Question

A point P is given on the circumference of a circle of radius r. The chord QR is parallel to the tangent line at P. The maximum area of the triangle PQR is –

• A. $\frac { 3 \sqrt { } 2 } { 4 }$r²
• B. $\frac { 3 \sqrt { } 3 } { 4 }$r²
• C. $\frac { 3 } { 8 }$ r
• D. ) None of these

Answer 1

Answer: (B)

$\frac { 3 \sqrt { } 3 } { 4 }$r²

Answer 2

A = $\frac { 1 } { 2 }$ . QR . PN

= $\frac { 1 } { 2 }$ 2r sin q . (r + r cos q)

A = r² $\left( \sin \theta + \frac { 1 } { 2 } \sin 2 \theta \right)$ and it will be maximum

When cos q + cos 2q = 0

or cos 2q = – cos q = cos (p – q) or when

q = p/3. In this case the triangle will be equilateral and its area will be

r² $\left( \sin \frac { \pi } { 3 } + \frac { 1 } { 2 } \sin \frac { 2 \pi } { 3 } \right)$ = $\frac { 3 \sqrt { } 3 } { 4 }$r²