# The PERIMETER of isosceles trapezoid ABCD is equal to 80cm. The length of the line AB is 4 times bigger than lenght of a CD line which is 2/5 the lenght of the line BC (or the lines which are the same in lenght). What is the area of the trapezoid?

## I'm sorry for the grammar mistakes etc. Thanks!

Dec 18, 2016

Area of trapezium is $320$ $c {m}^{2}$.

#### Explanation:

Let the trapezium be as shown below:

Here, if we assume smaller side $C D = a$ and larger side $A B = 4 a$ and $B C = \frac{a}{\frac{2}{5}} = \frac{5 a}{2}$.

As such $B C = A D = \frac{5 a}{2}$, $C D = a$ and $A B = 4 a$

Hence perimeter is $\frac{5 a}{2} \times 2 + a + 4 a = 10 a$

But perimeter is $80$ $c m .$. Hence $a = 8$ cm. and two paallel sides shown as $a$ and $b$ are $8$ cm. and $32$ cm.

Now, we draw perpendiculars fron $C$ and $D$ to $A B$, which forms two identical right angled trianges, whose

hypotenuse is $\frac{5}{2} \times 8 = 20$ $c m .$ and base is $\frac{4 \times 8 - 8}{2} = 12$

and hence its height is $\sqrt{{20}^{2} - {12}^{2}} = \sqrt{400 - 144} = \sqrt{256} = 16$

and hence as area of trapezium is $\frac{1}{2} \times h \times \left(a + b\right)$, it is

$\frac{1}{2} \times 16 \times \left(32 + 8\right) = 8 \times 40 = 320$ $c {m}^{2}$.