Trapezoid Practice Problems for 9th Grade - Area & Properties

First, let's remind ourselves of the formula for the area of a trapezoid:

$A=\frac{\left(Base\text{ }+\text{ Base}\right)\text{ h}}{2}$

We substitute the given values into the formula:

(2.5+4)*6 =
6.5*6=
39/2 =
19.5

First, we need to remind ourselves of how to work out the area of a trapezoid:

$\frac{(Base+Base)\cdot h}{2}=Area$

Now let's substitute the given data into the formula:

(10+6)*5 =
2

Let's start with the upper part of the equation:

16*5 = 80

80/2 = 40

Given that the trapezoid is isosceles and the angles on both sides are equal, it can be argued that:

$∢C=∢D$

$∢A=∢B$

We know that the sum of the angles of a quadrilateral is 360 degrees.

Therefore we can create the formula:

$∢A+∢B+∢C+∢D=360$

We replace according to the existing data:

$120+120+2x+2x=360$

 $240+4x=360$

$4x=360-240$

$4x=120$

We divide the two sections by 4:

$\frac{4x}{4}=\frac{120}{4}$

$x=30$

Let's recall that in an isosceles trapezoid, the sum of the two angles on each of the trapezoid's legs equals 180 degrees.

In other words:

$A+C=180$

$B+D=180$

Since angle D is known to us, we can calculate:

$180-50=B$

$130=B$

The formula for the area of a trapezoid is:

We are given the following dimensions:

• Base $AB = 5$ cm
• Base $DC = 9$ cm
• Height $h = 7$ cm

Substituting these values into the formula, we have:

First, add the lengths of the bases:

Now substitute back into the formula:

Calculate the multiplication:

Then multiply by the height:

Thus, the area of the trapezoid is 49 cm$^2$.