Question

The base angle of an isosceles trapezium is 45°. if the shorter side and both the equal sides are 10 cm each, what is the area of the trapezium?

Answer 1

the area of the trapezium is 150 cm^2

Answer 2

Answer:

$\text{The area is }120.71 cm^2$

Step-by-step explanation:

Given the base angle of an isosceles trapezium is 45°. if the shorter side and both the equal sides are 10 cm each,

we have to find the area of the trapezium.

In ΔADE, by trigonometric ratios

$\sin 45\°=\frac{AE}{AD}=\frac{AE}{10}$

$AE=\frac{10}{\sqrt2}$

and also

$\tan 45\°=\frac{AE}{ED}=\frac{\frac{10}{\sqrt2}}{ED}$

$ED=\frac{10}{\sqrt2}$

Now, we will find the length of DC i.e the longer side

$DC=DE+EF+FC=\frac{10}{\sqrt2}+10+\frac{10}{\sqrt2}=\frac{20}{sqrt2}+10$

Area of trapezoid ABCD is

$\frac{1}{2}\times (\text{sum of parallel sides})\times height$

$=\frac{1}{2}\times [DC+AB]\times AE$

$=\frac{1}{2}\times [(\frac{20}{\sqrt2}+10)+10]\times \frac{10}{\sqrt2}$

$=(10\sqrt2+20)\times \frac{5}{\sqrt2}=120.71 cm^2$

$\text{The area is }120.71 cm^2$