Question

This figure was broken into a triangle and a rectangle as

Answer 1

Answer:

This figure was broken into a triangle and a rectangle as shown.

A figure can be broken into a rectangle and triangle. The rectangle has a base of 12 centimeters and height of 18 centimeters. The triangle has a base of 12 centimeters and height of 8 centimeters.

Use the figures to complete the statements.

The height of the triangle h is  

✔ 8

cm.

The area of the triangle is  

✔ 48

cm2.

The area of the rectangle is  

✔ 216

cm2.

The area of the irregular figure is  

✔ 264

cm2.

Step-by-step explanation:

Answer 2

Answer:

The height of the triangle is 8 centimeters.

The area of the rectangle is 48 square centimeters.

The area of the rectangle is 216 square centimeters.

The area of the irregular figure is 264 square centimeters.

Step-by-step explanation:

The height of the triangle is the total height of the figure minus the height of the rectangle, that is:

$h = 26\,cm - 18\,cm$

$h = 8\,cm$

The height of the triangle is 8 centimeters.

The area of the triangle ($A$), measured in square centimeters, is determined by the following formula:

$A = \frac{1}{2}\cdot b\cdot h$ (1)

Where:

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

If we know that $b = 12\,cm$ and $h = 8\,cm$, then the area of the triangle is:

$A = \frac{1}{2}\cdot (12\,cm)\cdot (8\,cm)$

$A = 48\,cm^{2}$

The area of the rectangle is 48 square centimeters.

The area of the rectangle ($A$), measured in square centimeters, is determined by the following formula:

$A = b\cdot H$ (2)

Where $H$ is the height of the rectangle, measured in centimeters.

If we know that $b = 12\,cm$ and $H = 18\,cm$, then the area of the rectangle is:

$A = (12\,cm)\cdot (18\,cm)$

$A = 216\,cm^{2}$

The area of the rectangle is 216 square centimeters.

The area of the irregular figure is the sum of the areas of the rectangle and the triangle. Then, the area of the irregular figure is 264 square centimeters.