2011 AMC 8 Problems/Problem 13

Problem

Two congruent squares, $ABCD$ and $PQRS$ , have side length $15$ . They overlap to form the $15$ by $25$ rectangle $AQRD$ shown. What percent of the area of rectangle $AQRD$ is shaded?

$[asy] filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); label("D",(0,0),S); label("R",(25,0),S); label("Q",(25,15),N); label("A",(0,15),N); filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black); label("S",(10,0),S); label("C",(15,0),S); label("B",(15,15),N); label("P",(10,15),N);[/asy]$

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25$

Solution

The overlap length is $2(15)-25=5$ , so the shaded area is $5 \cdot 15 =75$ . The area of the whole shape is $25 \cdot 15 = 375$ . The fraction $\dfrac{75}{375}$ reduces to $\dfrac{1}{5}$ or 20%. Therefore, the answer is $\boxed{ \textbf{(C)}\ \text{20} }$

Solution 2

The length of BP is 5. the ratio of the areas is $\dfrac{5\cdot 15}{25\cdot 15}=\dfrac{5}{25}=20\%$ -Megacleverstarfish15

Solution 3(similar to Solution 1)

To find the overlap length, we do the total length of the squares and subtract $25$ (side length of figure). $(15 + 15) - 25 = 5$ , so the overlap length is $5$ . To find what percentage of $AQRD$ is shaded, we divide the shaded part by the area of the $AQRD$ , so the percentage is $\dfrac{15 \cdot 5}{15 \cdot 25}$ = $\dfrac{5}{25}$ = $\dfrac{1}{5}$ = $\dfrac{20}{100}$ = $20$ %, so the answer is $\boxed{ \textbf{(C)}\ \text{20} }$ .

~NXC

Video Solution

https://www.youtube.com/watch?v=mYn6tNxrWBU

~==SpreadTheMathLove==

Video Solution by WhyMath

https://youtu.be/VLS29yiMHSw

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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