Question

A field in the shape of trapezium whose parallel sides are 25m and 10m the non parallel sides are 14m and 13m.find the area of the field of 9th class

Answer 1

Figure is in the attachment
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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
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Hence the area of field is 196m²

Answer 2

Answer:

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