2005 AMC 10A Problems/Problem 8

Problem

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

[asy] unitsize(4cm); defaultpen(linewidth(.8pt)+fontsize(10pt));  pair D=(0,0), C=(1,0), B=(1,1), A=(0,1); pair F=intersectionpoints(Circle(D,2/sqrt(5)),Circle(A,1))[0]; pair G=foot(A,D,F), H=foot(B,A,G), E=foot(C,B,H);  draw(A--B--C--D--cycle); draw(D--F); draw(C--E); draw(B--H); draw(A--G);  label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,NNW); label("$F$",F,ENE); label("$G$",G,SSE); label("$H$",H,WSW); [/asy]

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

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 obviously congruent, using the symmetry of the diagram). So $AH = 1$, and hence $HB = HE + BE = HE + 1$. Since $HE$ is one of the sides of the square whose area we want to find, we can now simply apply Pythagoras' theorem:

\begin{align*}1^2 + (HE+1)^2 = \left(\sqrt{50}\right)^2 &\iff 1 + (HE+1)^2 = 50 \\ &\iff (HE+1)^2 = 49 \\&\iff HE+1 = 7 \qquad \text{(as } HE \text{ is a length, so must be positive)} \\ &\iff HE = 6.\end{align*}

Thus the area of the square is $6^2 = \boxed{\textbf{(C) }36}$.

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
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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