Question

The Irregular figure can be broken into a triangle and a rectangle as shown with the dashed line.

Answer 1

Given:

Given that the irregular figure is broken into a triangle and a rectangle.

We need to determine the length b, the area of the triangle , the area of the rectangle and the area of the irregular figure.

Length of b:

The length of b , the base of the triangle is given by

$b=5-(2\frac{1}{3})$

Simplifying, we get;

$b=5-\frac{7}{3}$

$b=\frac{8}{3} \ ft$

Thus, the length of the base of the triangle is $\frac{8}{3} \ ft$

Area of the triangle:

The area of the triangle can be determined using the formula,

$A=\frac{1}{2}bh$

Substituting $b=\frac{8}{3} \ ft$ and $h=3 \ ft$, we get;

$A=\frac{1}{2}(\frac{8}{3})(3)$

$A=4 \ ft^2$

Thus, the area of the triangle is 4 square feet.

Area of the rectangle:

The area of the rectangle can be determined using the formula,

$A=lw$

where l = 5 ft, $w=1 \frac{1}{3}=\frac{4}{3} \ ft$, we get;

$A=5 \times \frac{4}{3}$

$A=\frac{20}{3} \ ft^2$

Thus, the area of the rectangle is $\frac{20}{3} \ ft^2$

Area of the irregular figure:

The area of the irregular figure can be determined by adding the area of the triangle and the area of the rectangle.

Thus, we have;

Area = Area of the triangle + Area of the rectangle

Substituting the values, we have;

$Area = 4+\frac{20}{3}$

$Area = 10.667 \ ft^2$

Thus, the area of the irregular figure is 10.667 square feet.