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Area of circle which has a triangle in ita

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by vishal.pathak » Mon Nov 21, 2011 9:44 am
An equilateral triangle that has an area of 93 is inscribed in a circle. What is the area of the circle?
A. π6 B. π9 C. π12 D. 9π √3 E. 18π √3

Any quick way of doing this? OA not available
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Source: — Problem Solving |

by kanwar86 » Mon Nov 21, 2011 10:10 am
vishal.pathak wrote:An equilateral triangle that has an area of 93 is inscribed in a circle. What is the area of the circle?
A. π6 B. π9 C. π12 D. 9π √3 E. 18π √3

Any quick way of doing this? OA not available
Kindly recheck the value of area. For one of the options to be correct, the area can't be 93.
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by rijul007 » Mon Nov 21, 2011 10:21 am
An equilateral triangle that has an area of 93 is inscribed in a circle. What is the area of the circle?
A. π6 B. π9 C. π12 D. 9π √3 E. 18π √3

Any quick way of doing this? OA not available
yep, kanwar86 is right..
please check your question again
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by vishal.pathak » Mon Nov 21, 2011 10:30 am
rijul007 wrote:
An equilateral triangle that has an area of 93 is inscribed in a circle. What is the area of the circle?
A. π6 B. π9 C. π12 D. 9π √3 E. 18π √3

Any quick way of doing this? OA not available
yep, kanwar86 is right..
please check your question again
This is what the question says, this was given to me by a friend so yes, source is not a standard one
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by kanwar86 » Mon Nov 21, 2011 10:37 am
vishal.pathak wrote:
rijul007 wrote:
An equilateral triangle that has an area of 93 is inscribed in a circle. What is the area of the circle?
A. π6 B. π9 C. π12 D. 9π √3 E. 18π √3

Any quick way of doing this? OA not available
yep, kanwar86 is right..
please check your question again
This is what the question says, this was given to me by a friend so yes, source is not a standard one
The answer to this question, considering area of equilateral triangle to be 93, is 124π/√3
Approach:
1) Find altitude of equilateral triangle.
2) Centroid divides the altitude (median) of equilateral triangle in the ratio 2:1 with the larger part towards the vertex.
3) Larger part, found in step 2, is the circumradius from which area of circumcircle can be calculated.
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by rijul007 » Mon Nov 21, 2011 10:41 am
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by vishal.pathak » Wed Nov 23, 2011 4:55 am
rijul007 wrote:Image
Incorrect: cos 30 = sqrt(3)/2

sqrt(3)/2 = a/2r

or r = a/sqrt(3)
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