1999 AMC 12 Problem 16

Below is the professionally curated solution for Problem 16 of the 1999 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.

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Concepts:rhombus incircle, incenter, and inradius Pythagorean Theorem triangle area

Difficulty rating: 1610

16.

What is the radius of a circle inscribed in a rhombus with diagonals of length $10$ and $24?$

$4$

$\dfrac{58}{13}$

$\dfrac{60}{13}$

$5$

$6$

Solution:

The half-diagonals are $5$ and $12,$ so each side of the rhombus is $\sqrt{5^2 + 12^2} = 13.$ One of the four right triangles formed by the diagonals has legs $5$ and $12$ and area $30.$

The altitude from the center to the side of length $13$ is $\dfrac{2 \cdot 30}{13} = \dfrac{60}{13},$ which is the inscribed circle's radius.

Thus, the correct answer is C.

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