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Question
The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle.
Solution
Let the radius of the inscribed circle be r cm.
Given:
Area of the circle = 154 cm2
We know ;
Area of the circle = πr2
`=> 154 = 22/7"r"^2`
`=> (154xx7)/22 = "r"^2`
⇒ r2 = 49
⇒ r = 7
In a triangle, the centre of the inscribed circle is the point of intersection of the medians and altitudes of the triangle. The centroid divides the median of a triangle in the ratio 2:1.
Here,
AO;OD = 2:1
Now,
Let the altitude be h cm
We have :
∠ADB = 90°
OD `=1/3"AD"`
OD `= "h"/3`
⇒ h = 3r
⇒ h = 21
Let each side of the triangle be a cm.
In the right - angled ΔADB, we have ;
AB2 = AD2 + DB2
`"a"^2 = "h"^2 + (a/2)^2`
`4"a"^2 = 4"h"^2 + "a"^2`
`a^2 = 4"h"^2`
`a^2 = (4"h"^2)/3`
`a = (2"h")/sqrt(3)`
`a = 42/sqrt(3) `
∴ Perimeter of the triangle = 3a
`= 3xx42/sqrt(3)`
`= sqrt(3)xx42`
= 72.66 cm
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