2011 AMC 8 Problems/Problem 13: Difference between revisions
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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>. | 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== | ==Video Solution== | ||
Revision as of 19:16, 18 November 2024
Problem
Two congruent squares,
and
, have side length
. They overlap to form the
by
rectangle
shown. What percent of the area of rectangle
is shaded?
Solution
The overlap length is
, so the shaded area is
. The area of the whole shape is
. The fraction
reduces to
or 20%. Therefore, the answer is
Solution 2
The length of BP is 5. the ratio of the areas is
-Megacleverstarfish15
Solution 3(similar to Solution 1)
To find the overlap length, we do the total length of the squares and subtract
(side length of figure).
, so the overlap length is
. To find what percentage of
is shaded, we divide the shaded part by the area of the
, so the percentage is
=
=
=
=
%, so the answer is
.
~NXC
Video Solution
https://www.youtube.com/watch?v=mYn6tNxrWBU
~==SpreadTheMathLove==
Video Solution by WhyMath
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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