Difference between revisions of "2005 AMC 10A Problems/Problem 8"

(Solution)
(See Also)
Line 25: Line 25:
 
[[Category:Area Problems]]
 
[[Category:Area Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
*[[2005 AMC 10A Problems]]
 +
 +
*[[2005 AMC 10A Problems/Problem 7|Previous Problem]]
 +
 +
*[[2005 AMC 10A Problems/Problem 9|Next Problem]]

Revision as of 15:15, 10 August 2017

Problem

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE$=1. What is the area of the inner square $EFGH$?

AMC102005Aq.png

$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42$

Solution

We see that side $BE$, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, $AH = 1$. Then $HB = HE + BE = HE + 1$, and $HE$ is one of the sides of the square whose area we want to find. So:

\[1^2 + (HE+1)^2=\sqrt{50}^2\]

\[1 + (HE+1)^2=50\]

\[(HE+1)^2=49\]

\[HE+1=7\]

\[HE=6\] So, the area of the square is $6^2=\boxed{36} \Rightarrow \mathrm{(C)}$.

See Also

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