Question

In $∆ABC,\angle A=\frac{\mathrm{\pi }}{2},b=4,c=3$, then the value of $\frac{R}{r}$ is

• A. $\frac{5}{2}$
• B. $\frac{7}{2}$
• C. $\frac{9}{2}$
• D. $\frac{35}{24}$

Answer 1

The correct option is A

$\frac{5}{2}$

Determine the value of $\frac{R}{r}$.

Step 1; Calculate the value of side $a$.

We know that by Pythagoras theorem for a right-angled triangle:

$$\begin{gathered} a^2=b^2+c^2 \\ =4^2+3^2 \\ =16+9 \\ =25 \\ a=5 \end{gathered}$$

Step 2: Calculate the value of area and semi-perimeter.

Semi-perimeter of triangle is given by, $s=\frac{(a+b+c)}{2}$

$$\begin{gathered} \therefore s=\frac{(5+4+3)}{2} \\ =6 \end{gathered}$$

Area of triangle $∆ABC$

$$\begin{gathered} \Rightarrow areaof(∆)=\frac{1}{2}bc\sin A \\ =\frac{1}{2}\times 4\times 3\times 1\left[\because \sin A=1ifA=\frac{\mathrm{\pi }}{2}\right] \\ =6 \end{gathered}$$

Step 3: Calculate the value of $\frac{R}{r}$

We know that radius of the outer circle of $∆ABC$ is given by, $R=\frac{abc}{4\times ar(∆)}$

$$\begin{gathered} R=\frac{abc}{4\times ar(∆)} \\ =\frac{5\times 4\times 3}{4\times 6} \\ =\frac{5}{2} \end{gathered}$$

Also, the radius of incircle of $∆ABC$ is given by $r=\frac{ar(∆)}{s}$

$$\begin{gathered} r=\frac{ar(∆)}{s} \\ =\frac{6}{6} \\ =1 \end{gathered}$$

Therefore $\frac{R}{r}=\frac{5/2}{1}\text{or}\frac{5}{2}$

Hence, option A is the correct answer.