These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3

Question 1.

The diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find the curved surface area.

Solution:

We have given that,

Diameter of the base of a cone = 10.5 cm

The radius of the base of a cone = 5.25 cm

and slant height of the cone (l) = 10 cm

∴ The curved surface area of cone = πrl

= \(\frac {22}{7}\) × 5.25 × 10

= 165 cm^{2}

Question 2.

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution:

We have given,

Slant height of the cone = 21 m

and diameter of its base = 24 m

∴ radius of its base = 12 m

∴ Total surface area of a cone = πr(l + r)

= \(\frac {22}{7}\) × 12 (21 + 12)

= \(\frac {22}{7}\) × 12 × 33

= 1244.57 m^{2} (approx)

Question 3.

The curved surface area of a cone is 308 cm^{2} and its slant height is 14 cm. Find

(i) radius of the base

(ii) Total surface area of the cone.

Solution:

(i) We have given that

The curved surface area of cone = 308 cm^{2}

and slant height of the cone = 14 cm.

We know that,

Curved surface area of cone = πrl

308 = \(\frac {22}{7}\) × r × 14

r = 7 cm

Therefore, radious of base (r) = 7 cm

(ii) Total surface area of cone = πr(l + r)

= \(\frac {22}{7}\) × 7(14 + 7)

= 462 cm^{2}

Question 4.

A conical tent is 10 m high and the radius of its base is 24 m. Find.

(i) Slant height of the tent

(ii) Cost of the canvas required to make the tent, if the cost of 1 m^{2} canvas is Rs. 70.

Solution:

(i) We have given that

height of the cone = 10 m

and radious of the cone = 24 m

By Pythagoras theorem we know that

h^{2} = p^{2} + b^{2}

⇒ AC^{2} = AB^{2} + BC^{2}

⇒ AC^{2} = (10)^{2} + (24)^{2}

⇒ AC^{2} = 100 + 576

⇒ AC^{2} = 676

⇒ AC = √676 = 26 m

Therefore slant height of the cone = 26 m.

(ii) Canvas required to make the tent = curved surface area of cone

or, Curved surface area of cone = πrl

= \(\frac {22}{7}\) × 24 × 26

= \(\frac{13728}{7}\) m^{2}

Cost of canvas = Rs. 70 per m^{2}

∴ Total cost of canvas required to make the tent = \(\frac{13728}{7} \times 70\) = Rs. 1,37,280.

Question 5.

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

Solution:

We have given that

The radius of base of cone = 6 m

Height of conical tent = 8 m

By Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

⇒ AC^{2} = (9)^{2} + (6)^{2}

⇒ AC^{2} = 64 + 36

⇒ AC^{2} = 100

⇒ AC = 10 cm

∴ Curved surface area of cone = πrl

= \(\frac {22}{7}\) × 6 × 10

= \(\frac{1320}{7}\) m^{2}

According to question

Curved surface area of conical tent = Area of rectangular tarpaulin.

\(\frac{1320}{7}\) = x × 3

x = \(\frac{1320}{7 \times 3}\) = 62.87 m (approx)

Again, stitching margins and wastage in cutting is 20 cm approximately.

Total length of tarpaulin = 62.87 + 0.20 = 63m (approx).

Question 6.

The slant height and base diameter of a Conical tomb is 25 m and 14 m respectively. Find the white-washing its curved surface at the rate of Rs. 210 per 100 m^{2}.

Solution:

We have given that

The slant height of cone = 25 m

Diameter of its base = 14 m

The radius of its base = 7 m

∴ Curved surface area of cone = πrl

= \(\frac {22}{7}\) × 7 × 25

= 550 m^{2}

Now, rate of white washing = Rs. 210 per 100 m^{2}

∴ Total cost of white washing the conical tomb = 550 × \(\frac {210}{100}\) = Rs. 1155.

Question 7.

A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution:

We have given that

Radius of conical cap = 7 cm

and height of conical cap = 24 cm

By Pythagoras theorem,

AC = \(\sqrt{(\mathrm{AB})^{2}+(\mathrm{BC})^{2}}\)

⇒ AC = \(\sqrt{(24)^{2}+(7)^{2}}\)

⇒ AC = \(\sqrt{576+49}\)

⇒ AC = √625

⇒ AC = 25 cm

∴ Curved surface area of one cap = πrl

= \(\frac {22}{7}\) × 7 × 25

= 550 cm^{2}

Sheet required to make one jocker cap = 550 cm^{2}

Sheet required to make ten such caps = 550 × 10 = 5500 cm^{2}.

Question 8.

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each one has a base diameter of 40 cm and height lm. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m^{2}, what will be the cost of painting all these cones?

(Use π = 3.14 and take √1.04 = 1.02)

Solution:

We have given that

Height of the cone = 1 m

and diameter of its base = 40 cm = 0.40 m

radius = \(\frac{0.40}{2}\) = 0.20 m

By Pythagoras theorem

AC = \(\sqrt{(\mathrm{AB})^{2}+(\mathrm{BC})^{2}}\)

= \(\sqrt{(1)^{2}+(0.20)^{2}}\)

= \(\sqrt{100400}\) (∴ √1.04 = 1.02 Given)

= 1.02m

Now, curved surface area of one cone = πrl

= 3.14 × 0.20 × 1.02

= 0.640 m^{2} (approx)

Therefore, the curved surface area of 50 such cones = 50 × 0.640

= 32.02 m^{2}

Rate of painting the cone = Rs. 12 per m^{2}

Total cost for painting fifty cones is 32.02 × 12 = Rs. 384.24 (approx)