In radius of a circle which is inscribed in a isosceles triangle one of whose angle is , is , then area of triangle (in sq units) is
- A.
- B.
- C.
- D.None of these
Show answer and solution
Correct option: C
Solution
Let the isosceles triangle have the apex angle 2π/3, so the two base angles are each π/6.
Let the equal sides be 'a'. Using the inradius formula: r = (Area)/(semi-perimeter).
The base of the triangle: using the cosine rule or direct geometry, base b = 2a·sin(π/3) = a√3.
Semi-perimeter s = (a + a + a√3)/2 = a(2 + √3)/2.
Area = (1/2)·a²·sin(2π/3) = (√3/4)a².
So r = (√3a²/4) / [a(2+√3)/2] = (√3a/4)·(2/(2+√3)) = √3a / (2(2+√3)).
Set r = √3:
√3 = √3·a / (2(2+√3))
1 = a / (2(2+√3))
a = 2(2+√3) = 4 + 2√3.
Area = (√3/4)·a² = (√3/4)·(4+2√3)².
(4+2√3)² = 16 + 16√3 + 12 = 28 + 16√3.
Area = (√3/4)·(28 + 16√3) = √3·7 + 4·3 = 7√3 + 12 = 12 + 7√3.
Let the equal sides be 'a'. Using the inradius formula: r = (Area)/(semi-perimeter).
The base of the triangle: using the cosine rule or direct geometry, base b = 2a·sin(π/3) = a√3.
Semi-perimeter s = (a + a + a√3)/2 = a(2 + √3)/2.
Area = (1/2)·a²·sin(2π/3) = (√3/4)a².
So r = (√3a²/4) / [a(2+√3)/2] = (√3a/4)·(2/(2+√3)) = √3a / (2(2+√3)).
Set r = √3:
√3 = √3·a / (2(2+√3))
1 = a / (2(2+√3))
a = 2(2+√3) = 4 + 2√3.
Area = (√3/4)·a² = (√3/4)·(4+2√3)².
(4+2√3)² = 16 + 16√3 + 12 = 28 + 16√3.
Area = (√3/4)·(28 + 16√3) = √3·7 + 4·3 = 7√3 + 12 = 12 + 7√3.
[AIEEE 2006]