Problem Solving

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??

This can be solved using the properties of 30-60-90 degrees right triangles. From the problem we can figure out that the side of the equilateral triangle is 6. Now if we draw three lines joining each vertex of the inscribed triangle to the center of the circle, the length of each of these lines will be equal to the radius of the circle and each of these lines will bisect the angle it runs through into two 30 degree angles. Consider now any of the 3 triangles formed by one side of the original triangle and two radii drawn from the two corners. The angle made at the center by this triangle will be 120 degree (since the angle subtending the same arc at the circumference is 60 degrees). Now if we drop an altitude from the center of circle (that is the 120 degree vertex of the triangle we are considering) then, this altitude will bisect the 120 degree angle into two 60 degrees angles and bisect the side of the equilateral triangle into two segments of 3 each forming two 30-60-90 degree right triangles. Using the properties of such triangles we can figure out that if the side opposite the 30 degree angle is a then the side opposite the 60 degree angle is sqrt3*a. Now we know that this side is 3 in length. Therefore sqrt3*a=3 or a= sqrt 3. Now the side opposite the 90 degree angle is 2a using the same property i.e. 2sqrt 3. This side is nothing but the radius of the circle. Therefore the radius of the circle is 2sqrt 3 and the area of the circle is 12 pi.

centroid is the centre point where all the medians of a tringle intersect.

the distance between one vertex of the triangle and the centroid is twice the distance between the centroid and the midpoint of one side.

AO = radius
therefore, radius/height = 2/3

now, the area is already given . area of an equilateral triangle = (root3/ 4) a^2 = 9 root3 (given)
a = 6
"a" is one side of the equilateral triangle.

area can also be written as: 1/2 * base * height = 1/2 * a * h = 1/2 * 6 * h = 9 root 3
we get h = 3 root3

r/h = 2/3 = r/ 3root3;
we get r = 2 root 3

Therefore, area of the circle = pi r square = pi* (2 root 3)^2 = 12pi
Ans: C
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No Need explanation to calculate side = 6

In an equilateral triangle, Circumcenter divide the altitude in the ratio 2:1

Height h = (root(3) / 2) * a = 3 root(3)

height is divided in ratio 2 root(3) : root(3)

In an 30-60-90 triangle x^2 = radius square = 3 + 9 = 12

so area of circle = 12 pi

Think the below link might put an end to all confusions :

https://www.mathopenref.com/trianglecircumcircle.html

C the answer.

12*(pi)

Every one remembers area of right angled triangle = 1/2*a*b

but it comes from 1/2*a*b*sin 90 its product of half & the sides containing the angle
so if you don't remember equilateral triangle area then simply
1/2*a*a*sin 60=1/2 a^2 *(sqrt3 /2)
so we find the area of equilateral triangle = 6

Now the radius of circle is 2/3 of total height & the other half is 1/3rd
so by pythagoras theorem
r^2 = (r/2)^2 + 3^2
so we have r = 2sqrt 3
so area = 12 pie

I looked at the responses and they seemed convoluted to me. I am a visual person so drawing diagrams helps me. We know that area = 9* sqrt(3) and we know area of equilateral triangle is s^2* sqrt(3)/4. We come up with s = 6 and we know, by virtue of equilateral triangle, that all angles will be equal 60 degrees. Inscribing my triangle in a circle and then connecting lines from each corner of triangle to center gives me 3 radii of equal length. If I bisect one of the triangles so I get a 30-60-90 triangle, I know one of my sides is 3. In the 30-60-90 triangle scenario, 3 will correspond to my 60 degrees side and sqrt (3) corresponds to the 30 degree side. Using Pythagorean theorem, I get last side to be 2 * sqrt (3). Now using my Area formula of pi* r^2, I can now calculate area of circle to be 12 pi.

The answer is 12pi. I didn't know any special formula for calculating the area of a circle inscribing an equilateral triangle; I had to come up with the answer. Obviously memorizing the formula is the fastest way to compute the answer.

For anyone curious about how I ended up with getting R = b / (root 3) where b = 6, these were my steps:

1. Determine that b = 6 by solving for A = 1/2*b*h = 9*(root 3) where h = (root 3)/2*b using the 30-60-90 triangle rule. Draw a line that bisects the 60 deg angle to form the 30-60-90 triangle, where the side across from the 30 deg angle is now 1/2*b. Note that in a 30-60-90 triangle, the sides have this relationship: side across from 30 deg angle is x, side across from 90 deg angle is 2x, and the side across from the 60 deg angle is (root 3) * x.

