Question

An equilateral triangle is inscribed in a circle of radius 6 CM then its side is

Answer 1

Suppose you have a triangle ABC. Let AD be the median where D lies on BC. Since it is an equilateral triangle, D will be the mid-point of side BC.

Now, for the time being, let us assume that the length of each side of the equilateral triangle is x.

=> AD = x/2

Also we know that all the angles of an equilateral triangle measure 60 degrees. Using trigonometry, sin60 = AD/AB

=> sqrt(3)/2 = AD/AB

=> AD = AB * sqrt(3)/2

We also know that the centroid of an equilateral triangle divides the median in two parts in the ratio 2:1 and bigger part of that median is the centre of the circle.

So the bigger part of the median (or the radius of the circle is)

        2/3 √(3)/2 * AB

 i.e., x/sqrt(3)   [AB=x]

i.e.,  6/sqrt(3)   [x=6, which is given]

=2√3

Answer 2

ANSWER:-

Given:

An equilateral ∆ is inscribed in a circle of radius is 6cm.

To find:

Find the side.

Solution:

Let ∆ABC be an equilateral ∆ inscribed in the circle with centre O.

Given,

OA= OB= OC =6 [radius of the circle]

Draw AD⊥BC

∠BAC = 60° [∆ABC is an equilateral ∆]

∠BOC = 2×∠BAC

[angle subtended by the arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle]

Therefore,

∠BOC = 2× 60° = 120°

In ∆BOC,

∠BOC + ∠OBC +∠OCB = 180°

120° + ∠OBC + ∠OBC = 180°

[OB=OC= ∠OBC =∠OCB]

=) 2∠OBC = 180° - 120°

=) 2∠OBC = 60°

=) ∠OBC = 60/2

=) ∠OBC=∠OCB= 30°

In ∆OBD,

$cos \angle OBD = \frac{BD}{OB} \\ \\ = > cos30 \degree = \frac{BD}{6} \\ \\ = > \frac{ \sqrt{3} }{2} = \frac{BD}{6} \\ \\ = > 2BD = 6 \sqrt{3} \\ \\ = > BD = \frac{6 \sqrt{3} }{2} \\ \\ = > BD = 3 \sqrt{3} cm$

Therefore,

BC= 2 × BD

=) 2× 3√3

=) 6√3 cm

So,

AB= BC = CA = 6√3 cm

Hence,

The length of each side of the quadrilateral ∆ is 6√3cm.

Hope it helps ☺️