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Difference between revisions of "2011 AMC 8 Problems/Problem 13"

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==Solution==
 
==Solution==
 
The overlap length is <math>2(15)-25=5</math>, so the shaded area is <math>5 \cdot 15 =75</math>. The area of the whole shape is <math>25 \cdot 15 = 375</math>. The fraction <math>\dfrac{75}{375}</math> reduces to <math>\dfrac{1}{5}</math> or 20%. Therefore, the answer is <math>\boxed{ \textbf{(C)}\ \text{20} }</math>
 
The overlap length is <math>2(15)-25=5</math>, so the shaded area is <math>5 \cdot 15 =75</math>. The area of the whole shape is <math>25 \cdot 15 = 375</math>. The fraction <math>\dfrac{75}{375}</math> reduces to <math>\dfrac{1}{5}</math> or 20%. Therefore, the answer is <math>\boxed{ \textbf{(C)}\ \text{20} }</math>
−
 
+
==Video Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=mYn6tNxrWBU
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=12|num-a=14}}
 
{{AMC8 box|year=2011|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:02, 17 December 2023

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 $\boxed{ \textbf{(C)}\ \text{20} }$

Video Solution 1 by SpreadTheMathLove

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

See Also

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

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