Math, asked by rishab3540, 1 year ago

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

Answers

Answered by nikitasingh79
5
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
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Answered by jevelin
2

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