3. Given an isosceles triangle, whose one angle is $120 ^ { \circ }$ and radius of its incircle $= \sqrt { 3 }$. Then the area of the triangle in sq. units is (A) $7 + 12 \sqrt { 3 }$ (B) $12 - 7 \sqrt { 3 }$ (C) $12 + 7 \sqrt { 3 }$ (D) $4 \pi$ Sol. (C) $$\Delta = \frac { \sqrt { 3 } } { 4 } \mathrm {~b} ^ { 2 }$$ Also $\frac { \sin 120 ^ { \circ } } { \mathrm { a } } = \frac { \sin 30 ^ { \circ } } { \mathrm { b } } \Rightarrow \mathrm { a } = \sqrt { 3 } \mathrm {~b}$ and $\Delta = \sqrt { 3 } \mathrm {~s}$ and $\mathrm { s } = \frac { 1 } { 2 } ( \mathrm { a } + 2 \mathrm {~b} )$ $\Rightarrow \quad \Delta = \frac { \sqrt { 3 } } { 2 } ( \mathrm { a } + 2 \mathrm {~b} )$ From (1) and (2), we get $\Delta = ( 12 + 7 \sqrt { 3 } )$.
Given an isosceles triangle, whose one angle is $120 ^ { \circ }$ and radius of its incircle $= \sqrt { 3 }$. Then the area of the triangle in sq. units is
3. Given an isosceles triangle, whose one angle is $120 ^ { \circ }$ and radius of its incircle $= \sqrt { 3 }$. Then the area of the triangle in sq. units is\\
(A) $7 + 12 \sqrt { 3 }$\\
(B) $12 - 7 \sqrt { 3 }$\\
(C) $12 + 7 \sqrt { 3 }$\\
(D) $4 \pi$
Sol. (C)
$$\Delta = \frac { \sqrt { 3 } } { 4 } \mathrm {~b} ^ { 2 }$$
Also $\frac { \sin 120 ^ { \circ } } { \mathrm { a } } = \frac { \sin 30 ^ { \circ } } { \mathrm { b } } \Rightarrow \mathrm { a } = \sqrt { 3 } \mathrm {~b}$\\
and $\Delta = \sqrt { 3 } \mathrm {~s}$ and $\mathrm { s } = \frac { 1 } { 2 } ( \mathrm { a } + 2 \mathrm {~b} )$\\
$\Rightarrow \quad \Delta = \frac { \sqrt { 3 } } { 2 } ( \mathrm { a } + 2 \mathrm {~b} )$
From (1) and (2), we get $\Delta = ( 12 + 7 \sqrt { 3 } )$.\\