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 .
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:
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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