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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− | + | Since we know 4 of the triangles both have side lengths a and b, we can create an equation. |
Revision as of 06:23, 25 July 2017
Contents
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 , and the other of length
. What is the value of
?
Solution 1
The total area of the four congruent triangles formed by the squares is . Therefore, the area of one of these triangles is
. The height of one of these triangles is
and the base is
. Using the formula for area of the triangle, we have
. Multiply by
on both sides to find that the value of
is
.
Solution 2
To solve this problem you could also use algebraic manipulation.
Since the area of the large square is , the sidelength is
.
We then have the equation .
We also know that the side length of the smaller square is , since its area is
. Then, the segment of length
and segment of length
form a right triangle whose hypotenuse would have length
.
So our second equation is .
Square both equations.
Now, subtract, and obtain the equation . We can deduce that the value of
is
.
Solution 3
Since we know 4 of the triangles both have side lengths a and b, we can create an equation.