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A point P is given on the circumference ...

A point `P` is given on the circumference of a circle of radius `rdot` Chords QR are parallel to the tangent at `Pdot` Determine the maximum possible area of triangle PQR.

Text Solution

Verified by Experts

The correct Answer is:
`(3sqrt(3))/(4)r^(2)` sq.units

Let O be the center and r the radius of the circle
Let QR be the chord parallel to the tangent at the point P on the circle
Let `angleQPR=theta Them angle QOD = angleROD=theta`
Area of `anglePQR=A =1/2 (QR)(PD)=QD(OP+OD)`
`=r sin theta (r+r cos theta)`
`=1/2 r^(2)(cos theta + cos 2 theta)`
`therefore (dA)/(d theta)=r^(2)(cos theta+cos 2 theta)`
Thus A is maximum when `theta = pi//3` the only critical point.Thus maximum (greatest) area A=`1/2r^(2)[ 2 sin (pi//3)+sin (2pi//3)]`
`=-/4(3sqrt(33)r^(2))`
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