Q. 224.3( 3 Votes )

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 .

Answer :


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


The vertical angle BAC = 2θ


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


Draw AM perpendicular to BC.


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


Let O be the circumcenter.


BOC = 2×2θ = 4θ (Using central angle theorem)


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


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


In Δ OMC,


CM = asin2θ


OM = acos2θ


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


BC = 2asin2θ


Height of Δ ABC = AM = AO + OM


AM = a + acos2θ



Differentiation this equation with respect to θ








Maxima or minima exists when:



Therefore,






Therefore,






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


Therefore, at θ = :




Both the sin values are positive. So the entire expression is negative. Hence there is a maxima at this point.


θ = will not form a triangle. Hence it is discarded.


There fore the maxima exits at:



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