Question

Figure ABCD has vertices A(−6, 3), B(2, 3), C(2, −4), and D(−6, −1). What is the area of Figure ABCD?

Answer 1

Answer:

Area of figure ABCD = 44 units square

Step-by-step explanation:

Plotting the given points:

$A(-6,3)$, $B(2,3)$, $C(2,-4)$ and $D(-6,-1)$

On plotting the points we find out that the figure is a trapezoid with sides

$AD\parallel BC$

From the graph Length of parallel sides can be found out as they are vertical lines:

$AD=4\ units$

$BC=7\ units$

Length of a non parallel side AB can be found out as it is a horizontal line.

$AB=8\ units$

Note: $AB\perp BC$, so $AB$ is the height of trapezoid as height is perpendicular distance between the parallel sides.

Area of trapezoid is given by

⇒ $\frac{Sum\ of\ parallel\ sides}{2} \times height$

⇒ $(\frac{AD+BC}{2})\times AB$

⇒ $(\frac{4+7}{2})\times 8$

⇒ $44$ units square

∴ Area of figure ABCD = 44 units square