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Difference between revisions of "2011 AMC 8 Problems/Problem 13"
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==Problem==
 
==Problem==
 
Two congruent squares, <math>ABCD</math> and <math>PQRS</math>, have side length <math>15</math>. They overlap to form the <math>15</math> by <math>25</math> rectangle <math>AQRD</math> shown. What percent of the area of rectangle <math>AQRD</math> is shaded?  
 
Two congruent squares, <math>ABCD</math> and <math>PQRS</math>, have side length <math>15</math>. They overlap to form the <math>15</math> by <math>25</math> rectangle <math>AQRD</math> shown. What percent of the area of rectangle <math>AQRD</math> is shaded?  
 +
 
<asy>
 
<asy>
 
filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black);
 
filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black);
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==Solution==
 
==Solution==
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The overlap length is <math>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>
 +
 
 +
==Solution 2==
 +
The length of BP is 5. the ratio of the areas is <math>\dfrac{5\cdot 15}{25\cdot 15}=\dfrac{5}{25}=20\% </math>
 +
-Megacleverstarfish15
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 +
==Solution 3(similar to Solution 1)==
 +
 
 +
To find the overlap length, we do the total length of the squares and subtract <math>25</math>(side length of figure). <math>(15 + 15) - 25 = 5</math>, so the overlap length is <math>5</math>. To find what percentage of <math>AQRD</math> is shaded, we divide the shaded part by the area of the <math>AQRD</math>, so the percentage is <math>\dfrac{15 \cdot 5}{15 \cdot 25}</math> = <math>\dfrac{5}{25}</math> = <math>\dfrac{1}{5}</math> = <math>\dfrac{20}{100}</math> = <math>20</math>%, so the answer is <math>\boxed{ \textbf{(C)}\ \text{20} }</math>.
 +
 
 +
~NXC
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=mYn6tNxrWBU
 +
 
 +
~==SpreadTheMathLove==
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/VLS29yiMHSw
  
 
==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}}

Latest revision as of 19:16, 18 November 2024

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

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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