# How do you find the area of the trapezoid below?

Nov 12, 2015

$190$

#### Explanation:

The quick way of finding the area of this trapezoid would be to simply use the know formula, which looks like this

$\textcolor{b l u e}{A = \frac{B + b}{2} \cdot h} \text{ }$, where

$B$, $b$ - the two bases of the trapezoid
$h$ - its height

In this case, you have a trapezoid that has the longer base equal to $20$, the shortest base equal to $18$, and the height equal to $10$.

This means that you would get

$A = \frac{20 + 18}{2} \cdot 10 = \frac{38}{2} \cdot 10 = \textcolor{g r e e n}{190}$

The more interesting way, assuming that you don't remember the formula, would be to try and find it "by hand", so to speak

Notice that you can form the area of the trapezoid by adding the area of the central rectangle, shown in light blue, and the areas of the two right triangles, shown in forest green.

Now, the central rectangle will have the dimensions of the short base and of the height, i.e. $18 \times 10$, and an area equal to

${A}_{\text{rectangle}} = 18 \cdot 10 = 180$

Now, you know that the difference between the long base and the short base is equal to $2$, since you have $20 - 18$.

This means that if you use $x$ to be the short base of the right triangle, and $2 - x$ to be the short base of the left triangle, you can say that their respective areas will be

${A}_{\text{left triangle}} = \frac{1}{2} \cdot \left(2 - x\right) \cdot 10$

and

${A}_{\text{right triangle}} = \frac{1}{2} \cdot x \cdot 10$

This means that the total area of the trapezoid will be

$A = {A}_{\text{left triangle" + A_"right triangle" + A_"rectangle}}$

$A = \frac{1}{2} \left(2 - x\right) \cdot 10 + \frac{1}{2} \cdot x \cdot 10 + 180$

$A = 10 - \textcolor{red}{\cancel{\textcolor{b l a c k}{5 x}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{5 x}}} + 180 = \textcolor{g r e e n}{190}$