2012 AMC 8 Problems/Problem 25
Contents
Problem
A square with area 4 is inscribed in a square with area 5, with each vertex of the smaller square on a 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 side length 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 (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.
which gives us the value of , which is .
Solution 4
First, observe that the given squares have areas and .
Then, observe that the 4 triangles with side lengths and have a combined area of .
We have, that is the total area of the 4 triangles in terms of and .
Since , we divide by two getting
Video Solution by Punxsutawney Phil
~sugar_rush
https://www.youtube.com/watch?v=QEwZ_17PQ6Q
Video Solution 2
https://youtu.be/MhxGq1sSA6U ~savannahsolver
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
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