Question

Find the area of a trapezium whose parallel sides are 12 cm and 20 cm and the distance between them is 10 cm.
pls answer fast someone

Answer 1

Answer:

A=a+b

2h

a Base

12

cm

b Base

20

Answer 2

Answer:

160cm²

Step-by-step explanation:

Let ABCD be a trapezium where AB || DC,

AB = 12cm, DC = 20cm

and height (diatance between AB and DC) = 10cm

$area \: of \: trapezium \: = \frac{1}{2}(AB + DC) \times height$

$⇒ \frac{1}{2} (12 + 20) \times 10 \\ = \frac{1}{2} \times 32 \times 10 \\ = 16 \times 10 = 160 {cm}^{2}$

Thus the area of the given trapezium is 160cm²