Question

Let $ABC$ be an acute angled triangle such that $2{\sec }^{4}A={\sec }^{4}B+{\sec }^{4}C;$ $\sec A\sec B\sec C=8$ and ${\log }_{\sec B}\sec A,{\log }_{\sec C}\sec B,{\log }_{\sec A}\sec C$ form a geometric progression. If the circumradius of $△ABC$ is $3\sqrt{3}$ units, then

• A. the length of one of the sides of triangle is $9$ units
• B. area of triangle is $\frac{27\sqrt{3}}{4}$ sq. units
• C. semi-perimeter of triangle is $27$ units
• D. inradius of triangle is $\frac{3\sqrt{3}}{2}$ units

Answer 1

Answer: (A), (D)

inradius of triangle is $\frac{3\sqrt{3}}{2}$ units

Since, $ABC$ is an acute angled triangle,
$\therefore A,B,C\in (0,\frac{\pi }{2})$
$2{\sec }^{4}A={\sec }^{4}B+{\sec }^{4}C...\left(1\right)$
$\sec A\sec B\sec C=8...\left(2\right)$


Consider the given G.P.
$\therefore ({\log }_{\sec C}\sec B{)}^2=({\log }_{\sec B}\sec A)\cdot ({\log }_{\sec A}\sec C)$
$\Rightarrow \frac{(\log \sec B{)}^2}{(\log \sec C{)}^2}=\frac{\left(\log \sec A\right)}{\left(\log \sec B\right)}\cdot \frac{\left(\log \sec C\right)}{\left(\log \sec A\right)}$
$\Rightarrow (\log \sec B{)}^3=(\log \sec C{)}^3$
$\Rightarrow \sec B=\sec C$
Putting this in the equation $\left(1\right)$, we get
$2{\sec }^{4}A=2{\sec }^{4}C$
$\Rightarrow \sec A=\sec C$
$\Rightarrow A=C$
$\Rightarrow A=B=C$

From $\left(2\right),$ $\sec A=2$
$\Rightarrow A=\frac{\pi }{3}=B=C$
$\Rightarrow A,B,C$ form an equilateral triangle.
$\Rightarrow$ Side $=2R\sin \frac{\pi }{3}=2\cdot 3\sqrt{3}\cdot \frac{\sqrt{3}}{2}=9$ units

$\therefore$ Area $=\frac{a^2\sqrt{3}}{4}=\frac{81\sqrt{3}}{4}$ sq. units

Semi-perimeter $=\frac{3a}{2}=\frac{27}{2}$ units

Inradius $=\frac{\Delta }{s}=\frac{81\sqrt{3}}{4}\cdot \frac{2}{27}=\frac{3\sqrt{3}}{2}$ units