2007 AMC 8 Problem 14

Attempt Problem 14 of the 2007 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 2007 AMC 8 solutions, or check the answer key.

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

14.

The base of isosceles △ABC\triangle ABC is 2424 and its area is 60.60. What is the length of one of the congruent sides?
55
88
1313
1414
1818
Answer: C
Concepts:isosceles triangletriangle areaPythagorean Theorem
Small hint:
Draw the altitude from the top vertex to the base.
Big hint:
In an isosceles triangle, that altitude splits the base into two equal parts.
Solution:
Construct BD‾\overline{BD} as the altitude from BB to AC‾.\overline{AC}.
Then 60=12⋅BD⋅24, 60 = \dfrac{1}{2} \cdot BD \cdot 24, which gives us that BD=5.BD = 5.
From this, we apply the Pythagorean Theorem on △ABD:\triangle ABD: AB2=52+122=169=132. AB^2 = 5^2 + 12^2 = 169 = 13^2. This gives us that AB=13.AB = 13.
Thus, C is the correct answer.
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