2002 AMC 10A Problem 13 — Solution | LIVE by Po-Shen Loh

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2002 AMC 10A Problem 13

Below is the professionally curated solution for Problem 13 of the 2002 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10A solutions, or check the answer key.

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Concepts:right triangle altitude triangle area

Difficulty rating: 1280

13.

The sides of a triangle have lengths of $15,$ $20,$ and $25.$ Find the length of the shortest altitude.

$6$

$12$

$12.5$

$13$

$15$

Solution:

Since $15^2+20^2=225+400=625=25^2,$ the triangle is right with legs $15$ and $20,$ and area $\dfrac{1}{2}(15)(20)=150.$

The shortest altitude falls to the longest side $25,$ and equals $\dfrac{2\cdot 150}{25}=12.$

Thus, the correct answer is B.

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