Question

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

Answer 1

Answer:

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Answer 2

Given :

• The parallel sides of trapezium are 25m and 10m

• The non-parallel sides are 14m and 13m

To find :

• Area of field=?

Solution :

Let ABCD be a trapezium with

AB|| CD

•AB=25m

•CD=10m

• BC=14m

•AD=13m

Draw CE || DA .So ADCE is a parallelogram with,

•CD=AE=10m

•CE=AD=13m

•BE=AB-AE=25-10m=15m

$\bf{ In \triangle \: BCE \: , semi \: perimeter \: will \: be, }$

$\\ \implies \sf \: s = \frac{a + b + c}{2} \\ \\ \implies \sf \: s = \frac{14 + 13 + 15}{2} \\ \\ \implies \sf \: s = \frac{27 + 15}{2} \\ \\ \implies \boxed{ \sf{s = 21m}}$

•$\bf{Area \: of \: \triangle BCE}$

$\\ \implies \sf \: area = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \implies \sf \: \sqrt{21(21 - 14)(21 - 13)(21 - 15)} \\ \\ \implies \sf \: \sqrt{21(7)(8)(6)} \\ \\ \implies \sf \: \sqrt{7056} \\ \\ \implies \boxed{ \sf \: 84 {m}^{2} }$

•$\bf{ Also, \: area \: of \: \triangle BCE \: is }$

$\\ \implies \sf \: area \: = \frac{1}{2} \times base \times height$

$\\ \implies\sf{84 =\dfrac{1}{2} \times 15 \times CL}$

$\\ \\ \implies\sf{ \dfrac{84 \times 2}{15} = CL}$

$\\ \\ \implies\boxed{\sf{ CL=\dfrac{56}{5}m}}$

•$\bf{Now , \: Area \: of \: trapezium \: is }$

$\\ \implies \sf Area \: = \frac{1}{2} (sum \: of \: parallel \: side) \times (height) \\ \\ \implies \sf \: Area \: of \: trapezium = \frac{1}{2} \times (25 + 10) \times \frac{56}{5} \\ \\ \implies \boxed{ \sf{ Area \: = 196 {m}^{2} }}$

➡ Therefore the Area of trapezium is 196m²