Difference between revisions of "2011 AMC 8 Problems/Problem 13"
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==Solution== | ==Solution== | ||
− | 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> |
==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 00:52, 27 December 2022
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
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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