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

LOOK AT THIS OK

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2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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 AJHSME/AMC 8 Problems and Solutions

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