Find the area of a triangle whose vertices are$(a, c + a), (a, c)$ and $(-a, c – a)$

Given:

Vertices of a triangle are $(a, c + a), (a, c)$ and $(-a, c – a)$.

To do:

We have to find the area of the given triangle.

Solution:

Let $A(a, c + a), B(a, c)$ and $C(-a, c – a)$ be the vertices of a $\triangle ABC$.

We know that,

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, 

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( ABC=\frac{1}{2}[a(c-c+a)+a(c-a-c-a)+(-a)(c+a-c)] \)

\( =\frac{1}{2}\left(a^{2}-2 a^{2}-a^{2}\right) \)

\( =\frac{1}{2}\left(-2 a^{2}\right) \)

\( =a^{2} \) sq. units

The area of the given triangle is $a^{2}$ sq. units.