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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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
