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In the figure above, FG = 4, and FH is a diameter of the cir
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31 Aug 2020, 10:02
31 Aug 2020, 10:02
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91% (00:44) correct
8% (00:50) wrong based on 23 sessions
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GRE In the figure above, FG = 4, and FH.png [ 25.2 KiB | Viewed 2948 times ]
In the figure above, FG = 4, and FH is a diameter of the circle. What is the area of the circle?
A. \(4 \pi\)
B. \(8 \pi\)
C. \(12 \pi\)
D. \(16 \pi\)
E. \(20 \pi\)
Re: In the figure above, FG = 4, and FH is a diameter of the cir
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04 Sep 2020, 22:09
04 Sep 2020, 22:09
1
This relies on knowing your special triangles, which include 30-60-90, 45-45-90, and 60-60-60.
If it helps, redraw the triangle so FG is at the bottom. For a 30-60-90, if the short side FG is 4, the hypoteneuse FH is 8. The radius would then be 4, and the area would be pi*r^2 or 16*pi.
If you forget the dimensions of 30-60-90 triangles, imagine doubling this triangle by reflecting FGH along the hypoteneuse FH. Then you would have an equilateral triangle, and each side would be 8.
In the figure above, FG = 4, and FH is a diameter of the circle. What is the area of the circle?
A. \(4 \pi\)
B. \(8 \pi\)
C. \(12 \pi\)
D. \(16 \pi\)
E. \(20 \pi\)
If it helps, redraw the triangle so FG is at the bottom. For a 30-60-90, if the short side FG is 4, the hypoteneuse FH is 8. The radius would then be 4, and the area would be pi*r^2 or 16*pi.
If you forget the dimensions of 30-60-90 triangles, imagine doubling this triangle by reflecting FGH along the hypoteneuse FH. Then you would have an equilateral triangle, and each side would be 8.
Carcass wrote:
Attachment:
GRE In the figure above, FG = 4, and FH.png
In the figure above, FG = 4, and FH is a diameter of the circle. What is the area of the circle?
A. \(4 \pi\)
B. \(8 \pi\)
C. \(12 \pi\)
D. \(16 \pi\)
E. \(20 \pi\)
Re: In the figure above, FG = 4, and FH is a diameter of the cir
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09 Sep 2020, 01:59
09 Sep 2020, 01:59
1
Since the triangle is a 30:60:90 angle then the hypotenuse(diameter) will equal 2*smallest side
Hence D=2*4=8
Area equal 4*4*π=16πD
Posted from my mobile device
Hence D=2*4=8
Area equal 4*4*π=16πD
Posted from my mobile device
