# 6.3: Area of Parallelograms

**At Grade**Created by: CK-12

## Learning Objectives

- Understand the basic concepts of the meaning of area.
- Use formulas to find the area of parallelograms, including rectangles.

The **area** of a shape is the space *inside* the perimeter.

You can think of the **perimeter** as a line you may draw with a pen of the outer border of a shape. The **area** is what you would paint with a paintbrush in order to fill in the entire shape, painting inside the perimeter line.

For example,

In a fenced field, the *fence* would be the **perimeter** and the *grass in the field* would be the **area**.

In a basketball court, the *sidelines* would be the **perimeter** and the *wooden court surface* would be the **area**.

In a pool, the ________________ would be the **perimeter** and the ____________ would be the **area**.

## Area of a Rectangle

If a rectangle has **base** \begin{align*}b\end{align*} units and **height** \begin{align*}h\end{align*} units, then the **area**, \begin{align*}A\end{align*}, is \begin{align*}bh\end{align*} square units.

\begin{align*}\text{Area} & = \text{base} \cdot \text{height}\\ A & = bh\end{align*}

## Area of a Parallelogram

**Example 1**

*How could we find the area of this parallelogram?*

Make it into a rectangle by moving the triangular part:

The rectangle is made of the same parts as the parallelogram, so their areas are the same. The area of the rectangle is \begin{align*}bh\end{align*}, so the area of the parallelogram is also \begin{align*}bh\end{align*}.

**Warning:** Notice that the height \begin{align*}h\end{align*} of the parallelogram is the *perpendicular distance between two parallel sides of the parallelogram,* **not** a side of the parallelogram (unless the parallelogram is also a rectangle, of course).

If a parallelogram has base \begin{align*}b\end{align*} units and height \begin{align*}h\end{align*} units, then the area, \begin{align*}A\end{align*}, is \begin{align*}bh\end{align*} square units.

\begin{align*}\text{Area} & = \text{base} \cdot \text{height}\\ A & = bh\end{align*}

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