**Telangana**** SCERT Solution Class IX (9) Math Chapter 10 Surface Areas and Volumes Exercise 10.3**

**(1) The base area of a cone is 38.5 cm ^{2} . Its volume is 77 cm3 . Find its height**

Given,

Area base of cone = 38.5 cm^{2}

Let,

The radius of cone be r

The height of lone be h

** (2) The volume of a cone is 462 m ^{3} . Its base radius is 7 m. Find its height.**

**Solution:** Given,

Volume of a cone = 462 m^{3}

Radius of base = 7 m

Let, the radius of the cone be r

The height of the cone be h

∴ πr^{2}h = 462

**(4) The cost of painting the total surface area of a cone at 25 paise per cm ^{2} is `176. Find the volume of the cone, if its slant height is 25 cm.**

**Solution:** Given,

Cost of painting total surface area of cone at 25 paisa

Per cm^{2} is ₹176

Slant height l of cone = 25 cm

Let, the radius of base of cone be r

The height of cone be h

∴ The radius of circle is 9 cm

∴ Cone is a right angled solid

∴ By Pythagoras theorem

r^{2} + h^{2} = l^{2}

Or, 9^{2} + h^{2} = 15^{2}

Or, h = √225-81

Or, h = √144

= 12 cm

∴ Volume of the cone is = 1/3πr^{2}h

= 1/3 x 22/7 x 9 x 9 x 12

= 7128/7

= 1018.28 cm^{3}

**(6) The height of a tent is 9 m. Its base diameter is 24 m. What is its slant height? Find the cost of canvas cloth required if it costs `14 per sq.m.**

Given,

∴ The height of a tent (conical) ‘h’ 9 m

∴ The diameter of a tent = 24 m

The radius of a tent = (r)

∴ Let, the slant height of the tent be l

∴ By Pythagoras theorem,

r^{2} + h^{2} = l^{2}

Or, 12^{2} + 9^{2} = l^{2}

Or, l^{2} = √(144+81)

= √225

= 15 m

∴ Curved surface area of the tent = πrl

= 22/7 x 12 x 15

= 3970/7

= 565.71 m^{2}

Cost of canvas = ₹ 14/m^{2}

∴ Cost of total canvas for the tent = 14 x 565.71

= ₹7920

**(7) The curved surface area of a cone is 1159-5/7 cm2 . Area of its base is 254-4/7 cm ^{2}.Find its volume.**

**Solution:**

**(8) A tent is cylindrical to a height of 4.8 m. and conical above it. The radius of the base is 4.5m. and total height of the tent is 10.8 m. Find the canvas required for the tent in square meters**

**Solution:** Given,

The tent is cylindrical for 4.8 m

Then it is a core as shown in the adjacent figure

∴Let, The radius of cylinder & coins base be r

The height of cylinder be H

The height of cone be h

The slant height of cone be l [H = 4.8m, r = 4.5m, h = 1.8 – 4.8 = 6m]

∴Curved surface area of the cylindrical part of tent

= 2πrH

= 2 x 22/7 x 4.5 x 4.8

= 950.4/7

= 135.77 m^{2}

Total height of tent = 10.8 m

Height of cone part of tent h = 10.8 – 4.8

= 6m

By Pythagoras theorem,

r^{2} + h^{2} = l^{2}

Or, 4.5^{2} + 6^{2} = l^{2}

Or, 2.25 + 36 = l^{2}

Or, l = √56.25

= 7.5 m

Curved surface area of the cone = πrl

= 22/7 x 4.5 x 7.5

= 742/5/7

= 106.07 m^{2}

Total canvas required = total area of tent

= sum of curved surface area of the cylinder and cone

= 135.77 + 106.07

= 241.84 m^{2} (approx)

** (9) What length of tarpaulin 3 m wide will be required to make a conical tent of height 8m and base radius 6m? Assume that extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (use π = 3.14)**

**Solution:**

Given,

Width of tarpaulin = 3m

Height of conical tent ‘h’ = 8 m

Radius of conical tent ‘r’ = 6m

Let, start height of tent be l

∴ By Pythagoras theorem

r^{2} + h^{2} = l^{2}

Or, 6^{2} + 8^{2} = l^{2}

Or l = √(36+64)

= √100

= 10 m

∴ Given, π = 3.14

Now, Curved surface area of tent = πrl

= 3.14 x 6 x 10

= 31.4 x 6

= 188.4 m^{2}

Extra cutting material as wastage = 20 cm

= 0.2 m

∴ Total tarpaulin required = (curved surface area of tent)/(width of tarpaulin-waste cloth)

= 188.4/3 + 0.2

= 62.8 + 0.2

= 63 m

**(10) A Joker’s cap is in the form of a right circular cone of base radius 7 cm and height 27 cm. Find the area of the sheet required to make 10 such caps.**

**Solution:**

Given,

Height ‘h’ of joker cap = 27 cm

Radius of base of joker cap = 7 cm

Let the slant height of the joker cap b ‘l’

By Pythagoras theorem,

r^{2} + h^{2} = l^{2}

Or, 7^{2} + 27^{2} = l^{2}

Or, l = √(49+729)

Or, l = √778

Now, to calculate this was long division method

Curved surface area of cap = πrl

= 22/7 x 7 x 27.89

= 613.58 cm^{2}

Total paper required to make 10 such caps

= 613.58 x 10 cm^{2}

= 6135.8 cm^{2}

**(11) Water is pouring into a conical vessel of diameter 5.2m and slant height 6.8m (as shown in the adjoining figure), at the rate of 1.8 m ^{3} per minute. How long will it take to fill the vessel?**

**Solution:**

Thank you so much

I just want questions because I forgot my text book in school thanks a lot