Question

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Answer 1

Question:-

Find the area of the following field (fig iv).

Given:-

• The figure (fig iv).

To Find:-

• The area of the following field (fig iv).

Solution:-

We apply formula for area of triangle and area of trapezium.

${ \boxed{ \sf \large \color{red}Area \: of \: triangle = \frac{1}{2} \times base \times height. }}$

${ \boxed{ \sf \large \color{red}Area \: of \: trapezium = \frac{1}{2} \times sum \: of \: parallel \: sides \times height. }}$

$\sf \large Area \: of \: part \: I = \frac{1}{2} \times 150 \times 120 = 9,000 \: m {}^{2} .$

$\sf \large Area \: of \: part \: II = \frac{1}{2} \times 40 \times 50 = 1,000 \: m {}^{2} .$

$\sf \large Area \: of \: part \: III = \frac{1}{2} \times (50 + 70) \times 30 = 15 \times 120 = 1,800 \: m {}^{2} .$

$\sf \large Area \: of \: part \: IV = \frac{1}{2} \times 70 \times 25 = 875 \: m {}^{2} .$

$\sf \large Total \: area = 9,000 + 1,000 + 1,800 + 875 = 12,675 \: m {}^{2} .$

Answer:-

${ \boxed{ \sf \large \color{red} \therefore The \: area \: of \: the \: field \: = 12,675 \: m {}^{2}. }}$

Hope you have satisfied. ⚘