EMC Q.39
0. The triangle of maximum area inscribed in a circle of radius r is
- Option : B
- Explanation :
Let ABC be a triangle inscribed in the circle with centre 0 and radius r. If area of this triangle is maximum, then vertex C should be at a maximum distance from the base AB i.e. , CD must be perpendicular to AB. Hence ABC is an isosceles triangle.
If ∠ BCD = 0, where D is the mid.-point of BC, then ∠ BOD = 2ϑtherefore, AB= 2BD = 2r sin 2ϑ CD = CO + OD = r + r cos 2ϑ If S be the area of the triangle ABC, then S = 1/2 AB x CD = 1/2 x 2r sin 2 ϑ(r + r cos 2ϑ)ds /dϑ = r2 [sin 2ϑ (-2 sin 2ϑ) + (1 + cos 2ϑ) (2 cos 2ϑ)] = 2r2 [cos 22ϑ - sin2 2ϑ + cos 2ϑ] = 2r2 (cos 4ϑ + cos 2ϑ) For maximum and minimum, ds / dϑ = 0 cos 4ϑ + cos 2ϑ = 0 2 cos 3ϑ cos ϑ = 0Hence either cos 3ϑ = 0, or cos ϑ = 0, which is impossibleif cos ϑ = 0, then ϑ = Π / 2 if cos 3ϑ = 0, then 3 ϑ = Π / 2 ϑ = Π / 6 ( a2s / dϑ 2) 6 = Π/6 is negative therefore S is maximum for ϑ = 1 / 6 * Π ∠ ACB = 2ϑ = 2( Π / 6 ) = Π / 3 = ∠ ABC = ∠ BAC Hence ABC is an equilateral triangle.
