# How to Find the Area of a Trapezoid Part 2

In the previous post, we have learned the formula for **finding the area of a trapezoid**. We derived that the formula for the area of a trapezoid with base and (the base are the parallel sides), and height is

In this post, which is the second part of **Finding the Area of a Trapezoid Series**, we are going to continue with some examples. We will not only find the area of a trapezoid, but other missing dimensions such as base and height. Now, get your paper and pencils and try to solve the problems on your own before reading the solution.

We have already discussed one example in the previous post, so we start with the second example.

**Example 2**

What is the area of a trapezoid whose parallel sides measure 6 cm and 8 cm and whose altitude is 2.5 cm?

*Solution*

In this example, the parallel sides are the base, so we can substitute them to and . Since we are looking for the sum of and , we can substitute them interchangeably. The term *altitude *is also another term for height. So, , and .

We now substitute.

So the area is square centimeters.

**Be Careful!**: Again, remember that if we talk about area, we are talking about square units, and in this case square centimeters. If you choose an option which is **17.5 centimeters**, then it is WRONG. It should be **7.5 square centimeters!**

**Example 3**

Find the height of a trapezoid whose base lengths are 5 and 8 units and whose area is 18 square units.

*Solution*

In this problem, we look for the height. But don’t worry, we will still use the same formula, and manipulate the equation later to find . So here, we have , , and .

Multiplying 1/2 and 12, we have

.

We are looking for , so to eliminate 6, we divide both equations by 6. That is,

.

So, the height of the trapezoid is 3 units (not square units!).

In the **next post**, we will have two more examples.

**Note**: If you have a hard time understanding the solution, or particularly solving equations, you should read the series on **solving equations**.