Q. 64.0( 8 Votes )

# In the adjoining figure, ABCD is a trapezium in which AB ‖ DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD.

Answer :

Given

AB = 7 cm

AD = BC 5 cm

AL = BM = 4cm (height)

DC = ?

Here in the given figure AB = LM

LM = 7 cm ------------1

Now Consider ALD

By Pythagoras theorem

AD2 = AL2 + DL2

52 = 42 + DL2

DL2 = 5 2 - 42 = 25 – 16 = 9

DL = 3 cm --------------2

Similarly in BMC

By Pythagoras theorem

BC2 = BM2 + MC2

52 = 42 + MC2

MC2 = 5 2 - 42 = 25 – 16 = 9

MC = 3 cm --3

from 1 2 and 3

DC = DL + LM + MC = 3 + 7 + 3 = 13 cm

We know that area of trapezium is x (sum of parallel sides) x height

Area of trapezium = x (AB + DC) x AL

= x (7 + 13) x 4 = 40 cm2

Area of trapezium ABCD = 180cm2

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