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2012 AMC 8 Problems/Problem 25

Problem

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 $ab$? $[asy] draw((0,2)--(2,2)--(2,0)--(0,0)--cycle); draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle); label("a",(-0.1,0.15)); label("b",(-0.1,1.15));[/asy]$ $\textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4$

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 $a \cdot b$ is $\boxed{\textbf{(C)}\ \frac{1}2}$.

Solution 2

To solve this problem you could also use algebraic manipulation.

Since the area of the large square is $5$, the sidelength is $\sqrt{5}$.

We then have the equation $a + b = \sqrt{5}$.

We also know that the side length of the smaller square is $2$, since its area is $4$. Then, the segment of length $a$ and segment of length $b$ form a right triangle whose hypotenuse would have length $2$.

So our second equation is $\sqrt{{a^2}+{b^2}} = 2$.

Square both equations. $a^2 + 2ab + b^2 = 5$ $a^2 + b^2 = 4$

Now, subtract, and obtain the equation $2ab = 1$. We can deduce that the value of $a \cdot b$ is $\boxed{\textbf{(C)}\ \frac{1}2}$.

Solution 3 (similar to solution 1)

Since we know 4 of the triangles both have side lengths a and b, we can create an equation, which the area of the inner square plus the sum of the 4 triangles being the area of the outer square. $4 + 2ab = 5$

which gives us the value of $a \cdot b$, which 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 $a \cdot b=\boxed{\textbf{(C) }\frac{1}{2}}$

Solution 5

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 