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.

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25.

A square with area 44 is inscribed in a square with area 55, 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 aa and the other of length bb. What is the value of ab? ab ?

15 \dfrac{1}5

25 \dfrac{2}5

12 \dfrac{1}{2}

1 1

4 4

Answer: C
Concepts:square (geometry)congruence (geometry)area decomposition
Difficulty rating: 1790
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 aa as we have. The total area of the triangles is 54=1,5-4 = 1, so we have 4 4 congruent triangles with a combined area of 1.1. This means the area of each triangle is 14. \dfrac{1}{4}. The area of each triangle is also ab2, \frac {ab}2, so ab2=14. \frac{ab}2 = \dfrac{1}{4}. This means ab=12.ab = \dfrac{1}{2} .

Thus, the correct answer is C .

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