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.
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=
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,
BC= 2BM [perpendicular from the center bisects the chord]
height of the triangle ΔABC = AM = AO+OM
AM= r +rcos2a ..............(2)
area of the ΔABC A =
since r (the radius of the circle ) is fixed.
differentiating (3) wrt a :
differentiating again wrt a:
for maximum value of area equating dA/da = 0
if 2a = π it will not form the triangle.
for a=π/6, 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.