Question

Find the area of triangle whose vertices are $\left(3,2\right),\left(11,8\right)$ and $\left(8,12\right)$.

Answer 1

Calculation for the area of a triangle:

The area of the triangle with vertices $A\left(x_1,y_1\right),B\left(x_2,y_2\right),C\left(x_3,y_3\right)$ is given by

$\text{Area of}△ABC=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

Given vertices of triangle are $A\left(3,2\right),B\left(11,8\right)$ and $C\left(8,12\right)$.

Here $x_1=3,y_1=2,x_2=11,y_2=8,x_3=8,y_3=12$.

$\begin{array}{rcl}\text{Area of}△ABC& =& \frac{1}{2}\left|3\left(8-12\right)+11\left(12-2\right)+8\left(2-8\right)\right|\\ & \Rightarrow & \text{Area of}△ABC=\frac{1}{2}\left|3\left(-4\right)+11\left(10\right)+8\left(-6\right)\right|\\ & \Rightarrow & \text{Area of}△ABC=\frac{1}{2}\left|-12+110-48\right|\\ & \Rightarrow & \text{Area of}△ABC=\frac{1}{2}\left|110-60\right|\\ & \Rightarrow & \text{Area of}△ABC=\frac{1}{2}\left|50\right|\\ & \Rightarrow & \text{Area of}△ABC=\frac{50}{2}\\ \therefore \text{Area of}△ABC& =& 25\end{array}$

Final answer: The area of triangle whose vertices are $\left(3,2\right),\left(11,8\right)$ and $\left(8,12\right)$ is 25 sq. units.