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 that each of the triangles have side lengths and , we can create an equation: the area of the inner square plus the sum of the triangles equals 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 ~David
Video Solution 2
https://youtu.be/MhxGq1sSA6U ~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=2
~ pi_is_3.14
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.