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What is the area of the shaded region in the figure above, i
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13 Sep 2020, 08:30
13 Sep 2020, 08:30
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Question Stats:
58% (03:26) correct
41% (02:36) wrong based on 29 sessions
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GRE What is the area of the shaded region.png [ 12.61 KiB | Viewed 4879 times ]
What is the area of the shaded region in the figure above, in terms of a, b, and c ?
A. \(\sqrt{3} (a^2+b^2+c^2)\)
B. \(\frac{\sqrt{3}}{2}\) \((a^2-b^2-c^2)\)
C. \(\frac{\sqrt{3}}{2}\) \((a^2-b^2+c^2)\)
D. \(\frac{\sqrt{3}}{2}\) \((a^2+b^2-c^2)\)
E. \(\frac{\sqrt{3}}{2}\) \((a^2+b^2+c^2)\)
Re: What is the area of the shaded region in the figure above, i
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14 Sep 2020, 21:48
14 Sep 2020, 21:48
3
by using pythagoras theorem for triangle ABC
\((2a)^2=x^2+((a√3)^2)\) (let x be unknown side of triangle ABC)
then x=a
area of big triangle =(1/2)*a*a√3=\((√3/2)*a^2\)
we need to remove the area of unshaded region to get area of shaded region
lets find area of unshaded region
from ΔDGH ∠D=30 and HG=b-c
then DG=√3(b-c) (by applying tan formula)
area of ΔDGH=\((√3(b-c)^2 /2)\)------(2)
now area of ▭DEFG = √3(b-c)*c
we can write this as=\((√3/2)*(2bc-2c^2)\) -----(3)
total area of unshaded region=\((√3/2)*(b^2-c^2)\)
now total area of shaded region will be eqn(1)-eqn(3)=\((√3/2)*(a^2-b^2+c^2)\)
option C is correct
\((2a)^2=x^2+((a√3)^2)\) (let x be unknown side of triangle ABC)
then x=a
area of big triangle =(1/2)*a*a√3=\((√3/2)*a^2\)
we need to remove the area of unshaded region to get area of shaded region
lets find area of unshaded region
from ΔDGH ∠D=30 and HG=b-c
then DG=√3(b-c) (by applying tan formula)
area of ΔDGH=\((√3(b-c)^2 /2)\)------(2)
now area of ▭DEFG = √3(b-c)*c
we can write this as=\((√3/2)*(2bc-2c^2)\) -----(3)
total area of unshaded region=\((√3/2)*(b^2-c^2)\)
now total area of shaded region will be eqn(1)-eqn(3)=\((√3/2)*(a^2-b^2+c^2)\)
option C is correct
Attachments
pc2.jpg [ 9.67 KiB | Viewed 4756 times ]
Re: What is the area of the shaded region in the figure above, i
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24 Jan 2021, 08:50
24 Jan 2021, 08:50
