2012 AMC 8 Problems/Problem 25

Problem

A square with area $4$ is inscribed in a square with area $5$ , with each vertex of the smaller square on a 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 $ab$ ?

$[asy] draw((0,2)--(2,2)--(2,0)--(0,0)--cycle); draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle); label("$a$",(-0.1,0.15)); label("$b$",(-0.1,1.15));[/asy]$

$\textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4$

Solution 1

The total area of the four congruent triangles formed by the squares is $5-4 = 1$ . Therefore, the area of one of these triangles is $\frac{1}{4}$ . The height of one of these triangles is $a$ and the base is $b$ . Using the formula for area of the triangle, we have $\frac{ab}{2} = \frac{1}{4}$ . Multiply by $2$ on both sides to find that the value of $ab$ is $\boxed{\textbf{(C)}\ \frac{1}2}$ .

Solution 4

First, observe that the given squares have areas $4$ and $5$ .

Then, observe that the 4 triangles with side lengths $a$ and $b$ have a combined area of $5-4=1$ .

We have, that $4\cdot\frac{ab}{2}=2ab$ is the total area of the 4 triangles in terms of $a$ and $b$ .

Since $2ab=1$ , we divide by two getting $a \cdot b=\boxed{\textbf{(C) }\frac{1}{2}}$

Video Solution by Punxsutawney Phil

https://youtu.be/RyKWp2YDHJM

~sugar_rush

https://www.youtube.com/watch?v=QEwZ_17PQ6Q ~David

Video Solution 2

https://youtu.be/MhxGq1sSA6U ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2

~ pi_is_3.14

2012 AMC 8 (Problems • Answer Key • Resources)
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