**The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of rs 7.50 per m ^{2}.**

**Answer**

length of the room = 5m

breadth of the room = 4m

height of the room = 3m

Area of four walls including the ceiling = 2(l+b)×h + (l×b)

= 2(5+4)×3 + (5×4) m^{2 }

^{ }= (54 + 20) m^{2 }

= 74^{ }m^{2 }Cost of white washing = ���7.50 per m^{2}

Total cost = rs^{ }(74×7.50) = rs 555

**The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of rs 10 per m ^{2} is rs15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]**

**Answer**

Perimeter of rectangular hall = 2(l + b) = 250 m

Total cost of painting = ���15000

Rate per m^{2 }=^{ }rs10

Area of four walls = 2(l + b) h m^{2 }= (250×h) m^{2}

A/q,

(250×h)×10 = rs15000

⇒ 2500×h = rs15000

⇒ h = 15000/2500 m

⇒ h = 6 m

Thus the height of the hall is 6 m.

**The paint in a certain container is sufficient to paint an area equal to 9.375 m ^{2}. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?**

**Answer**

Volume of paint = 9.375 m^{2 }=^{ }93750 cm^{2}

Dimension of brick = 22.5 cm×10 cm×7.5 cm

Total surface area of a brick = 2(lb + bh + lh) cm^{2}

= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm^{2}

= 2(225 + 75 + 168.75) cm^{2}

= 2×468.75 cm^{2} = 937.5^{ }cm^{2}

**Number of bricks can be painted** = 93750/937.5 = 100

A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.

(i) Which box has the greater lateral surface area and by how much?

(ii) Which box has the smaller total surface area and by how much?

(i) Lateral surface area of cubical box of edge 10cm = 4×10^{2} cm^{2} = 400 cm^{2}

Lateral surface area of cuboid box = 2(l+b)×h

= 2×(12.5+10)×8 cm^{2}

= 2×22.5×8 cm^{2} = 360 cm^{2}

Thus, lateral surface area of the cubical box is greater by (400 – 360) cm^{2} = 40 cm^{2}

(ii) Total surface area of cubical box of edge 10 cm =6×102cm^{2}=600cm^{2}

Total surface area of cuboidal box = 2(lb + bh + lh)

= 2(12.5×10 + 10×8 + 8×12.5)cm^{2}

= 2(125+80+100) cm^{2}

= (2×305) cm^{2 }= 610 cm^{2}

Thus, total surface area of cubical box is smaller by 10 cm^{2}