fontsize(10));
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real r = 3.5;
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pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0),
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R = intersectionpoint(B--P,C--Q);
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draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7));
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dot("$A$",A,S);
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dot("$B$",B,S);
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dot("$C$",C,N);
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dot("$D$",D,N);
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dot("$Q$",Q,S);
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dot("$P$",P,W);
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dot("$R$",R,1.3*S);
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label("$7$",(P+R)/2,NE);
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label("$6$",(R+B)/2,NE);
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</asy>
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<math>\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125</math>
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==Solution==
 
==Solution==
  

Revision as of 11:30, 23 November 2021

Problem

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?

[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]

$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$

Solution

Note that $\triangle APB \cong \triangle BQC.$ Then, it follows that $\overline{PB} \cong \overline{QC}.$ Thus, $QC = PB = PR + RB = 7 + 6 = 13.$ Define $x$ to be the length of side $CR,$ then $RQ = 13-x.$ Because $\overline{BR}$ is the altitude of the triangle, we can use the property that $QR \cdot RC = BR^2.$ Substituting the given lengths, we have \[(13-x) \cdot x = 36.\] Solving, gives $x = 4$ and $x = 9.$ We eliminate the possibilty of $x=4$ because $RC > QR.$ Thus, the side lengnth of the square, by Pythagorean Theorem, is \[\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.\] Thus, the area of the sqaure is $(\sqrt{117})^2 = 117.$ Thus, the answer is $\boxed{(\textbf{D}.)}.$

~NH14

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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