2015 AMC 8 Problem 21

Attempt Problem 21 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).

21.

In the given figure hexagon ABCDEFABCDEF is equiangular, ABJIABJI and FEHGFEHG are squares with areas 1818 and 3232 respectively, △JBK\triangle JBK is equilateral and FE=BC.FE=BC. What is the area of △KBC?\triangle KBC?
62 6\sqrt{2}
99
1212
929\sqrt{2}
3232
Answer: C
Concepts:equiangular polygonequilateral triangletriangle area
Small hint:
Find JBJB and BCBC from the two square areas.
Big hint:
In the diagram, BKBK is perpendicular to BC.BC.
Video solution:
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Written solution:
The square with area 1818 has side length JB=18=32.JB=\sqrt{18}=3\sqrt2. Since △JBK\triangle JBK is equilateral, BK=32.BK=3\sqrt2.
The square with area 3232 has side length FE=32=42.FE=\sqrt{32}=4\sqrt2. Since FE=BC,FE=BC, we have BC=42.BC=4\sqrt2.
At B,B, the equilateral triangle, square, and equiangular hexagon contribute angles of 60∘,60^\circ, 90∘,90^\circ, and 120∘,120^\circ, respectively. Hence ∠KBC=360∘−60∘−90∘−120∘=90∘.\begin{aligned} \angle KBC &= 360^\circ-60^\circ-90^\circ\\ &\qquad-120^\circ\\ &=90^\circ. \end{aligned} Therefore [△KBC]=12(32)(42)=12.[\triangle KBC]=\dfrac12(3\sqrt2)(4\sqrt2)=12.
Thus, C is the correct answer.
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