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Difference between revisions of "2012 AMC 8 Problems/Problem 25"

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Now, subtract, and obtain the equation <math> 2ab = 1 </math>. We can deduce that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
 
Now, subtract, and obtain the equation <math> 2ab = 1 </math>. We can deduce that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
  
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{{AMC8 box|year=2012|num-b=24|after=Last Problem}}
 
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Since we know 4 of the triangles both have side lengths a and b, we can create an equation.

Revision as of 05:23, 25 July 2017

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 $ab$ 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 $ab$ is $\boxed{\textbf{(C)}\ \frac{1}2}$.

Solution 3

Since we know 4 of the triangles both have side lengths a and b, we can create an equation.

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