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

m (Solution 1)
m (Solution 2)
Line 20: Line 20:
 
We then have the equation <math> a + b = \sqrt{5} </math>.
 
We then have the equation <math> a + b = \sqrt{5} </math>.
  
We also know that the sidelength of the smaller square is  <math> 2 </math>, since its area is <math> 4 </math>. Then, the segment of length <math> a </math> and segment of length <math> b </math> form a right triangle whose hypotenuse would have length <math> 2 </math>.  
+
We also know that the side length of the smaller square is  <math> 2 </math>, since its area is <math> 4 </math>. Then, the segment of length <math> a </math> and segment of length <math> b </math> form a right triangle whose hypotenuse would have length <math> 2 </math>.  
  
 
So our second equation is <math> \sqrt{{a^2}+{b^2}} = 2 </math>.
 
So our second equation is <math> \sqrt{{a^2}+{b^2}} = 2 </math>.

Revision as of 23:07, 10 November 2016

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

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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

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