# The area of ∆ABC with vertices A(3, 0), B(7, 0) and C(8, 4) is

Question:

The area of ∆ABC with vertices A(3, 0), B(7, 0) and C(8, 4) is

(a) 14 sq units
(b) 28 sq units
(c) 8 sq units
(d) 6 sq units

Solution:

(c) 8 sq units

The given points are $A(3,0), B(7,0)$ and $C(8,4)$.

Here, $\left(x_{1}=3, y_{1}=0\right),\left(x_{2}=7, y_{2}=0\right)$ and $\left(x_{3}=8, y_{3}=4\right)$

Therefore,

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

$=\frac{1}{2}[3(0-4)+7(4-0)+8(0-0)]$

$=\frac{1}{2}[-12+28+0]$

$=\left(\frac{1}{2} \times 16\right)$

$=8$ sq units