# Area of a Triangle using Radius and Perimeter

Area of a triangle can be calculated using its perimeter and the radius of the circle which is inscribed in the triangle.

### Worked Examples of Area of a Triangle using Radius and Perimeter

(a)    Find an expression for the area of $\triangle LOM$.

\begin{aligned} \displaystyle \text{Area of } \triangle LOM &= \frac{1}{2} \times \text{base} \times \text{height} \\ &= \frac{1}{2} kr \\ \end{aligned} \\

(b)    Show that the area of $\triangle KLM$ is given by $\displaystyle A = \frac{1}{2} Tr$, where $T$ is the perimeter of $\triangle KLM$.

\begin{aligned} \displaystyle A &= \frac{1}{2} kr + \frac{1}{2} \ell r + \frac{1}{2} mr \\ &= \frac{1}{2} (k+\ell +m)r \\ \therefore A &= \frac{1}{2} Tr &\color{red} k\color{red}+\color{red}\ell \color{red}+\color{red}m \color{red}= \color{red}T \\ \end{aligned} \\

(c)    Find how far from the foot of the fence the board touches the ground.

$\text{Let the base of the triangle by } 2+b.$
\begin{aligned} \displaystyle \require{color} \text{Area of the triangle using its base and height } &= \frac{1}{2} \times 8 \times (2+b) \color{red} \cdots (1) \\ \text{Area of the triangle using } \frac{1}{2} Tr &= \frac{1}{2} \times (6+6+2+2+b+b) \times 2 \color{red} \cdots (2) \\ \frac{1}{2} \times 8 \times (2+b) &= \frac{1}{2} \times (6+6+2+2+b+b) \times 2 &\color{red} (1) = (2) \\ 16 + 8b &= 32 + 4b \\ 4b &= 16 \\ b &= 4 \\ 2+b &= 6 \\ \end{aligned} \\
$\text{Therefore the board touches the ground at a point 6 m from the fence.}$

(d)    Find the radius of the second circle.

\begin{aligned} \displaystyle \require{color} MP &= \sqrt{8^2 + 6^2} \\ &= 10 \\ MQ &= \sqrt{8^2 + (6+9)^2} \\ &= 17 \\ \text{Area of } \triangle MPQ \text{ using its base and height } &= \frac{1}{2} \times 9 \times 8 \color{red} \cdots (3) \\ \text{Area of } \triangle MPQ \text{ using } \frac{1}{2} Tr &= \frac{1}{2} \times (10+17+9) \times r \color{red} \cdots (4) \\ \frac{1}{2} \times (10+17+9) \times r &= \frac{1}{2} \times 9 \times 8 &\color{red} (4) = (3) \\ 18 r &= 36 \\ \therefore r &= 2 \\ \end{aligned} \\ 