2002 AMC 10A Problem 13

Attempt Problem 13 of the 2002 AMC 10A 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 2002 AMC 10A solutions, or check the answer key.

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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

Check whether

15,

20,

25

is a right triangle

The shortest altitude is drawn to the longest side

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.

Problem 13 in other years