Question

Find the cost of watering a trapezoidal field whose parallel sides are 10 m and 25 m respectively, the perpendicular distance between them is 15 om and the rate of watering is 34 m. per​

Answer 1

Appropriate Question :-

Find the cost of watering a trapezoidal field whose parallel sides are 10 m and 25 m respectively, the perpendicular distance between them is 15 m and the rate of watering is Rs 34 sq. m.

$\large\underline{\sf{Solution-}}$

Given a trapezoidal field, whose parallel sides are 10 m and 25 m respectively and distance between parallel sides is 15 m.

We know,

$\boxed{ \sf{ \:Area_{(Trapezium)} = \frac{1}{2} \times (Sum \: of \: \parallel \: sides) \times height \: }}$

So, on substituting the values, we get

$\sf \: Area_{(Trapezium)} = \frac{1}{2}(10 + 25) \times 15$

$\sf \: Area_{(Trapezium)} = \frac{1}{2} \times 35 \times 15$

$\sf \:  \implies \: \sf \: Area_{(Trapezium)} = \frac{525}{2} \: {m}^{2}$

Now,

$\sf \: Cost \: of \: watering \: 1 \: {m}^{2} = Rs \: 34$

So,

$\sf \: Cost \: of \: watering \: \frac{525}{2} \: {m}^{2} = 34 \times \frac{525}{2} = 525 \times 17$

$\sf \:  \implies \: Cost \: of \: watering \: \frac{525}{2} \: {m}^{2} = Rs \: 8925$

$\rule{190pt}{2pt}$

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

☯ Question :-

Find the cost of watering a trapezoidal field whose parallel sides are 10 m and 25 m respectively, the perpendicular distance between them is 15 m and the rate of watering is Rs. 34 sq. m

☯ Solution :-

As the rate of watering is Rs. 34 sq. m

So, we have to find the area of the trapezoidal field.

As, we know

$\pink{ \boxed{ \bf ✯\: Area _{(trapazium)} \: = \frac{1}{2} (Sum \: of \: parallel \: sides ) (Distance \: between \: them )}}$

Here-

✞ 2 Parallel sides = 10m and 25 m

✞ Distance between them = 15 m

So

$\sf Area _{(trapezium)} = \frac{1}{2} (10 + 25) \times 15 \\ \\ \\ ➼\:\sf Area _{(trapezium)} = \frac{35 \times 15}{2} \\ \\ \\ \bf{ \color{brown}➼\:Area _{(trapazium)} = { \frac{525}{2} \: {m}^{2} }{} }$

So,

$\sf \: Cost \: of \: watering = \cancel{34} \times \frac{525}{ \cancel2} \\ \\ \\ ➺\sf \: Cost \: of \: watering = \: 17 \times 525 \\ \\ \\ \bf{ \color{aqua}➺\: Cost \: of \: watering = 8325}$

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❒ Cost of watering the field is Rs. 8325

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