Question

A square with area $4$ is inscribed in a square with area $5$ , with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$ and the other of length $b$ . What is the value of $ab ?$

$\dfrac{1}5$

$\dfrac{2}5$

$\dfrac{1}{2}$

$1$

$4$

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Written solution:

Since all the triangles can be made from each other by rotating them around, they are all congruent. Therefore, we can place the $a$ as we have. The total area of the triangles is $5-4 = 1,$ so we have $4$ congruent triangles with a combined area of $1.$ This means the area of each triangle is $\dfrac{1}{4}.$ The area of each triangle is also $\frac {ab}2,$ so $\frac{ab}2 = \dfrac{1}{4}.$ This means $ab = \dfrac{1}{2} .$

Thus, the correct answer is C .

2012 AMC 8 Problem 25

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Problem 25 in Other Years