Problem Solving

Equilateral triangle ABC is inscribed in a circle with center O, as shown. If the radius of the circle is 2, what is the area of triangle ABC?

A. 3
B. 2sqrt(3)
C. 3sqrt(2)
D. 3sqrt(3)
E. 4sqrt(2)

D

I have seen people make use of this formula :

(Area of triangle inscribed in circle)/(Area of circle) = 3sqrt(3)/(4pi)

But it is not working. Can you please explain why?

Sorry the formula works !!!

nirmal019 wrote:Equilateral triangle ABC is inscribed in a circle with center O, as shown. If the radius of the circle is 2, what is the area of triangle ABC?

A. 3
B. 2sqrt(3)
C. 3sqrt(2)
D. 3sqrt(3)
E. 4sqrt(2)

D

I have seen people make use of this formula :

(Area of triangle inscribed in circle)/(Area of circle) = 3sqrt(3)/(4pi)

But it is not working. Can you please explain why?

Hi, I think you just need to know a few equations,

Area of an equilateral triangle with side "a" is a^2*sqrt(3)/4

so if you are given "a" and want to find the area, then it is a direct application of the above formula.

The height of the triangle with side "a" is a*sqrt(3)/2.

Now you can find the height using the above equation.

2/3 of this height is the radius of the circle that is circumscribed around the triangle.

To solve this problem you can use the reverse of these steps.

You know what radius "r" is so find side "a" of the triangle

==> a*sqrt(3)/2*(2/3) = 2
==> a = 6/sqrt(3) = 2sqrt(3)

with this value of side "a" proceed to find area of the equilateral triangle

area = a^2*sqrt(3)/4 = 4*3*sqrt(3)/4 = 3*sqrt(3) = D

Please let me know if this helps. Seems long but once you have the equations it wont take long.

Hi Deepsun, how do you get that the radius is 2/3 of the height... I understand everything else but I cannot get that. Is this always true from an equilateral triangle inscribed in a circle?

Thanks[/quote]

rodalvarezz wrote:Hi Deepsun, how do you get that the radius is 2/3 of the height... I understand everything else but I cannot get that. Is this always true from an equilateral triangle inscribed in a circle?

Thanks

[/quote]

Hi rodalvarezz, I believe it has to do with the Geometry. Some relationship with the centroid and the radius. If it helps please refer the following link for an explanation from Sudhir3127.

https://www.beatthegmat.com/equilateral- ... 16199.html

Thanks

The formula is correct, but what is not obvious here is that neither the height nor base length of the triangle are given.

The centroid divides the median in the ratio 2:1. Hence the height is 3 and the base length is 2*sqrt(3). You can find this using the 30-60-90 rule. The hypotenuse will have length equal to the radius, and the height of one of smaller triangle will be 1. (The medians of the triangle create 3 isosceles triangles )

Therefore area of the equilateral triangle is 1/2*2*sqrt(3)*3 = 3*sqrt(3)

nirmal019 wrote:Equilateral triangle ABC is inscribed in a circle with center O, as shown. If the radius of the circle is 2, what is the area of triangle ABC?

A. 3
B. 2sqrt(3)
C. 3sqrt(2)
D. 3sqrt(3)
E. 4sqrt(2)

D

I have seen people make use of this formula :

(Area of triangle inscribed in circle)/(Area of circle) = 3sqrt(3)/(4pi)

But it is not working. Can you please explain why?

Look for what the circle and the triangle have IN COMMON:



The RADIUS of the circle is the HYPOTENUSE of the 30-60-90 triangle shown above, in which the sides are proportioned x : x√3 : 2x.
The drawing indicates that each side of the equilateral triangle has a length of 2√3.
Area of an equilateral triangle = (s²/4)√3.
Thus, area = (2√3)²/4 * √3 = 3√3.

The correct answer is D.

radius of curcumscribed circle = s (sqrt3/3) (this is the formula where s stands for the lenght of side and is only for equilateral triangle)
radius= 2 thus by putting the value in the formule we will get s = 6/sqrt3
we know area of equilateral triangle = s^2 (sqrt3/4)
putting the value and we will get 3sqrt(3)...D

as Mitch mentioned We can make a small triangle with ratio as 30:60:90 as x:y:2

x & y comes out to be 1 & root 3

side of triangle 2 root 3

area of triangle = s^2 *(root 3)/4 = 4*3 (root 3)/ 4 = 3 root 3 is answer.


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