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2005 AMC 10A Problems/Problem 8: Difference between revisions

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==Solution==
==Solution==
We see that side <math>BE</math>, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So:
We see that side <math>BE</math>, which we know is <math>1</math>, is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So:


<cmath>1^2 + (HE+1)^2=\sqrt{50}^2</cmath>
<cmath>1^2 + (HE+1)^2=\sqrt{50}^2</cmath>
Line 18: Line 18:


<cmath>HE=6</cmath>  
<cmath>HE=6</cmath>  
So, the area of the square is <math>6^2=\boxed{36} \Rightarrow \mathrm{(C)}</math>.
So, the area of the square is <math>6^2=\boxed{\textbf{(C) }36}</math>


==See also==
==See also==

Revision as of 10:58, 13 December 2021

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 $EFGH$?

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

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 won’t 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 $6^2=\boxed{\textbf{(C) }36}$

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

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