2012 AMC 8 Problem 25
Attempt Problem 25 of the 2012 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
25.
A square with area is inscribed in a square with area , 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
Answer: C
Small hint:
The area between the squares is split into four congruent right triangles.
Big hint:
Each small triangle has area .
Video solution:
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Written solution:
Since all the triangles can be made from each other by rotating them around, they are all congruent. Therefore, we can place the as we have. The total area of the triangles is so we have congruent triangles with a combined area of This means the area of each triangle is The area of each triangle is also so This means
Thus, the correct answer is C .
Problem 25 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
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