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An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. Show that the area of triangle is maximum when θ = π6

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Question

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

Sum
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Solution


Let ABC be an isosceles triangle inscribed in the circle with radius a such that AB = AC.

AD = AO + OD = a + a cos2θ and BC = 2BD = 2a sin2θ (see fig. 16.4)

Therefore, area of the triangle ABC

i.e. ∆ = `1/2` BC . AD

= `1/2 2"a" sin2theta * ("a" + "a" cos2theta)`

= a2sin2θ (1 + cos2θ)

⇒ ∆ = `"a"^2sin2theta + 1/2 "a"^2 sin4theta`

Therefore, `("d"∆)/("d"theta)` = 2a2cos2θ + 2a2cos4θ

= 2a2(cos2θ + cos4θ)

`("d"∆)/("d"theta)` = cos2θ = –cos4θ = cos (π – 4θ)

Therefore, 2θ = π – 4θ

⇒ θ = `pi/6`

`("d"^2∆)/("d"theta)` = 2a2 (–2sin2θ – 4sin4θ) < 0 `("at"  theta = pi/6)`.

Therefore, Area of triangle is maximum when θ = `pi/6`.

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Chapter 6: Application Of Derivatives - Solved Examples [Page 131]

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NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 6 Application Of Derivatives
Solved Examples | Q 18 | Page 131

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