An isosceles triangle of vertical angle 2a is inscribed in a circle of radius r Show that the area of - Maths - Application of Derivatives

Question

An isosceles triangle of vertical angle 2a is inscribed in a
circle of radius r Show that the area of triangle is maximum when
a=(pi)/6

Ajanta Trivedi · Accepted Answer

given: ABC is an isosceles triangle such that AB=AC. the vertical angle ∠BAC =2a. and triangle is inscribed in the circle with center O and radius r.

TPT: area(ΔABC) is maximum when a= $π / 6$

since the triangle is an isosceles triangle . the circum center of the circle will lie on the perpendicular from A to BC.

let it be AM.

∠BOC = 2*2a= 4a [angle subtended at the center is twice the angle subtended at the circle]

∠BOM = 2a [since ΔOMB and OMC are congruent triangles]

OA=OB=OC =r [radius of the circle]

now in the triangle OMB,

$B M = r \sin 2 a a n d O M = r \cos 2 a$

BC= 2BM [perpendicular from the center bisects the chord]

BC=2r sin2a..........(1)

height of the triangle ΔABC = AM = AO+OM

AM= r +rcos2a ..............(2)

area of the ΔABC A = $\frac{1}{2} B C * A M$

$A = \frac{1}{2} * 2 r \sin 2 a * (r + r \cos 2 a) A = r^{2} \sin 2 a (1 + \cos 2 a) A = r^{2} (\sin 2 a + \sin 2 a \cos 2 a) A = r^{2} (\sin 2 a + \frac{1}{2} \sin 4 a)$ ..............(3)

since r (the radius of the circle ) is fixed.

differentiating (3) wrt a :

$\frac{d A}{d a} = r^{2} (2 \cos 2 a + \frac{1}{2} * 4 \cos 4 a) \frac{d A}{d a} = r^{2} (2 \cos 2 a + 2 \cos 4 a) = 2 r^{2} (\cos 2 a + \cos 4 a)$ ........(4)

differentiating again wrt a:

$\frac{d^{2} A}{d a^{2}} = 2 r^{2} (- 2 \sin 2 a - 4 \sin 4 a) = - 4 r^{2} (\sin 2 a + 2 \sin 4 a)$ .........(5)

for maximum value of area equating dA/da = 0

$2 r^{2} (\cos 2 a + \cos 4 a) = 0 \cos 2 a + \cos 4 a = 0 \cos 2 a + 2 \cos^{2} 2 a - 1 = 0 2 \cos^{2} 2 a + 2 \cos 2 a - \cos 2 a - 1 = 0 (2 \cos 2 a - 1) (\cos 2 a + 1) = 0$

$\cos 2 a = \frac{1}{2} o r \cos 2 a = - 1 2 a = \frac{π}{3} o r 2 a = π a = \frac{π}{6} o r a = \frac{π}{2}$

if 2a = π it will not form the triangle.

for a=π/6, $\frac{d^{2} A}{d a^{2}}$ is negative therefore for a=π/6 area is maximum.

thus for the maximum area triangle must be an equilateral triangle.

hope this helps you.

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