Question

The Height of an equilateral triangle is 9cm what is the measure of its in-radius?

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Answer 1

${\underline{\boxed{\bf\red{Question}}}}$

the height of an equilateral triangle is 9cm what is the measure of its in-radius?

${\underline{\boxed{\bf\red{Answer}}}}$

Triangle ABC is an equilateral triangle such that AB = BC = CA = 9cm.

Construction:- We have drawn perpendicular bisectors AD , BF and CE of the corresponding three sides of the triangle. ‘O’ is the point of intersection of these perpendicular bisectors.

We know that in an equilateral triangle, all sides and all angles are equal.

Measure of each angle = 60∘

∵ AD is the perpendicular bisector of the side BC. So, we can say that:

BD = DC = 4.5cm.

We know that the perpendicular bisector divides the triangle into two congruent triangles.

$\bf{ ∴∠EBO=∠DBO= \frac{60∘}{2} =30∘}$

We have point ‘O’ which is the intersection of these perpendicular bisectors and also the centre of the circumcircle.

So, we will now find the radius of this circle using trigonometry.

△OBD is a right triangle and ∠OBD=300.

On applying trigonometry ratio, we get:

$\bf{cos30∘= \frac{BD}{OB} }$

$\bf{⇒OB= \frac{BD}{cos30∘} }$

Putting the value in above equation, we get:

$\bf{⇒OB= \frac{4.53}{√3/2} = \frac{9}{1.732} =5.196}$

Therefore, the radius of the circle = OB = 5.196cm.