<u>Ⲁⲛ⳽ⲱⲉⲅ</u><u> </u><u>:</u>
<h3>
<u>Ⲋⲟⳑⳙⲧⳕⲟⲛ :</u></h3>
<u>We </u><u>are </u><u>given </u><u>that:</u>
- The length of the room is three time it's breadth.
- Volume of the room is 240m³
- Cost of plastering 4 walls is Rs.8000
- Cost of plastering 1 wall is Rs.50
<u>Some </u><u>assumptions:</u>
- Let the breadth be x
- Then length will be 3x
- Let the height be h
<u>Using </u><u>Formula:</u>
➙ V = l × b × h
➙ V = x × 3x × h
➙ 240 = 3x² × h ㅤㅤ⸻ ( 1 )
Now, the cost of plastering it's 4 walls is Rs 8000 at the rate of 50 /m². So, we can find the area of the walls of the room by dividing 8000 by 50 i.e:
➙ A = Total cost / cost per m²
➙ 8000 / 50
➙ 800 / 5
➙ 160m²
In a cuboid , there are a pair of two opposite and equal rectangular sides, and area of rectangle is given by :
So, area of 2 walls along length :
➙ 2 × length × height
➙ 2 × 3x × h
➙ 6xh
Area of 2 walls along breadth:
➙ 2 × breadth × height
➙ 2 × x × h
➙ 2xh
Sum of areas of these four walls will be equal to the area of the walls, that we have find earlier;
➙ ( 6xh ) + ( 2xh ) = 160
➙ 8xh = 160
➙ xh = 160 / 8
➙ xh = 20 ㅤㅤㅤ⸻ ( 2 )
- <u>Using </u><u>equation </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u>
➙ 240 = 3x²h
➙ x²h = 240 / 3
➙ x × xh = 80
➙ x × 20 = 80 ㅤ[ <em>from </em><em>equation </em><em>(</em><em> </em><em>2</em><em> </em><em>)</em><em> </em>]
➙ x = 80 / 20
➙ x = 4
- <u>Using </u><u>equation </u><u>(</u><u> </u><u>2</u><u> </u><u>)</u>
➙ xh = 20
➙ 4 × h = 20
➙ h = 20 / 4
➙ h = 5
ㅤㅤㅤㅤㅤ~<u>H</u><u>e</u><u>n</u><u>c</u><u>e</u><u>,</u><u> </u><u>the </u><u>height </u><u>of </u><u>the </u><u>ro</u><u>o</u><u>m </u><u>is </u><u>5</u><u>m</u><u>.</u>