Question

A field is in the the shape of trapezium whose parallel sides are 35 m and 10 m the non parallel sides are 14m and 13m find the area of the field

Answer 1

Given:

AB || CD

AB =25 m

CD= 10m

BC=14m

AD= 13m

From point C draw CE || DA . Hence ADCE is a ||grm having CD || AE & AD||CE

AE= CD= 10m

CE= AD= 13m

BE= AB- AE =25- 10=15 m

BE= 15m

In ∆BCE

BC= 14m, CE= 13m, BE= 15m

Semiperimeter (s)= (a+b+c)/2

Semiperimeter(s) =( 14+13+15)/2

s= 42/2= 21m

s= 21m

Area of ∆BCE= √ s(s-a)(s-b)(s-c)

Area of ∆BCE=√ 21(21-14)(21-13)(21-15)

Area of ∆BCE= √ 21×7× 8×6

Area of ∆BCE= √ 7×3× 7× 4×2×2×3

Area of ∆BCE=√7×7×3×3×2×2×4

Area of ∆BCE= 7×3×2×2= 21× 4= 84m²

Area of ∆BCE= 84m²

Area of ∆BCE= 1/2 × base × altitude

Area of ∆BCE= 1/2 × BE ×CL

84= 1/2×15×CL

84×2= 15CL

168= 15CL

CL= 168/15

CL= 56 /5m

Height of trapezium= 56/ 5m

Area of trapezium= 1/2( sum of || sides)( height)

Area of trapezium=1/2(25+10)(56/5)

Area of trapezium= 1/2(35)(56/5)

Area of trapezium= 7×28= 196m²