1999 AMC 8 Problem 14

Attempt Problem 14 of the 1999 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 1999 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

In trapezoid ABCD,ABCD, the sides ABAB and CDCD are equal. The perimeter of ABCDABCD is
2727
3030
3232
3434
4848
Answer: D
Concepts:trapezoidPythagorean Theorem
Small hint:
Drop an altitude from BB
Big hint:
The two side right triangles have horizontal leg 44
Solution:
Let HH be where the altitude from BB to AD‾\overline{AD} intersects AD‾.\overline{AD}.
Then we have that AH=16−82=4, AH = \dfrac{16 - 8}{2} = 4, since AB=CD.AB = CD.
We then have that AB=42+32=25=5. AB = \sqrt{4^2 + 3^2} = \sqrt{25} = 5.
Then the perimeter is 8+16+2⋅5=34. 8 + 16 + 2 \cdot 5 = 34.
Thus, D is the correct answer.
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