1995 AMC 8 Problem 24

Attempt Problem 24 of the 1995 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 1995 AMC 8 solutions, or check the answer key.

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24.

In parallelogram ABCD,ABCD, DE‾\overline{DE} is the altitude to the base AB‾\overline{AB} and DF‾\overline{DF} is the altitude to the base BC‾.\overline{BC}. (Both pictures represent the same parallelogram.) If DC=12,DC = 12, EB=4,EB = 4, and DE=6,DE = 6, then DF=DF =
6.46.4
77
7.27.2
88
1010
Answer: C
Concepts:parallelogramPythagorean Theorem
Small hint:
Opposite sides are equal, so AB=12;AB = 12; find AE,AE, then ADAD with the right triangle ADEADE
Big hint:
The area equals base ×\times height for either base: AB⋅DE=BC⋅DFAB \cdot DE = BC \cdot DF
Solution:
Since AB=DC=12,AB = DC = 12, we get AE=12−4=8.AE = 12 - 4 = 8. In right triangle ADE,ADE, AD=82+62=10,AD = \sqrt{8^2 + 6^2} = 10, so BC=AD=10.BC = AD = 10.
The area is AB⋅DE=12⋅6=72,AB \cdot DE = 12 \cdot 6 = 72, and also BC⋅DF=10⋅DF.BC \cdot DF = 10 \cdot DF. So DF=7210=7.2.DF = \dfrac{72}{10} = 7.2.
Thus, the correct answer is C .
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