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

AD^{2} = AL^{2} + DL^{2}

5^{2} = 4^{2} + DL^{2}

DL^{2} = 5 ^{2} - 4^{2} = 25 – 16 = 9

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

Similarly in BMC

By Pythagoras theorem

BC^{2} = BM^{2} + MC^{2}

5^{2} = 4^{2} + MC^{2}

MC^{2} = 5 ^{2} - 4^{2} = 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 cm^{2}

Area of trapezium ABCD = 180cm^{2}

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