## Book: RS Aggarwal - Mathematics

### Chapter: 18. Area of Circle, Sector and Segment

#### Subject: Maths - Class 10th

##### Q. No. 29 of Exercise 18B

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29
##### The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle. [Take √3 = 1.73.]

Consider the figure shown above, AF, BE and CD are perpendicular bisector.

Now we know that the point at which all three perpendiculars meet is called incentre, so O is the incentre, thus O divides all three perpendiculars in a ratio 2:1.

Let AB = BC = CA = a cm

Therefore let AF = h cm

AFC = 90° and OF = 1/3 × AF

OF = h/3 cm (putting value of OF)

h = 3×OF eqn1

And we can see from figure that OF = radius of circle

Now let radius of circle be = r cm

Area of circle = πR2

where R = radius of circle

Given area of circle = 154 cm 2

πr2 = 154

r2 = 49

r = 7 cm

Therefore OF = 7 cm

h = 3×7 (from eqn 1)

h = 21 cm

we know area of an equilateral triangle =

where a = side of triangle

Also, Area of triangle = 1/2 ×base×height

Equating both the areas we get,

Put the values of BC and AF

(putting value of h = 21 cm)

(rationalize it)

a = 14√3 cm

Perimeter of equilateral triangle = 3×side of triangle

Perimeter of ∆ ABC = 3×14√3 (put √3 = 1.73)

Perimeter of ∆ABC = 42×1.73

Perimeter of ∆ABC = 72.66 cm

The perimeter of triangle is 72.66 cm

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