Problem Solving

There are definitely multiple ways... just adding another method to solve the problem..

Area of Eq. Triangle = 9√3 = side²*(√3)/4
side = 6

Considering the side of the eq. triangle and the diameter (and other side being the perpendicular bisector extended to the circumference forming a 90° at its base with the side of the eq. triangle) as two sides of a right angle triangle, formed, we get the diameter, d, as follows,

sin 60° = 6/d
=> √3/2 = 6/d
=> d = 12/√3
radius = r = 6/√3

Area = pi * (6/√3)^2 = 12*pi

Answer C.

Hi folks,

I honestly doubt this kind can be asked in GMAT. No GMAT TUTOR replied, I guess.

Regards,
Harish

12pi:)

vishubn wrote:An equilateral triangle that has an area of 9 3^1/2 is inscribed in a circle. What is the
area of the circle?
A . 6pi
B. 9pi
C. 12 pi
D. 9pi 3^1/2
E. 18pi 3^1/2

i was able to dril down to side of eaxch triangle beign 6 !! further/? i guess i am blacked out !

Any comments please??

In an equilateral triangle, A = b²/4 * √3.

Thus, in the problem above:
9√3 = b²/4 *√3
36 = b²
b = 6.

To determine the radius of the circle, draw a 30-60-90 triangle:

In a 30-60-90 triangle, the sides are proportioned x: x√3 : 2x.
In the 30-60-90 triangle shown above, x√3 = 3.
Thus, x = 3/√3 and 2x = 6/√3.

The hypotenuse of the 30-60-90 triangle is also the radius of the circle.
Thus, r=6/√3.
A = πr² = π(6/√3)² = 12π.

The correct answer is C.

Another way to achieve this:-

From the area of the triangle, we can deduce the side of the triangle(using the formula sqrt(3)/4*(side)^2 i.e.6.

Let the center of the circle be O and the equilateral triangle be ABC. A line segment from O to A bisects the angle(BAC). Since the angle(BAC) = 60, so angle(OAC) = 30, similarly angle(ACO) = 30. Therefore angle(AOC) = 120.

Using the sine formula i.e. sin A/a = sin B/b = sin C/c , we can deduce the measure of line segment OA. This line segment is the radius of the circle.
The figure in the post by GMATGuruNY can be taken a s reference

It comes out to be 2* sqrt(3).
Thus we can find out the area using the formula pi * (radius)^2.

Best Regards,
John

12 TT

Late to the party but I had a "simple" explanation that may be repaeating what has already been said. Maybe I just find my own explanation to myself easier than reading equations. Anyway I hope with helps someone who is still puzzled.

Getting the side=6 should be a piece of cake for everyone. It is figuring out the radius from that, that can be tricky.

I drew the triangle and bisected its angles. The bisectors of triangles are called medians.

There was a point in the triangle at which all the meidans met.(intuition told me this must be the center of the circle since it was the centroid of the equilateral triangle)

Then I remembered a property of triangles i.e 2/3 of the median measured from the vertex is the centroid of the triangle.

Since we had already established that the medians were 3*(sqrt of 3) we can then say that the distance from the vertex to the centroid is 2/3 of that, which is 2*(sqrt of 3)

2*(sqrt of 3) is your radius. The rest sould be very simple.

area of the triangle = sqrt(3)/4 * a^2 = 9 sqrt (3).
So, side of triangle, a = 6.

Now, each side of the inscribed triangle in the circle is bisected by the perpendicular drawn on it from center of the circle. Take center of circle as O and side AB of the triangle being cut by perpendicular from the center at D. Then in the right angled triangle OAD, Angle OAD = 30 deg and AD = 3.

Cos 30 = base/hypotenuse = AD/OA.
so, sqrt (3)/2 = 3/OA
This gives the radius of circle, OA = 6/sqrt(3).

Area of circle = pi * r ^ 2 = 12 pi.

the ans is 12 pi C

vishubn wrote:An equilateral triangle that has an area of 9 3^1/2 is inscribed in a circle. What is the
area of the circle?
A . 6pi
B. 9pi
C. 12 pi
D. 9pi 3^1/2
E. 18pi 3^1/2

i was able to dril down to side of eaxch triangle beign 6 !! further/? i guess i am blacked out !

Any comments please??

parul9 wrote:area of the triangle = sqrt(3)/4 * a^2 = 9 sqrt (3).
So, side of triangle, a = 6.

Now, each side of the inscribed triangle in the circle is bisected by the perpendicular drawn on it from center of the circle. Take center of circle as O and side AB of the triangle being cut by perpendicular from the center at D. Then in the right angled triangle OAD, Angle OAD = 30 deg and AD = 3.

Cos 30 = base/hypotenuse = AD/OA.
so, sqrt (3)/2 = 3/OA
This gives the radius of circle, OA = 6/sqrt(3).

Area of circle = pi * r ^ 2 = 12 pi.

Do we even have trigonometry in GMAT?

a^2 sqrt(3)/4 = area triangle where a is each side of equilateral triangle

Now a/2 is a side of rt triangle with hypotenuse = radius and angle between hypotenuse and (a/2) as 30 degree. Therefore r = (a/2)/cos 30 = 2sqrt(3)
area circle = pi * 12

vishubn wrote:An equilateral triangle that has an area of 9 3^1/2 is inscribed in a circle. What is the
area of the circle?
A . 6pi
B. 9pi
C. 12 pi
D. 9pi 3^1/2
E. 18pi 3^1/2

i was able to dril down to side of eaxch triangle beign 6 !! further/? i guess i am blacked out !

Any comments please??

If this triangle is inscribed in a circle, does that mean all 3 points are touching the circumfrence of the circle or just 1 is?

giatch wrote:what formula is that?

https://en.wikipedia.org/wiki/Equilateral_triangle

vishubn wrote:An equilateral triangle that has an area of 9 3^1/2 is inscribed in a circle. What is the
area of the circle?
A . 6pi
B. 9pi
C. 12 pi
D. 9pi 3^1/2
E. 18pi 3^1/2

i was able to dril down to side of eaxch triangle beign 6 !! further/? i guess i am blacked out !

Any comments please??

D?

A= 1/2 diagonal square

9sg rt of 3 = 1/2 * dia sq

18 sq rt 3 = diagonal sq

radius = 2d

so divived by 2

= 9 sg rt 3

Not sure

Area of triangle is 9*[3^(1/2)].

applying area formula of triangle [{3^(1/2)}* (side^2)]/4 = given area.

side = 6

now apply pythagoras theorem -:

(half of base ^ 2) + (altitude ^ 2) = (side ^ 2)

(3 ^ 2) + (altitude ^ 2) = (6 ^ 2)

altitude = (36-9) ^ 1/2 = 3 * (3 ^ 1/2)

Now, as we all now altitude of an equilateral triangle is median too.

And (2/3) * median = radius of circle circumscribed

Therefore, (2/3) * {3 * (3 ^ 1/2)} = 2 * (3 ^ 1/2)

Hence, area of circle will 12 pi option (C).