Question

In the accompanying figure, triangle ABC has coordinates A(0,3), B(7,3), C(7,7). What is the area of triangle ABC in square units?

Answer 1

We know area of triangle = $\frac{1}{2} * base * height$

In triangle ABC , AB is the base and BC is the height

Now we find the distance between A and B

Distance formula is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

A(0,3), B(7,3)

Distance AB = $\sqrt{(7-0)^2 + (3-3)^2}$

= $\sqrt{(7)^2 }$ = 7

B(7,3), C(7,7)

Distance BC = $\sqrt{(7-7)^2 + (7-3)^2}$

= $\sqrt{(4)^2 }$ = 4

area of triangle = $\frac{1}{2} * base * height$

area of triangle = $\frac{1}{2} * AB * BC$

= $\frac{1}{2} * 7 * 4$ = 14 $units ^2$