Question

find area of a field which is in shape of trapezium having parallel sides are 20m and 42m and non parallel sides are 21m and 23m.

Answer 1

Refer diagram from attachment please follow it.$\uparrow\:\uparrow$

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Given : ABCD is a trapezium whose parallel sides are AB||CD :

AB = 20 m ; CD = 42 m

Non parallel sides are :

AD = 21 m ; BC = 23 m

Construction : Draw a parallel line from B to O so that DA||BO.

As notified the ABDO is a parallelogram and AD = BO = 21 m

Solution :

OC = DC - DO

$\implies 42 - 20 \\\implies 22\: m$

In ∆BOC

$S = \frac{a+b+c}{2} \\\implies \frac{22+21+23}{2}\\\implies \frac{66}{2} \\\implies 33 m$

$Area \: of\: \triangle BOC = \sqrt{S(S-a)(S-b)(S-c)}\\\implies \sqrt{33(33-22)(33-21)(33-23)} \\\implies \sqrt{33\times 11\times 12\times 10} \\\implies \sqrt{ 11\times3\times 11\times 4\times 3 \times 10} \\\implies 3\times 11\times 2 \sqrt{10} \\\implies 66\times 3.162\\\implies 208.56\: m^{2}$

$Area \:of \:\triangle{BOC}= \frac{1}{2} \times Base \times Height\\ \implies 2 \times 208.56 = OC\times BF \\\implies BF(h) = \frac{208.56\times2}{22} \\\implies 18.97\: m$

$Area\: of\: trapezium\:ABCD= \frac{1}{2} \times(sum \:of \:parallel\: sides)\times height\\\implies \frac{1}{2} \times (42+20)\times 18.97\\\implies \frac{\cancel{66}}{\cancel{2}}\times 18.97 \\\implies 588.09 \: m^{2}$

$\red{\underline{Answer}}$

$\pink{\bold{\boxed{ Area \: of\: trapezium = 588.09 \: m^{2}}}}$