2. Determine the diameter of the circle inscribing the equilateral triangle by drawing a 30 deg, 30 deg, and 120 deg triangle that shares a side with the equilateral triangle. The diameter of the circle should be SUM of the height of the first equilateral triangle (root 3)/2*b and the height of the new isoceles triangle you drew.

The new isoceles triangle you drew, when bisected by the line that is part of the circle's diameter, forms two congruent 30-60-90 triangles. You can determine the height of the isoceles triangle again using the 30-60-90 rule I just mentioned. This time, it is the side across from the 60 deg angle that is 1/2*b (remember it shares a side with the equilateral). You will find out h for this triangle is (root 3)/6*b.

The diameter is the sum of the equilateral triangle's height and the isocele's triangles height. That is D = (root 3)/2*b + (root 3)/6*b = 2*b/(root 3).

Take half the diameter to get R = b/(root 3); square R and multiply it by pi to get pi*b^2/3. Fill in for b = 6 and get 12 pi.

Here's a quick estimation method:

In conjunction with the attached illustration

T = area of inscribed equilateral triangle
H = area of hexagon = 2T
C = area of circle = H + blue edges

and the blue edges sum to approximately H/6

So C ≈ 7H/6 = 7T/3 where T = 9√3

Hence C ≈ 21√3 ≈ 36 ≈ 12pi (using pi ≈ 3)

ANSWER (C) 12 pi

SueD wrote:I looked at the responses and they seemed convoluted to me. I am a visual person so drawing diagrams helps me. We know that area = 9* sqrt(3) and we know area of equilateral triangle is s^2* sqrt(3)/4. We come up with s = 6 and we know, by virtue of equilateral triangle, that all angles will be equal 60 degrees. Inscribing my triangle in a circle and then connecting lines from each corner of triangle to center gives me 3 radii of equal length. If I bisect one of the triangles so I get a 30-60-90 triangle, I know one of my sides is 3. In the 30-60-90 triangle scenario, 3 will correspond to my 60 degrees side and sqrt (3) corresponds to the 30 degree side. Using Pythagorean theorem, I get last side to be 2 * sqrt (3). Now using my Area formula of pi* r^2, I can now calculate area of circle to be 12 pi.

Hi there, I thought you might like the visual estimation approach posted today Thu Dec 05, 2013 3:51 pm

From the given data we can will that side of equilateral triangle is 6 unit and we know radius of circle circumscribing equilateral triangle is S/(3^1/2). Therefor radius of circle is 2*(3^1/2) which gives us area of 12pi.

In the picture above the triangle ABC is divided into 2 30-60-90 triangles by the perpendicular AD, which passes through the centre of the circle O.

The height (AD) will be a * sqrt(3) /2, where a is the side of the triangle

Area(ABC) = Area(ABD) + Area(ADC)

9 * sqrt(3) = 1/2[ (a / 2) * (a * sqrt(3) / 2) ] + 1/2[ (a / 2) * (a * sqrt(3) / 2) ]

a= 6

AD = 3 * sqrt(3)

Now draw the radius (r) from the centre O to the vertex B.



In triangle BOD height OD can be calculated by the formula OD = AD- 0A

OD = 3 * sqrt(3) - r

where r is the radius of the circle.

Pythogrean theorem tells us that OB^2 = BD ^2 + OD ^2

r ^2 = 9 + (3 * sqrt(3) - r) ^2

r = 2 * sqrt(3)

Area of the circle will be Pi * r ^2

Area of the circle. = 12 Pi.

Hence, Option C is correct.

@logitech The formula is applicable for equilateral triangle or any form of traingle....

Pretty challenging. I correctly got (C) but just by logical reasoning. Since the base of the equilateral triangle has side 6, and that's a chord of the circle, the radius must be greater than 3. The largest chord is the diameter and the base of the triangle would not be a diameter (the diameter would be formed if the vertex of the triangle was 90 degrees, not 60 degrees). Thus, answer choices (A) and (B) are too small. Answer choices (D) and (E) are silly. How are you going to have a square root if you are squaring the radius to find area? Even if the radius has a square root, once squared it would be gone. This would require the radius to contain a fourth root... Thus, only answer choice (C) could possibly be the correct answer.

But thanks for the derivations. Will definitely be more prepared if I see something like this on test day.

C
Area of the equilateral triangle = (3^(1/2)/4) *a^2 (a is the equilateral side)
--> 9*3^(1/2)=(3^(1/2)/4) *a^2
--> a = 36^(1/2) = 6

Circumscribed circle radius=[3^(1/2)/3]*a= 2*3^(1/2)

Area of the circle = pi*a^2=12pi