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

Formula for the area of a trapezoid:

$\frac{(base+base)}{2}\times altura$

We substitute the data into the formula and solve:

$\frac{9+12}{2}\times5=\frac{21}{2}\times5=\frac{105}{2}=52.5$

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:

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

$x=30$

In order to calculate the perimeter of the trapezoid we must add together the measurements of all of its sides:

7+10+7+12 =

36

And that's the solution!