Question

Two parallel sides of an isosceles trapezium are 31 cm and 15 cm. Its non-parallel sides are each equal to
17 cm. Find the area of the trapezium.​

Answer 1

Given:

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

•Its non-parallel sides are each equal to 17 cm

To Find:

•Area of Trapezium.

Step-by-step 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.

●Refer the attachment

AB=EF=15 cm

So, CE= FD=(31-15)/2=8 cm

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

•Applying Pythagoras theorem in ΔACE,

${ac}^{2} = {ae}^{2} + {ce}^{2} \\ {17}^{2} = {ae}^{2} + {8}^{2} \\ 282 = {ae}^{2} + 64 \\ ae = \sqrt{225} \\ ae = 15$

Hence,the height of trapezium is 15 cm

Area of Trapezium:

$area = \frac{1}{2} \times (sum \: of \: parallel \: sides ) \times hieght \\ area = \frac{1}{2} \times (31 + 15) \times 15 \\ area = \frac{1}{2} \times 46 \times 15 \\ area = \frac{690}{2} \\ area = 345 \: {cm}^{2}$

Hence,the area of trapezium is 345 cm²

Therefore, the area of the trapezium is 345 cm².