**Telangana SCERT Solution Class IX (9) Math Chapter 10 Surface Areas and Volumes Exercise 10.1**

**(1) Find the later surface area and total surface area of the following right prisms**

Solution:

** (2.) The total surface area of a cube is 1350 sq.m. Find its volume.**

Solution: Given,

Total surface area of a cube is 1350 m^{2}

∴ Let, be the length of each side of the cube

Total surface are of cube = 6a^{2}

Or, 1350 = 6a^{2}

Or, a^{2} = 1350/6

Or, a = √225

= 15 cm

∴ Volume of the cube = a^{3}

= 15^{3}

= 3375 m^{3}

**(3) Find the area of four walls of a room (Assume that there are no doors or windows) if its length 12 m., breadth 10 m. and height 7.5 m.**

Solution: Given,

Length of the room l = 12 m

Breadth of the room b = 10 m

Height of the room h = 7.5 m

Now, 4 wall of the room is the lateral surface area of the room

**∴ **Lateral surface area = 2h(l +b)

= 2 x 7.5 (12 +10)

= 15 x 2

= 330 m^{2}

∴ The area of the four walls of the room is 330 m^{2}

**(4) The volume of a cuboid is 1200 cm ^{3} . The length is 15 cm. and breadth is 10 cm. Find its height.**

Given,

Volume of the cuboid = 1200 cm^{2}

Length of the cuboid l = 15 cm

Breadth of the cuboid b = 10 cm

Let, height of the cuboid be h

∴ We know,

Volume of a cuboid = l x b x h

Or, 1200 = 15 x 10 x h

Or, h = 1200/(15 x 10)

= 8 cm

∴ Height of the cuboid is 8 cm

**(5) How does the total surface area of a box change if**

**(i) Each dimension is doubled? (ii) Each dimension is tripled?**

**Express in words. Can you find the total surface area of the box if each dimension is raised to n times?**

Let,

The breadth of the box be l

The breadth of the box be b

The height of the box be h

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

(i) ∴When dimensions are doubled

Length = 2l

Breadth = 2b

Height = 2h

∴ Total surface area of the box when dimensions are doubled = 2 (2l x 2b + 2b x 2h + 2l x 2h)

= 2 (4lb + 4bh + 4lh)

= 4 x 2 (lb + bh + lh)

= 4 x total surface area of the box

∴ If dimensions are doubled the total surface one of the box becomes 4 times its initial surface area

(ii) When dimensions are tripled

Length l = 3l

Breadth b = 3b

Height h = 3h

Total surface area when dimensions are tripled

= 2 (3l x 3b + 3b + 3h + 3l x 3h)

= 2 (9lb + 9bh + 9lh)

= 9 x 2 (lb + bh + lh)

9 x total surface area of the box

When dimensions are tripled the total surface area of the box becomes of times the initial surface area of the box.

(iii) when dimensions are raised on times,

Length = nl

Breadth = nb

Height = nh

The total surface area of the box when dimensions are raised n times = 2 (nl x nb + nb x nh + nl x nh)

= 2 (n^{2}lb + n^{2}bh + n^{2}lb)

= n^{2} x 2 (lb + bh + lh)

= n^{2} x total surface area of the box

When dimensions are raised by n times the total surface area of the box becomes n^{2} times the initial surface area of the box.

**(6) The base of a prism is triangular in shape with sides 3 cm., 4 cm. and 5 cm. Find the volume of the prism if its height is 10 cm.**

Solution:

**(7) A regular square pyramid is 3 m. height and the perimeter of its base is 16 m. Find the volume of the pyramid**

Given, Height of the pyramid = 3 m

Perimeter of the base of square pyramid = 16 m

∴ Let, the side of the base of pyramid be a

The height be h

∴ Perimeter of base = 4a

Or, 16 = 4a

Or, a = 16/4

= 4 m

∴ Volume of square pyramid = a^{2} h/3

= 4^{2} x 3/5

= 16 m^{3}

**(8) Solution:** Given,

Length of the swimming pool is 50 m

Breadth of the swimming pool is 25 m

Depth or height of the swimming pool is 3 m

Given, 1m^{3} =1000 litres

∴ Volume of the swimming pool = l x b x h

= 50 x 25 x 3

= 3750 m^{3}

∴ Amount of water it holds = volume x 1000 litres

= 3750 x 1000

= 3750000 litres

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