Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as (i) the sum of the areas of two triangles and one rectangle. (ii) the difference of the area of a rectangle and the sum of the areas of two triangles.

Question

Philoid · Accepted Answer

Area of a trapezium ABCD
= area (∆DFA) + area (rectangle DFEC) + area (∆CEB)
= (1/2 × AF × DF) + (FE × DF) + (1/2 × EB × CE)
= (1/2 × AF × h) + (FE × h) + (1/2 × EB × h)

= 1/2 × h × (AF + 2FE + EB)
= 1/2 × h × (AF + FE + EB + FE)
= 1/2 × h × (AB + FE)

= 1/2 × h × (AB + CD) [Opposite sides of rectangle are equal]

= 1/2 × 6 × (15 + 10)

= 1/2 × 6 × 25 = 75

Area of trapezium = 75 cm²

Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as(i) the sum of the areas of two triangles and one rectangle.(ii) the difference of the area of a rectangle and the sum of the areas of two triangles.

Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as
(i) the sum of the areas of two triangles and one rectangle.

(ii) the difference of the area of a rectangle and the sum of the areas of two triangles.