Question

Square $EFGH$ is inside square $ABCD$ so that each side of $EFGH$ can be extended to pass through a vertex of $ABCD.$ Square $ABCD$ has side length $\sqrt{50},$ $E$ is between $B$ and $H,$ and $BE = 1.$ What is the area of the inner square $EFGH?$

$25$

$32$

$36$

$40$

$42$

Answer: C

Concepts:square (geometry)Pythagorean Theorem congruence (geometry)

Difficulty rating: 1460

Solution:

By the symmetry of the figure, triangles $ABH,$ $BCE,$ $CDF,$ and $DAG$ are congruent right triangles. Hence $\begin{aligned} &BH = CE = \sqrt{BC^2 - BE^2} \\ &= \sqrt{50 - 1} = 7. \end{aligned}$

Since $E$ lies between $B$ and $H,$ the side of the inner square is $EH = BH - BE = 7 - 1 = 6.$

Therefore the area of $EFGH$ is $6^2 = 36.$

Thus, the correct answer is C.

2005 AMC 12A Problem 7

Answer: C

Difficulty rating: 1460

Solution:

Problem 7 in Other Years