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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 r. Chord QR is parallel to the tangent at P. Determine the maximum possible area of the 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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