Question

In acute $△ABC$, length of side $BC$ is $5$, inradius and circumradius of $△ABC$ are $\frac{5}{2}(1+\sqrt{2})(\sqrt{3}-\sqrt{2})$ and $5$ respectively. Then, which of the following is /are correct?

• A. Length of internal angle bisector through $A$ is $\frac{5}{2}(2+\sqrt{3})$
• B. Area of $△ABC$ is $\frac{25}{2}(2+\sqrt{3})$
• C. Length of altitude through vertex $A$ is $5(2+\sqrt{3})$
• D. Perimeter of $△ABC$ is greater than $20$

Answer 1

Answer: (A), (D)

The correct options are
A Length of internal angle bisector through $A$ is $\frac{5}{2}(2+\sqrt{3})$
D Perimeter of $△ABC$ is greater than $20$


From the property of triangle and circumradius we have
$\frac{5}{\sin A}=2\times 5\Rightarrow A={30}^{\circ }$
Now from relation between inradius and circumradius we have
$\cos A+\cos B+\cos C=1+\frac{r}{R}\Rightarrow \cos B+\cos C=1+\frac{(1+\sqrt{2})(\sqrt{3}-\sqrt{2})}{2}-\frac{\sqrt{3}}{2}\Rightarrow \cos B+\cos C=\frac{\sqrt{2}(\sqrt{3}-1)}{2}=2\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)\Rightarrow \cos B+\cos C=2\left(\cos {75}^{\circ }\right)\Rightarrow \angle B=\angle C={75}^{\circ }$
Now $\frac{b}{\sin B}=\frac{a}{\sin A}\Rightarrow \frac{b}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=\frac{5}{\frac{1}{2}}\Rightarrow b=5\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right),c=5\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right)$
Length of bisector from the vertex
$A=\frac{2bc}{b+c}\cos \frac{A}{2}=c\cos \frac{A}{2}=5\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)=\frac{5}{2}(2+\sqrt{3})$
$Ar\left(△ABC\right)=\frac{1}{2}bc\sin A=\frac{25}{4}(2+\sqrt{3})$
length of altitude through vertex A $=\frac{2\left(Ar\right(△ABC\left)\right)}{a}=\frac{5}{2}(2+\sqrt{3})$
Now Perimeter $=a+b+c=5(\sqrt{6}+\sqrt{2}+1)$