Difference between revisions of "2011 AMC 10A Problems/Problem 11"
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<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math> | <math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == Solution == | ||
Let <math>8s</math> be the length of the sides of square <math>ABCD</math>, then the length of one of the sides of square <math>EFGH</math> is <math>\sqrt{(7s)^2+s^2}=\sqrt{50s^2}</math>, and hence the ratio in the areas is <math>\frac{\sqrt{50s^2}^2}{(8s)^2}=\frac{50}{64} = \boxed{\frac{25}{32} \ \mathbf{(B)}}</math>. | Let <math>8s</math> be the length of the sides of square <math>ABCD</math>, then the length of one of the sides of square <math>EFGH</math> is <math>\sqrt{(7s)^2+s^2}=\sqrt{50s^2}</math>, and hence the ratio in the areas is <math>\frac{\sqrt{50s^2}^2}{(8s)^2}=\frac{50}{64} = \boxed{\frac{25}{32} \ \mathbf{(B)}}</math>. | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/GrCtzL0S-Uo?t=292 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
== See Also == | == See Also == | ||
{{AMC10 box|year=2011|ab=A|num-b=10|num-a=12}} | {{AMC10 box|year=2011|ab=A|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 03:42, 23 January 2023
Problem 11
Square has one vertex on each side of square
. Point
is on
with
. What is the ratio of the area of
to the area of
?
Solution
Let be the length of the sides of square
, then the length of one of the sides of square
is
, and hence the ratio in the areas is
.
Video Solution by OmegaLearn
https://youtu.be/GrCtzL0S-Uo?t=292
~ pi_is_3.14
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.