Question

in the adjoining figure ABCD is a trapezium in which ab parallel DC.ab=7cm ad=bc=5cm and the distance between Ab and DC is 4 cm find the length of dC and hence find the area of trapezium ABCD

Answer 1

answer is 27.5 .
you got answer briefly.

Answer 2

Area of Trapezium ABCD = 40 cm²

AB = 7 cm  (Given)

AD = BC = 5 cm  (Given)

AL = BM = 4 cm (i.e Height)

DC = ?

AB = LM

∴ LM = 7 cm (Eq.1)

In ΔALD, using Pythagoras theorem

AD² = AL² + DL²

5²= 4² + DL²

DL² = 25 – 16 ⇒ 9

∴ DL = 3 cm (Eq.2)

Similarly in ΔBMC, using Pythagoras theorem

BC² = BM² + MC²

5² = 4² + MC²

MC² = 25 – 16 ⇒ 9

∴ MC = 3 cm (Eq.3)

Equating 1, 2 & 3

DC = LM + DL + MC = 7 + 3 + 3 ⇒ 13 cm

Area of trapezium = 1/2 x (AB + DC) x AL

= 1/2 x (7 + 13) x 4

= 40 cm²