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.

Answer

`( 3sqrt3)/(4) r ^(2) sq`. units

Explanation

Since, the chord QR is parallel to the tangent at P
`therefore " " ON bot QR`
Consequently, N is the mid-point of chord QR.
`therefore " " Q R = 2 QN = 2 r sin theta `
Also, ` " " ON = r cos theta `
` therefore" " PN = r + r cos theta `

Let A denotes the area of `Delta P QR `
Then , ` " " A = (1)/(2) *2 r sin theta ( r + r cos theta)`
`rArr " " A = r ^(2) (sin theta + sin theta cos theta)`
`rArr " "A = r ^(2) (sin theta + (1)/(2) sin 2 theta )`
`rArr " " ( dA)/( d theta) = r ^(2)( cos theta + cos 2 theta ) `
` and " " ( d^(2)A) /( d theta ^(2)) = r ^(2) (- sin theta - 2 sin 2theta)`
For maximum and minimum values of `theta `, we put `(dA)/( d theta ) =0`
`rArr cos theta + cos 2 theta = 0 rArr cos 2 theta = - cos theta `
`rArr " " cos theta = cos ( pi- 2 theta )rArr theta = (pi)/(3)`
Clearly, ` (d^(2) A)/( d theta ^(2)) lt 0 " for " theta = (pi)/(3)`
Hence, the area of ` Delta PQR` is maximum when `theta = (pi)/(3)`.
The maximum area of ` Delta P QR` is given by
`A = r ^(2) ( sin ""(pi )/(3) + (1)/(2) sin ""( 2pi )/(3) ) = r^(2) ( ( sqrt3)/( 2) + (sqrt 3)/( 4)) `
`" " = ( 3 sqrt 3)/( 4) r ^(2) ` sq units.