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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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
Small hint:
The two right triangles use legs 15,20,2515,20,25 in opposite roles.
Big hint:
Drop an altitude in each isosceles triangle.
Solution:
Since these triangles are isosceles, the altitude shown from the apex of each triangle bisects its base and creates two congruent right triangles.
Using the Pythagorean theorem, we get that the altitude of the triangle with area AA equals 252−152=20.\sqrt{25^2 - 15^2} = 20. Similarly, we get that the altitude of the triangle with area BB equals 252−202=15.\sqrt{25^2 - 20^2} = 15.
With these altitudes, we can calculate the areas of the triangles. We get that A=12⋅20⋅30=300. A = \dfrac{1}{2} \cdot 20 \cdot 30 = 300. Similarly, B=12⋅15⋅40=300. B = \dfrac{1}{2} \cdot 15 \cdot 40 = 300.
Therefore, A=B.A = B.
Thus, C is the correct answer.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.