Question

In $\Delta ABC$, the lengths of sides $AC$ and$AB$ are $12cm$and $5cm$, respectively. If the area of $\Delta ABC$is $30c{m}^2$ and $R$and $r$ are respectively the radii of circumcircle and incircle of$\Delta ABC$, then the value of$2R+r(\text{in}cm)$ is equal to

Answer 1

Step 1: Finding the angle $A$ and side BC of the triangle

We know that area of $∆ABC$ is

$$\begin{gathered} \text{Area of}△ABC=\frac{1}{2}\times AB\times AC\times \sin A \\ \Rightarrow 30=\frac{1}{2}\times 5\times 12\times \sin A;\left[\because △=30c{m}^2,AB=5\&AC=12\text{are given}\right] \\ \Rightarrow \sin A=1 \\ \Rightarrow A=\frac{\mathrm{\pi }}{2} \end{gathered}$$

We know that the Pythagoras theorem for a right-angled triangle is,

$$\begin{gathered} \Rightarrow \left(\text{Hypotenuse}\right)²=\left(\text{Perpendicular}\right)²+\left(\text{Base}\right)² \\ \Rightarrow {\left(BC\right)}^2={\left(AB\right)}^2+{\left(AC\right)}^2 \\ \Rightarrow {\left(BC\right)}^2=5^2+{12}^2 \\ \Rightarrow {\left(BC\right)}^2=169 \\ \Rightarrow B{C}^{}=13 \end{gathered}$$

Step 2: Finding the value of $2R+r$

We know that if the angle subtended to the perimeter of a circle is a right angle then the base of the triangle is the diameter of the circumcenter.

So, $R=\frac{BC}{2}=\frac{13}{2}...\left(i\right)\left[\because BC=13\right]$

Semiparametric of the triangle is

$$\begin{gathered} s=\frac{AB+BC+CA}{2} \\ =\frac{5+12+13}{2} \\ =15 \end{gathered}$$

We know that the incircle of the triangle is,

$$\begin{gathered} \Rightarrow r=\frac{∆}{s} \\ \Rightarrow r=\frac{30}{15}\left[\because ∆=30c{m}^2\text{and}s=15\right] \\ \Rightarrow r=2....\left(ii\right) \end{gathered}$$

$$\begin{gathered} 2R+r=2\left(\frac{13}{2}\right)+2[\text{from}equation(i)\&(ii\left)\right] \\ =13+2 \\ =15 \end{gathered}$$

Hence the value of $2R+r=15cm$