Question

two parellel sides of an isosceles trapezium are 31cm and 15 cm.its non-parellel sides are reach equal to 17 cm.find the area of the trapezium​

Answer 1

The area of the trapezium is 345 cm².

Step-by-step explanation:

Given : Two parallel sides of an isosceles trapezium are 31 cm and 15 cm it's non parallel side are each equal 17 cm.

To find : The area of the trapezium ?

Solution :

Let the trapezium be ABCD, where AB║CD.

Two parallel sides of an isosceles trapezium are 31 cm and 15 cm.

So, AB = 15 cm and CD = 31 cm.

We draw two perpendiculars from A,B on CD point as E and F.

AB = EF = 15 cm

So, CE = FD = $\frac{(31-15)}{2}$ = 8 cm

The length of non-parallel side i.e. AC = BD = 17 cm.

Applying Pythagoras theorem in Δ ACE,

$AC^{2} = AE^{2} + CE^{2}$

$17^{2} = AE^{2} + 8^{2}$

$289^{2} = AE^{2} + 64$

$AE = \sqrt{225}$

$AE = 15$

The height of trapezium is 15 cm.

The area of the trapezium is

$A = \frac{1}{2}$ × $Sum of parallel side$ × $Height$

$A = \frac{1}{2}$ × $(31 + 15)$ × $15$

$A =$  $\frac{1}{2}$ × $46$ × $15$

$A = 345$

Therefore, the area of the trapezium is 345 cm².