Question

The height of a room is 40% of its semi - perimeter. It costs 260 rupees to paper the walls of the room with paper 50 cm wide at 2 rupees per metre allowing an area of 15 m^2 for doors and windows. The height of the room is ?​ Plz don't copy.​

Answer 1

Given :-

• The height of a room is 40% of its semi - perimeter. It costs 260 rupees to paper the walls of the room with paper 50 cm wide at 2 rupees per metre allowing an area of 15 m^2 for doors and windows.

To find:-

• The height of the room is ?

Solution:-

Let,

length = l

breadth  = b

height  = h

Then,

=>  h = 0.4 (l + b)

→Area of the four walls = 2 (l + b)h

= 2 (l + b) × 0.4 (l + b) = 0.8 (l+b) ²

→Required area where paper has to be pasted

= 4/5 (l+b)² - 15

→Area of paper = area of wall

⇒ Length × breadth = 4/5 (l+b)² - 15

⇒Length = 4/5(l+b)²−15 ÷ 1 / 2

Given,

⇒ 2 [ 4 / 5(l+b)²−15] × 2 = 260

=> 16/5(l+b)² - 60 = 260

=> 16/5(l+b)² = 320

=> (l+b)² = 100

=> (l+b) = 10

∴ h = 0.4 × 10 = 4 m.

Hence, height of the room is 4m.

_______________________________

Answer 2

Question:

The height of a room is 40% of its semi-perimeter. It costs 260 rupees to paper the walls of the room with paper 50 cm wide at rs.2/meter allowing an area of 15 m² for doors and windows. The height of the room is?

Solution:

Here,

Let, the length, breadth and height of the room be l,b and h respectively.

Then,

h = 0.4(l + b)

Now, we know that,

➩$\underline{\boxed{ ➢ \rm \color{purple}c.s.a\: of \: cuboid =2(l + b)h }}$

where,

• h = 0.4(l + b)

Now,

➩$\rm area \: of \: four \: walls =2(l + b)h$

➩$\rm area \: of \: four \: walls =2(l+b)\times 0.4(l+b)$

➩$\rm area \: of \: four \: walls = 0.8 {(l + b)}^{2}$

$\rm \therefore Area \: which \: is \: Papered$

$\rm = 0.8 {(l + b)}^{2} - 15$

Area of paper = Area of wall

$\rm \implies Length = \frac{0.8 {(l + b) }^{2} - 15 }{0.5} .... \\ \rm(∵width = 50cm = \frac{5 \cancel0}{10\cancel0} = 0.5m)$

Given, Rs.

$\rm = \frac{(0.8 {(l + b) }^{2} - 15 )}{0.5} \times 2 = Rs. 260$

$\rm \implies 0.8 {(l + b) }^{2} - 15 = \frac{260 \times 0.5}{2}$

$\rm \implies 0.8 {(l + b) }^{2} - 15 = 65$

$\rm \implies 0.8 {(l + b) }^{2} = 65 + 15$

$\rm \implies 0.8 {(l + b) }^{2} = 80$

$\rm \implies {(l + b) }^{2} = \frac{ 80}{0.8} = 100$

$\rm \implies {(l + b) }^{2} = {10}^{2}$

$\rm \implies l + b = 10$

$\rm \therefore h = 0.4(l + b) = 0.4 \times 10$

$\rm \implies 4m$

Hence, The height of the room will be 4m