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:

Answer:

4m

room be l, b and h respectively.

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

= 45 (l+b)2 - 15

Now, area of paper = area of wall

Length × breadth = 45 (l+b)2 - 15

Length = 45(l+b)2−1512

Given, 2 [ 45(l+b)2−15] × 2 = 260

=> 165(l+b)2 - 60 = 260

=> 165(l+b)2 = 320

=> (l+b)2 = 100

=> (l+b) = 10

∴ h = 0.4 × 10 = 4 m

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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.