Question

Abcd is an isosceles trapezium such that ad||bc, ab = 5 cm, ad = 8 cm and bc = 14 cm. what is the area (in cm2 ) of trapezium?

Answer 1

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….