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

Answer:

Here is your answer

Step-by-step explanation:

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

Then, h=0.4(l+b)

Area of four walls =2(l+b)h

=2(l+b)×0.4(l+b)

=0.8(l+b)

2

∴ Area which is papered =0.8(l+b)

2

−15

Area of paper = Area of wall

⇒ Length =

0.5

0.8(l+b)

2

−15

...(∵Width=50cm=0.5m)

Given, Rs.

0.5

(0.8(l+b)

2

−15)

×2=Rs260

⇒0.8(l+b)

2

−15=

2

260×0.5

⇒0.8(l+b)

2

−15=65

⇒0.8(l+b)

2

=65+15=80

⇒(l+b)

2

=

0.8

80

=100

⇒l+b=10

∴h=0.4×10=4m

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Answer 2

Answer:

$\huge\underline\bold {Answer:}$

Let the length, breadth and height of the room be l, b and h respectively. Then, h = 0.4(l + b)

Area of four walls = 2(l + b)h

= 2(l + b) × 0.4(l + b)

= 0.8(l + b)^2

Therefore area which is prepared

= 0.8(l + b)^2 – 15

Now, area of paper = Area of wall

=> Length × breadth

= 0.8(l + b)^2 – 15

=> Length

$= \frac{0.8(l + b) {}^{2} - 15 }{0.5} \: \: since \: \: width \: \: = \: \: 50 \: \: cm \\ = 0.5m$

Given,

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

$= > 0.8(l + b) {}^{2} - 15 = \frac{260 \times 0.5}{2} \\ = > 0.8(l + b) {}^{2} - 15 = 65 \\ = > 0.8(l + b) {}^{2} = 65 + 15 = 80 \\ = > (l + b) {}^{2} = \frac{80}{0.8} = 100 \\ = > l + b = 10$

Therefore h = 0.4 × 10 = 4 m.