Question

ABCD is a trapezium with AB = 10 cm, AD = 5
cm, BC = 4 cm and DC = 7 cm Find the area
of the ABCD in cm2.
​

Answer 1

Answer:

formula = 1/2(sum of parallel sides) height

Answer 2

Answer:

ISOSCELES TRAPEZIUM : In an isosceles trapezium the two non-parallel sides are equal.

GIVEN : AB = 5cm, AD = 8cm, BC = 14cm

AB = CD(8 cm ) & BE= CF (ABCD is an isosceles trapezium)

SOLUTION :

AD = EF = 8 cm

BC = BE + EF + FC

BC = BE+ 8 + BE (BE = CF)

14 = 2BE +8

14 -8 = 2BE

6 = 2BE

BE = 6/2= 3 cm

BE = 3cm

In ∆ABE ,

AB² = BE² + AE²

5² = 3² + AE²

25 = 9 + AE²

AE² = 25 - 9

AE² = 16

AE = √16 = 4 cm

AE = 4 cm

Area of trapezium = ½(sum of parallel sides)× height

ar(ABCD) = ½(AD + BC ) × AE

ar(ABCD) = ½(8 + 14) × 4

ar(ABCD) = ½ (22)× 4

ar(ABCD) = 22 × 2 = 44 cm²

Hence, area of trapezium ABCD is 44 cm².

HOPE THIS WILL HELP YOU….