Question

The area of a triangle whose vertices are $(a,a),(a+1,a+1),(a+2,a)$ is

• A. $a^2$
• B. $2a$
• C. $1$
• D. $\sqrt{2}$

Answer 1

The correct option is C $1$

R.E.F image

Foundation chapter & LO

Given vertices of a triangle $(a,a),(a+1,a+1),(a+2,a)$

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

$=\frac{1}{2}\left(|a\right(a+1-a)+(a+1\left)\right(a-a)+(a+2\left)\right(a-a-1\left)|\right)$

$=\frac{1}{2}(|a\times 1+0+(a+2\left)\right(-1\left)|\right)$

$=\frac{1}{2}(|a-a-2|)$

$=\frac{1}{2}\times 2=1$ sq.units.