Question

An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when θ = π/6.

Answer 1

Δ ABC is an isosceles triangle such that AB = AC.

The vertical angle of BAC = 2θ

Triangle is inscribed in the circle with center O and radius a.

Draw the AM perpendicular to BC.

Here, Δ ABC is an isosceles triangle, the circumcenter of the circle will lie on the perpendicular from A to BC.

Suppose O be the circumcenter.

BOC = 2 × 2θ = 4θ (On using central angle theorem)

COM = 2θ (here, Δ OMB and Δ OMC are congruent triangles)

OA = OB = OC = a (radius of the circle)

In the Δ OMC,

CM = asin2θ

OM = acos2θ

BC = 2CM (Perpendicular from the center bisects the chord)

BC = 2asin2θ

Height of  the Δ ABC = AM = AO + OM

AM = a + acos2θ

Area of Δ ABC = (1/2) x AM x BC

Differentiate this eqaution with respect to θ

So,

To check the whether which point has a maxima, we have to double differentiate it

So, θ = π/6 

Both the value of sin are positive, therefore the entire expression is negative.

Thus, there is a maxima at this point

θ = π/2 will not form a triangle. Thus it is discarded.

So, the maxima the exits at;

θ = π/6