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

Revision as of 19:06, 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)

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
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

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

@@ Line 22: / Line 22: @@
 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
+==Solution 3(similar to Solution 1)==
 ==Video Solution==

Page

Toolbox

Search