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.

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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
Difficulty rating: 1350
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.

Problem 13#13
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