JEE Advanced · Mathematics · 17. Properties of Triangles
Inradius of a circle which is inscribed in an isosceles triangle one of whose angle is \(2 \pi / 3\), is \(\sqrt{3}\), then area of triangle is
- A
\(4 \sqrt{3}\)
- B
\(12-7 \sqrt{3}\)
- C
\(12+7 \sqrt{3}\)
- D
None of these
Answer & Solution
Correct Answer
(C)
\(12+7 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Let \(A B=A C\) and \(\angle A=120^{\circ}\)
\(\therefore\) Area of \(\Delta=\frac{1}{2} a^2 \sin 120^{\circ}\)
where, \(a=A D+B D\)
\(=\sqrt{3} \tan 30^{\circ}+\sqrt{3} \cot 15^{\circ}\)
\(=1+\sqrt{3}\left(\frac{1+\tan 45^{\circ} \tan 30^{\circ}}{\tan 45^{\circ}-\tan 30^{\circ}}\right)\)
\(=1+\sqrt{3}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)\)
\(\therefore \quad a=4+2 \sqrt{3}\)
\(\therefore\) Area of \(\Delta=\frac{1}{2}(4+2 \sqrt{3})^2\left(\frac{\sqrt{3}}{2}\right)=12+7 \sqrt{3}\)

\(\therefore\) Area of \(\Delta=\frac{1}{2} a^2 \sin 120^{\circ}\)
where, \(a=A D+B D\)
\(=\sqrt{3} \tan 30^{\circ}+\sqrt{3} \cot 15^{\circ}\)
\(=1+\sqrt{3}\left(\frac{1+\tan 45^{\circ} \tan 30^{\circ}}{\tan 45^{\circ}-\tan 30^{\circ}}\right)\)
\(=1+\sqrt{3}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)\)
\(\therefore \quad a=4+2 \sqrt{3}\)
\(\therefore\) Area of \(\Delta=\frac{1}{2}(4+2 \sqrt{3})^2\left(\frac{\sqrt{3}}{2}\right)=12+7 \sqrt{3}\)

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