# 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