The parallel sides of trapezium are 65 cm and 49 cm and each of the non parallel sides is 10 cm. Find the area of Trapezium.

Draw CF || AD, intersecting AB in F and CE ⊥ BF.

CF || AD and AF || CD.

∴ AFCD is a parallelogram.

AF = CD = 49 cm    (Opposite sides of parallelogram are equal)

AD = CF = 10 cm  (Opposite sides of parallelograms are equal)

BF = AB – AF = 65 cm – 49 cm = 16 cm

ΔBCF is an isosceles triangle.

∴EF = EB= 16/2 = 8 cm  (In an isosceles triangle, altitude from the vertex bisects the base)

In ΔCFE,

CE² + FE² = CF²

∴ CE² = CF² – FE² = (10 cm)² – (8 cm)² = 100 cm² – 64 cm² = 36 cm²

⇒ CE = 6 cm

Area of trapezium 

= ½ × (AB + CD) × CE

= ½ × (65 + 49) × 6

= ½ × 114 × 6

= 114 × 3

= 342 cm²