Question

A field is in the shape of a trapezium of parallel sides 11 m and 25 m and of non-
parallel sides 15 m and 13 m. Find the cost of watering the field at the rate of 5 paise
per 500 cm?
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.
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. solve the question step by step​

Answer 1

$\LARGE{\color{teal}{\textsf{\textbf{ Qυєѕтiση }}}}$

A field is in the shape of a trapezium of parallel sides 11 m and 25 m and of non- parallel sides 15 m and 13 m. Find the cost of watering the field at the rate of 5 paise per 500 cm?

$\LARGE{\color{teal}{\textsf{\textbf{ αnѕwєr }}}}$

Area of trapezium

$\sf = \frac{1}{2} \times Sum \: of \: parallel \: sides \times Height \\ \\ \sf = \frac{1}{2} \times (AB+ DC) \times BF$

• To calculate BF, we do the following steps

Step 1:

• Draw BE || AD
• Note AB || DE (AS AB || DC)

Since, opposite sides are parallel,

ABED is a parallelogram.

In parallelogram,

opposite sides are equal

• BE = AD = 13 m
• ED = AB = 10 m

Also, EC = DC - ED

= 25-10

= 15 m

Finding Area ∆ BEC by Herons formula.

$\sf \boxed{ \sf \pink{ Area \: of \: triangle = \sqrt{s(s-a)(s - b) (s - c)} }}$

Here, s is the semi-perimeter, and a, b, c are the sides of the triangle.

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Here, a 14 m, b= 15m, c= 13 m

$\sf \: S= \frac{a+b+c}{2} \\ \\ \sf = \frac{14+15+13}{2} \\ \\ \sf = \frac{42}{2} \\ \\ \sf = 21 \: m$

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$\sf \boxed{ \sf \pink{ Area \: of \: triangle = \sqrt{s(s-a)(s - b) (s - c)} }}$

Putting a = 14 m, b= 15m, c = 13 m & s= 21 m

$\sf \: = \sqrt{21(21−14)(21−13)(21−15)} \\ \\ \sf = \sqrt{21(7)(8)(6)} \\ \\ \sf = \sqrt{7056} \\ \\ \sf=84 \: {m}^{2}$

Since ABEC has height BF and base EC, we use base height formula to find area

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$\sf \: Area \: of \: \triangle \: BEC = \frac{1}{2} \times Base \times Height \\ \\ \sf \: Area \: of \: \triangle \: BEC = \frac{1}{2} \times CE \times BF \\ \\ \sf \: 84 \: m² = \frac{1}{2} \times 15 \: m \times BF \\ \\ \sf \: BF = 84 \times \frac{2}{15} m \\ \\ \sf \: BF = 11.2 \: m$

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$\sf \: Area \: of \: trapezium = \frac{1}{2} \times Sum \: of \: parallel \: sides \times Height \\ \\ \sf \: = \frac{1}{2} \times (AB+ DC) \times BF \\ \\ \sf \: = \frac{1}{2} \times (10 + 25) × 11.2 \\ \\ \sf = \frac{1}{2} \times (35) \times 11.2 \\ \\ \sf= \boxed{ \red{ \sf196 \: m²}}$

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$\sf \therefore \: Area \: of \: trapezium \: is \: \bold{ 196 \: {m}^{2} }$

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Answer 2

$\underline{\underline{\maltese\: \: \textbf{\textsf{ Question }}}}$

A field is in the shape of a trapezium of parallel sides 11 m and 25 m and of non- parallel sides 15 m and 13 m. Find the cost of watering the field at the rate of 5 paise per 500 cm?

$\underline{\underline{\maltese\: \: \textbf{\textsf{Answer}}}}$

Sides of the trapenzium are in cm.

$\bf \: 11 \: m = 1100 \: cm \\ \\ \bf25 \: m = 2500 \: cm \\ \\ \bf15 \: m= 1500 \: cm \\ \\ \bf \: 13 \: m = 1300 \: cm$

$\bf So, \pink{the \: perimeter \: of \: trapezium = sum \: of \: all \: sides} \\ \\ \bf \: = 1100 + 2500 + 1500 + 1300 \\ \\ \bf= 6400 \: cm$

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$\bf \: it \: costs \: 5 \: paise \: per \: 500 \: cm. \\ \\ \bf \: so, per \: 1 \: cm \: it \: will \: be = \frac{5}{500} \: paise \\ \\ \bf \: for \:6400 \: cm \: it \: will \: cost = \frac{5}{500} \times 6400 \\ \\ \bf = ₹ \: 64$

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