1995 AMC 8 Problem 24

Below is the professionally curated solution for Problem 24 of the 1995 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 AMC 8 solutions, or check the answer key.

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Concepts:parallelogram Pythagorean Theorem

Difficulty rating: 1150

24.

In parallelogram $ABCD,$ $\overline{DE}$ is the altitude to the base $\overline{AB}$ (with $E$ on $\overline{AB}$ ) and $\overline{DF}$ is the altitude to the base $\overline{BC}.$ If $DC = 12,$ $EB = 4,$ and $DE = 6,$ then $DF =$

$6.4$

$7$

$7.2$

$8$

$10$

Solution:

Since $AB = DC = 12,$ we get $AE = 12 - 4 = 8.$ In right triangle $ADE,$ $AD = \sqrt{8^2 + 6^2} = 10,$ so $BC = AD = 10.$

The area is $AB \cdot DE = 12 \cdot 6 = 72,$ and also $BC \cdot DF = 10 \cdot DF.$ So $DF = \dfrac{72}{10} = 7.2.$

Thus, the correct answer is C .

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