Question

6. In the given figure, ABCD is a trapezium in which AB = 18 cm and CD = 10 cm. The area of AACD is 75 cm. Find the area of ABCD.​

Answer 1

Answer:

area ∆ACD = 1/2* 10* height=75

=>height= 15cm

it is the height (distance between 2 parallel lines of the trapezium)of the trapezium.

So, area of trapezium ABCD = 1/2* height *(sum of two parallel lines) = 1/2*15*(18+10) = 15*14=210 square cm