2011 AMC 8 Problem 16

Attempt Problem 16 of the 2011 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 8 solutions, or check the answer key.

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16.

Let AA be the area of the triangle with sides of length 25,25,25, 25, and 30.30. Let BB be the area of the triangle with sides of length 25,25,25, 25, and 40.40. What is the relationship between AA and B?B?

A=916BA = \dfrac{9}{16}B

A=34BA = \dfrac{3}{4}B

A=BA = B

A=43BA = \dfrac{4}{3}B

A=169BA = \dfrac{16}{9}B

Answer: C
Concepts:isosceles triangletriangle areaPythagorean Theorem
Difficulty rating: 1340
Solution:

Since these triangles are isosceles, we can drop altitudes to create two congruent right triangles as shown in the diagram.

Using the Pythagorean theorem, we get that the altitude of the triangle with area AA equals 252152=20.\sqrt{25^2 - 15^2} = 20. Similarly, we get that the altitude of the triangle with area BB equals 252202=15.\sqrt{25^2 - 20^2} = 15.

With these altitudes, we can calculate the areas of the triangles. We get that A=122030=300. A = \dfrac{1}{2} \cdot 20 \cdot 30 = 300. Similarly, B=121540=300. B = \dfrac{1}{2} \cdot 15 \cdot 40 = 300.

Therefore, A=B.A = B.

Thus, C is the correct answer.

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