2015 AMC 8 Problem 6

Attempt Problem 6 of the 2015 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 8 solutions, or check the answer key.

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

6.

In △ABC,\triangle ABC, AB=BC=29,AB=BC=29, and AC=42.AC=42. What is the area of △ABC?\triangle ABC?
100 100
420 420
500 500
609 609
701 701
Answer: B
Concepts:isosceles trianglePythagorean Theoremtriangle area
Small hint:
Drop the altitude from BB to AC.AC.
Big hint:
The altitude bisects AC,AC, giving a right triangle with hypotenuse 29.29.
Video solution:
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Written solution:
Drop the altitude from BB to AC,AC, meeting ACAC at X.X. Since AB=BC,AB=BC, point XX is the midpoint of AC,AC, so AX=21.AX=21.
In right triangle ABX,ABX, the altitude is BX=292−212=400=20. \begin{aligned} BX &= \sqrt{29^2-21^2} \\ &= \sqrt{400} = 20. \end{aligned}
The area of △ABC\triangle ABC is 12⋅42⋅20=420.\dfrac12\cdot42\cdot20=420.
Thus, B is the correct answer.
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