Question

The area of a triangle whose vertices are $(-4,0),(2,4)$ and $(4,0)$ is

• A. $8$
• B. $12$
• C. $16$
• D. $32$

Answer 1

The correct option is C $16$

Draw a triangle with given vertices
As we know that area of the triangle is given by
$A=\frac{1}{2}$ (Base) (Height)
So we have to find the base and height of a
triangle
Base, b $=4-(-4)=8$
For finding height drop a perpendicular from
(2, 4) on to the base of a triangle as shown below
Height $h=\sqrt{{(2-2)}^2+{(4-0)}^2}=\sqrt{16}=4$
$\therefore$ Area of given triangle $=\frac{1}{2}\times b\times h$
$=\frac{1}{2}\times 8\times 4=16$