Questions & Answers

Question

Answers

A. $2\sqrt 3 $

B. $3\sqrt 3 $

C. \[4\sqrt 3 \]

D. $5\sqrt 3 $

Answer
Verified

To solve this question , first of all we need to draw a rough diagram of an equilateral triangle of the side ( let it be $a$ ) $6cm$.

As all sides of the equilateral triangle are the same and all angles are of ${60^\circ }$ so we can draw the diagram as shown below.

Let $O$ be the centre and $A,B,C$ are the edges of the triangle.

To calculate $OA$ we drop a perpendicular to the line $AB$. In equilateral triangle the perpendicular dropped from the centroid to the side \[AB\] divides the side equally into two parts.

Therefore $AM = BM = \dfrac{{AB}}{2} = \dfrac{a}{2}$

$AM = \dfrac{6}{2} = 3cm$

Taking triangle $AOM$ into the consideration

To find $OA$ we use Cosine rule where,

Cosine$ = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$$ = \dfrac{{AM}}{{OA}}$

$\cos 30^\circ = \dfrac{{3cm}}{{OA}}$ , from here we can write

As we know the value of $\operatorname{Cos} {30^\circ } = \dfrac{{\sqrt 3 }}{2}$So substituting this value in the above equation we will get $OA = \dfrac{3}{{\cos 30^\circ }} = 2\sqrt 3 $ ,

Hence the value of circumradius in this case is $2\sqrt 3 $.