1.

The triangle of maximum area inscribed in a circle is :

Answer»

equilateral
isosceles
right-angled
none of these

Solution :The area of the triangle is maximum, when vertex C is at maximum distance from the base AB, or C is a point on diameter perpendicular to AO, OC = OB = a.
`THEREFORE Delta ABC` is isosceles triangle.

Area of `Delta ABC` is `Z=(1)/(2)xx AB xx CD`
`= a^(2)sin 2theta(1+cos 2theta)`
`therefore (dz)/(d THETA)=2a^(2)(cos 4theta + cos 2theta)=0`
`rArr 2 cos 3theta cos theta = 0`
`rArr theta = (pi)/(2)` or `theta = (pi)/(6)`
`therefore theta = (pi)/(2)` is impossible,
`therefore` at `theta = (pi)/(6)` area of triangle is maximum.


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