JEE Advanced · Mathematics · 17. Properties of Triangles
Consider a triangle \(A B C\) and let \(a, b\) and \(c\) denote the lengths of the sides opposite to vertices \(A, B\) and \(C\) respectively. Suppose \(a=6, b=10\) and the area of the triangle is \(15 \sqrt{3}\). If \(\angle A C B\) is obtuse and if \(r\) denotes the radius of the incircle of the triangle, then \(r^2\) is equal to
- A 3
- B 5
- C 8
- D 9
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
\(\sin C=\frac{\sqrt{3}}{2}\) and \(C\) is given to be obtuse
\[
\begin{aligned}
& \Rightarrow \quad C=\frac{2 \pi}{3}=\sqrt{a^2+b^2-2 a b \cos C} \\
& =\sqrt{6^2+10^2-2 \times 6 \times 10 \times \cos \frac{2 \pi}{3}}=14 \\
& \therefore \quad r=\frac{\Delta}{s} \Rightarrow r^2=\frac{225 \times 3}{\left(\frac{6+10+14}{2}\right)^2}=3
\end{aligned}
\]
\[
\begin{aligned}
& \Rightarrow \quad C=\frac{2 \pi}{3}=\sqrt{a^2+b^2-2 a b \cos C} \\
& =\sqrt{6^2+10^2-2 \times 6 \times 10 \times \cos \frac{2 \pi}{3}}=14 \\
& \therefore \quad r=\frac{\Delta}{s} \Rightarrow r^2=\frac{225 \times 3}{\left(\frac{6+10+14}{2}\right)^2}=3
\end{aligned}
\]
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