Question

In a circle of radius r, an isosceles triangle ABC is inscribed with AB = AC. If the ∆ABC has perimeter p = 2 [$\sqrt{2hr - h^{2}}$+$\sqrt{2hr}$] and area A = h$\sqrt{2hr - h^{2}}$, where h is the altitude from A to BC, then$\frac{A}{p^{3}}$ is –

• A. 128 r
• B. $\frac{1}{128r}$
• C. $\frac{1}{64r}$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{128r}$

Answer 2

$\frac { \mathrm { A } } { \mathrm { p } ^ { 3 } }$ = $\frac { h \sqrt { 2 h r - h ^ { 2 } } } { 8 \left[ \sqrt { 2 h r - h ^ { 2 } } + \sqrt { 2 h r } \right] ^ { 3 } }$

=

= $\frac { \sqrt { 2 \mathrm { r } - \mathrm { h } } } { 8 [ \sqrt { 2 \mathrm { r } - \mathrm { h } } + \sqrt { 2 \mathrm { r } } ] ^ { 3 } }$

= $\frac { \sqrt { 2 r } } { 8 ( 2 \sqrt { 2 r } ) ^ { 3 } }$ = $\frac { 1 } { 128 \mathrm { r } }$ .

Hence (2) is the correct answer.