Question

In $\Delta ABC,a-6,b-10$ and area of triangle is $15\sqrt{3}$. If $\angle ACB$ is obtuse and $r$ is inradius then $r^2=$

• A. $4$
• B. $5$
• C. $6$
• D. $3$

Answer 1

The correct option is D $3$

$a=6,b=10$

$s=\frac{a+b+c}{2}$

$=\frac{16+10+c}{2}$

$=\frac{16+c}{2}$

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}=15\sqrt{3}$

$\left(\frac{16+c}{2}\right)\left(\frac{4+c}{2}\right)\left(\frac{c-a}{2}\right)\left(\frac{16-c}{2}\right)=225\times 3$

$(16+c)(16-c)(c+4)(c-4)=16\times 675$

$(256-c^2)(c^2-16)=16\times 675$

put $c^2=p$

$-p^2+16p+256p-16\times 256=16\left(675\right)$

$p^2+16(256+675)-272p=0$

$\Rightarrow p=\frac{272\pm \sqrt{(272{)}^2-4\times 16(931)}}{2}$

$\Rightarrow p=\frac{272\pm 140}{2}$

$=76$ or $196$

Given, $\angle c=$obtuse

$\Rightarrow c=$ longest side

$\Rightarrow c>10$

$c^2>100$

$p>100$

$\Rightarrow p=196$

$\Rightarrow c^2=196$

$\Rightarrow c=14$cm

$\therefore S=\frac{a+b+c}{2}$

$=\frac{16+14}{2}={15}^o$

$\Delta =rs$

$\Rightarrow 15\sqrt{3}=r\times 15$

$\Rightarrow r=\sqrt{3}$

$r^2=(\sqrt{3}{)}^2=3$

$\therefore r^2=3$.