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If the circle with center O has area 9π, what is the area of
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11 Dec 2018, 13:34
11 Dec 2018, 13:34
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Question Stats:
72% (02:23) correct
27% (02:17) wrong based on 83 sessions
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If the circle with center O has area 9π, what is the area of equilateral triangle ABC?
#GREpracticequestion If the circle with center O has area 9π, what is the area of equilateral triangle ABC.jpg [ 11.68 KiB | Viewed 14676 times ]
A. \(9 \sqrt{3}\)
B. \(18\)
C. \(12 \sqrt{3}\)
D. \(24\)
E. \(16 \sqrt{3}\)
Attachment:
#GREpracticequestion If the circle with center O has area 9π, what is the area of equilateral triangle ABC.jpg [ 11.68 KiB | Viewed 14676 times ]
A. \(9 \sqrt{3}\)
B. \(18\)
C. \(12 \sqrt{3}\)
D. \(24\)
E. \(16 \sqrt{3}\)
Re: If the circle with center O has area 9π, what is the area of
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17 Dec 2018, 21:06
17 Dec 2018, 21:06
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the clue here is that we have an equilateral triangle, which means each triangle is a 30,60,90 triangle.
keep that in mind.
the area of the circle is 9pi, which means the radius is 3 and the diameter of the circle is 6.
this diameter is the height of your triangle, now we need to find the base of it. It's already split in two for us, so no need to use 1/2bh for the area, just use bh
since it's a 30, 60, 90 triangle, just divide the height by sqrt(3)
The answer is C
so h =6/Sqrt(3)
then B*H=6/sqrt(3)*6=36/sqrt(3)= 12sqrt(3)
keep that in mind.
the area of the circle is 9pi, which means the radius is 3 and the diameter of the circle is 6.
this diameter is the height of your triangle, now we need to find the base of it. It's already split in two for us, so no need to use 1/2bh for the area, just use bh
since it's a 30, 60, 90 triangle, just divide the height by sqrt(3)
The answer is C
so h =6/Sqrt(3)
then B*H=6/sqrt(3)*6=36/sqrt(3)= 12sqrt(3)
Re: If the circle with center O has area 9π, what is the area of
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12 Oct 2021, 21:21
12 Oct 2021, 21:21
2
This is an equilateral triangle, three sides of equal length is a and three equal 60 degrees angles.
Given that the area of the circle is 9π,
we know area of circle is πr^2=9π
so, r=3
Diameter AD=2r=6,
Now AD=h of triangle=6
we know The height h of an equilateral triangle is
h=a sin60 =6
=>a sqrt(3)/2 =6
=>base a=12/sqrt(3)
area is: 1/2*a*h =1/2*12/sqrt(3) * 6
=(12*3)/sqrt(3)
=12 sqrt(3)
Given that the area of the circle is 9π,
we know area of circle is πr^2=9π
so, r=3
Diameter AD=2r=6,
Now AD=h of triangle=6
we know The height h of an equilateral triangle is
h=a sin60 =6
=>a sqrt(3)/2 =6
=>base a=12/sqrt(3)
area is: 1/2*a*h =1/2*12/sqrt(3) * 6
=(12*3)/sqrt(3)
=12 sqrt(3)
