Question

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Answer 1

HELLO DEAR,


WE KNOW THAT THE AREA OF TRAPEZIUM=

Let ABCD be the trapezium whose parallel sides are AB=10mandCD=25m The non parallel sides are BC=14mandDA=13m

BE is drawn parallel to AD.It intersects CD at E.BF is the perpendicular drawn from B to CE . Hence BE is the height of the triangle BCE and trapezium ABCD

Now for triangle BCE

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

So semi perimeter of ΔBCE is 

$s = \frac{1}{2}(13 + 14 + 15) \\ = > \frac{1}{2} \times 42 \\ = > s = 21meter$
$\sqrt{21 \times (21 - 13) \times (21 - 14 ) \times (21 - 14)} \\ = > \sqrt{21 \times 8 \times 7 \times 6} \\ = > 84 {m}^{2} \\ = > \frac{1}{2} \times ce \times bf = 84 \\ = > \frac{1}{2} \times 15 \times bf = 84 \\ = > bf = \frac{84 \times 2}{15} \\ = > bf = 11.2meter$
So area of the trapezium ABCD

$\frac{1}{2} (AB +CD) \times BF \\ => \frac{1}{2} ( 10 + 25) \times 11.2 \\ = > \frac{1}{2} \times 35 \times 11.2 \\ = > 196 {m}^{2}$

Answer 2

hello..


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
_____________________________
Hence the area of field is 196m²