Question

$\large \; {\underline{\underline{\orange{\pmb{\frak{ \; Question \; :- }}}}}}$

$\dashrightarrow$ The floor of a Rectangular hall has a perimeter of 250 m .If the Cost of Painting the four walls is Rs.10 per m² is Rs.15000 .Find the Height of the Hall .



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$\large \; {\underline{\underline{\pmb{\color{darkblue}{\frak{ \; Note \; :- }}}}}}$

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Answer 1

Given that, The perimeter of the floor of the rectangular hall is 250m and the cost of painting the four walls at the rate of ₹10 per m² is ₹15000.

Hence, we can say that “ The area of the four walls of the cuboidal room will be the Lateral surface area of the cuboid. ”

$\boxed{ \color{blue} \textbf{\textsf{Lateral surface area = 2h(l + b)}}}$

The area of the four walls can also be obtained by dividing the total cost of the painting by the rate of painting per m².

Let the length, breadth, and height of the room be l , b and h respectively. The cost of painting the four walls is ₹15000.

The rate of painting is ₹ 10 per m²

Perimeter of the floor = 250 m

Therefore, 2(l + b) = 250m _____eq(1)

[Since, perimeter of a rectangle = 2(l + b) ]

Now, Finding Area of walls :-

$= \frac{15000}{10}$

$\rm = 1500 {m}^{2}$

2(l + b)h = 1500 m²  [From eq(1)]

$\rm 250 \times h = 1500$

$\rm h = \frac{1500}{250}$

$\color{magenta} \textbf{\textsf{h = 6m}}$

▫FINAL ANSWER :-

$\therefore \: \color{orange} \pmb{\textbf{\textsf{Height \:of \: Wall is 6m }}}$

Answer 2

$\huge \red \star \large \green \star \tiny \orange \star \huge \underline{ \boxed{ \red{ \frak{s}} \frak{ \orange{o} \blue{lu} \pink{t} \green{io} \red{n}}}} \tiny \orange \star \large \green \star \huge \red \star$

✪ Given :

• Perimeter = 250m
• Cost of painting 4 walls = rs10/m²
• Total Cost = rs 15000

✪ To Find :

• Height of the wall

✪ On Solving :

➜ Let Length be = l

➜ Let Breadth be = b

➜ Let height be = h

• Given Perimeter of rectangle = 250m

$\large \star{ \boxed{ \underline{ \pink{ \sf \: perimeter = 2(length + breadth)}}}} \star$

$\large \sf \implies \: 250m = 2(l + b)$

• On Transposing The Terms :

$\large \tt \implies \: l + b = \dfrac{250m}{2}$

$\large \red \mapsto \boxed{ \sf \: l + b = 125m} \longrightarrow \: (1)$

➜ Now, Area of 4 walls :

• Cost of painting 4 walls per m²= 10
• Total Cost = 15000

$\large \sf \implies \: area = \dfrac{1500 \cancel0}{1 \cancel0}$

$\large \therefore \boxed{ \bold{area = 1500 {m}^{2} }}$

• Finding Dimensions :

$\large \star{ \boxed{ \underline{ \pink{ \sf \: area = 2(lh + bh}}}} \star$

➜ Taking Common :

$\sf \large\implies \: area = 2(l + b)h$

➜ From (1) :

$\sf \large \red\implies \: area = 2(125)h$

• Area (Solved) = 1500m²

$\large \sf \red \implies \: 1500 {m}^{2} = 250 \times h$

$\large \sf \red \implies \: h = \dfrac{1500 {m}^{2} }{250m}$

$\\ \large \sf \red \implies \: h = \dfrac{ \cancel{1500} \: \: 6 \: {m}^{ \cancel2} }{ \cancel{250 } \: \cancel{m}}$

$\large \therefore \underline{ \boxed{ \bold{ \green{height = 6m}}}}$

$\large \sf \bigstar \: hope \: \: it \: \: helps \: \: uh \: : )$