Equilateral triangle ABC is inscribed in a circle with center O, as sh

Question

https://gmatclub.com/forum/download/file.php?id=29253 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. 2 \sqrt{3} C. 3 \sqrt{2} D. 3 \sqrt{3} E. 4 \sqrt{2} 2016-02-07_2130.png

sagarltl · Answer

answer : d
Divide the image in 3 triangle: larger ABC and two smaller COA and AOD
In triangle COA, angle OC =30 and the diameter CE bisects the 60 degree in half. Now oc=ac radius of the circle and this makes COA an issoceles triangle thus angle OCA= angle COA=30 and remaining angle COD=180-60=120

Again angle COA= angle ODA+angle AOD 
angle OAD=120-90=30
Angle AOD=60 
thus using 90-60-30, ad=\sqrt{} 3
and again refering to big 90-60-30 traingle ACB and using root 3 as base to find cd=3
and thus area= 1/2*2 \sqrt{} 3* 3=3 \sqrt{} 3

chetan2u · Answer

Hi,

another straight forward method is by circumcenter...
the circumcenter of the equilateral triangle is the radius and is given by  side/\sqrt{3} ..
so R= side/\sqrt{3} ..
or side= 2\sqrt{3}

thus  Area = \sqrt{3}* side^2/4 ..
 Area = \sqrt{3}* {2\sqrt{3}}^2/4 ..= 3\sqrt{3} 
D

davedekoos · Answer

A helpful property to know about triangles is that the intersection of the 3 medians (lines from a vertex to the center of the opposite side) all intersect at the centroid, and the centroid divides each median in the ratio of 2:1. More easily explained with a diagram:

Attachment:

Centroid.png [ 11.69 KiB | Viewed 18095 times ]

Here c is the centroid of triangle ABC, and c divides the median AD into two parts with ratio 2:1. That is, Ac:cD = 2:1. This is true of the other medians as well.

Ok, back to our problem:

Attachment:

2016-02-07_2130.png [ 7.89 KiB | Viewed 17983 times ]

In the problem, because we have an equilateral triangle, CO is the radius and CD is both the median and the altitude (height) of the triangle. So CO=2 (given), and OD=1 (from the ratio CO:OD = 2:1), therefore the height, CD = 2+1 = 3.

From here, we can use the property of a 30-60-90 triangle to find the length of the sides. In our 30-60-90 triangle CAD, we know that the ratio of the sides AD:CD:CA is 1:\(\sqrt{3}\):2. And we know that side CD is 3, so our side lengths are \(\sqrt{3}:3:2\sqrt{3}\)

Alternatively, if we know the height of an equilateral triangle, then the length of a side is \(\frac{2}{\sqrt{3}}*h\). So in our case, the length of a side is \(\frac{2}{\sqrt{3}}*3=2\sqrt{3}\)

CA = \(2\sqrt{3}\) which is the base of the triangle.

The area of the triangle is therefore \(\frac{bh}{2}=\frac{2\sqrt{3}*3}{2}=3\sqrt{3}\)

Answer: D

A shortcut method that requires memorizing a formula: For an equilateral triangle, if we know the height,

\(A=\frac{h^2}{\sqrt{3}}\)

\(A=\frac{3^2}{\sqrt{3}}=\frac{9}{\sqrt{3}}=3\sqrt{3}\)

gmatexam439 · Answer

Another straightforward answer:
O is the centre of the circle. Therefore Angle AOC = 2*Angle ABC.
Hence Angle AOC = 120.

Length of the line segment AC = 2r Sin(theta/2)
=> 2*2*sin(60)
=>AC=2*root3

Area=root3/4*(AC^2)
=> Area=3*root3

shashankism · Answer

Radius of circle = 2 
So , OC =2 
Now since triangle is equilateral and O is the centre of circle. It divides equilateral triangle in ratio 2: 1

So, Height of equilateral triangle = 3/2 * 2 = 3

Let side of equilateral triangle be a 
H = sqrt {3}/2 *a 
-> a= (2/sqrt{3} )* H
a = (2/sqrt{3} )* 3 = 2 sqrt {3}

so area =( sqrt{3}/4 )* a^2 =  3 \sqrt{3}

Answer D

Gprabhumir · Answer

Wouldn't the base be 4 cm and height be 2* sqrt(3) cm?
Then the area of ABC would be 3*2*0.5*4*2*sqrt(3)

Bunuel , can you share the official solution?

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Equilateral triangle ABC is inscribed in a circle with center O, as sh

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Equilateral triangle ABC is inscribed in a circle with center O, as sh

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