Question

In triangle ABC a=6, b=10 and area of the triangle is $15\sqrt{3}$. If $\angle ACB$ is obtuse and ‘r’ is inradius, then ${{r}^{2}}$ is
(a) 4
(b) 5
(c) 6
(d) 3

Answer 1

Hint: Use the sine formula $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=\dfrac{abc}{2\Delta }$, where $\Delta $ is the area of the triangle to calculate the value of $\sin \left( \angle ACB \right)$. Use the trigonometric identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ to calculate the value of $\cos \left( \angle ACB \right)$. Use the cosine formula $\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$ to calculate the measure of the length of side AB. Calculate the semi-perimeter of the triangle using the formula $s=\dfrac{a+b+c}{2}$. Use the formula $r=\dfrac{\Delta }{s}$ to calculate the inradius and thus, calculate ${{r}^{2}}$.

Complete step-by-step answer:
We know that in $\Delta ABC$, we have $a=6,b=10$ and the area of the triangle is $15\sqrt{3}$. We have to calculate the value of ${{r}^{2}}$, where ‘r’ is the inradius and $\angle ACB$ is obtuse.

We will use the sine formula for triangles $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=\dfrac{abc}{2\Delta }$, where $\Delta $ is the area of the triangle.
We can rewrite the equation $\dfrac{c}{\sin C}=\dfrac{abc}{2\Delta }$ as $\sin C=\dfrac{2\Delta c}{abc}=\dfrac{2\Delta }{ab}$.
Substituting $a=6,b=10,\Delta =15\sqrt{3}$ in the above formula, we have $\sin C=\dfrac{2\left( 15\sqrt{3} \right)}{6\times 10}=\dfrac{\sqrt{3}}{2}$.
We will now use the trigonometric identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ to calculate the value of $\cos \left( \angle ACB \right)$.
As we know $\sin \left( \angle ACB \right)=\dfrac{\sqrt{3}}{2}$, we have ${{\cos }^{2}}\left( \angle ACB \right)+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=1$.
Simplifying the above equation, we have ${{\cos }^{2}}\left( \angle ACB \right)+\dfrac{3}{4}=1$.
Rearranging the terms of the above equation, we have ${{\cos }^{2}}\left( C \right)=1-\dfrac{3}{4}=\dfrac{1}{4}$.
Taking the square root on both sides, we have $\cos C=\pm \dfrac{1}{2}$.
We know that $\angle ACB$ is obtuse and the cosine of obtuse angles is negative.
Thus, we will consider $\cos C=\dfrac{-1}{2}$.
We know the cosine formula $\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$.
Substituting $a=6,b=10,\cos C=\dfrac{-1}{2}$ in the above formula, we have $\dfrac{-1}{2}=\dfrac{{{6}^{2}}+{{10}^{2}}-{{c}^{2}}}{2\left( 6 \right)\left( 10 \right)}$.
Simplifying the above equation, we have $-1=\dfrac{36+100-{{c}^{2}}}{60}$.
Thus, we have $136-{{c}^{2}}=-60$.
Rearranging the terms, we have ${{c}^{2}}=136+60=196$.
Taking square root on both sides, we have $c=\sqrt{196}=14$ as the length is a positive quantity.
We will now calculate the semi-perimeter of the triangle using the formula $s=\dfrac{a+b+c}{2}$.
Substituting the values $a=6,b=10,c=14$ in the above formula, we have $s=\dfrac{6+10+14}{2}=\dfrac{30}{2}=15$.
We will now calculate the inradius ‘r’ using the formula $r=\dfrac{\Delta }{s}$.
Substituting $\Delta =15\sqrt{3},s=15$ in the above formula, we have $r=\dfrac{15\sqrt{3}}{15}=\sqrt{3}$.
Squaring the above equation, we have ${{r}^{2}}={{\left( \sqrt{3} \right)}^{2}}=3$.
Hence, the value of ${{r}^{2}}$ is 3, which is option (d).

Note: One needs to smartly use the formulas relating various parameters of triangles to calculate the value of the inradius of the circle. We must know that the point of intersection of internal angle bisectors is the incentre and perpendicular distance of incentre from any side is called inradius of the circle.