# NCERT Solutions Class 9 Maths Chapter 13 Surface area and Volume Exercise 13.3

## Introduction:

In this exercise/article we will learn about Surface Area And Volume. Surface area of a Right Circular Cone, Curved surface area of a cone and total surface area of cone. So far, we have been generating solids by stacking up congruent figures. Incidentally, such figures are called Prisms. Now let us look at another kind of solids which is not a prism. (These kinds of solids are called Pyramids).

Class 9 Maths Chapter 13 Surface Area And Volume :

- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.1
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.2
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.3
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.4
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.5
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.6
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.7
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.8

**Class 9 Maths Exercise 13.3 (Page-221)**

**Q1. **Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find the cuved surface area.

**Solution :**

According to the question,

Radius = \(\displaystyle \frac{d}{2}\)

Radius = \(\displaystyle \frac{{10.5}}{2}\)

Radius = 5.25 cm

Given, Radius = 5.25 cm

Slant height = 10 cm

So, Curved surface area of cone = πrl

= \(\displaystyle \frac{{22}}{7}\) × 5.25 × 10

= \(\displaystyle \frac{{22}}{7}\) × 52.5

= \(\displaystyle \frac{{1155}}{7}\)

= 165 cm2

∴ The curved surface area of a cone = 165 cm2 .

**Q2. **Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

**Solution :**

According to the question,

Radius = \(\displaystyle \frac{d}{2}\)

Radius = \(\displaystyle \frac{{24}}{2}\)

Radius = 12 m

Given, Radius = 12 m

Slant height = 21 m

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

= \(\displaystyle \frac{{22}}{7}\) × 12 ( 12 + 21 )

= \(\displaystyle \frac{{264}}{7}\) × ( 33 )

= \(\displaystyle \frac{{8712}}{7}\)

= 1244.57 m2

∴ Total surface area of a cone = 1244.57 m2 .

**Q3. **Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find

(i) radius of the base and (ii) total surface area of the cone.

**Solution :**

According to the question,

**(i)** Given, curved surface area of a cone = 308 cm2

Slant height = 14 cm

So, Curved surface area of a cone = πrl

⇒ 308 = \(\displaystyle \frac{{22}}{7}\) × r × 14

⇒ 308 × 7 = 22 × r × 14

⇒ 2156 = 308 r

⇒ \(\displaystyle \frac{{2156}}{{308}}\) = r

= 7 cm

∴ Radius of the base of cone = 7 cm

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

= \(\displaystyle \frac{{22}}{7}\) × 7 ( 7 + 14 )

= \(\displaystyle \frac{{22}}{7}\) × 7 ( 21 )

=

= 22 × 21

= 462 cm2

∴ Total surface area of a cone = 462 cm2 .

**Q4. **A conical tent is 10 m high and the radius of its base is 24 m. Find

(i) slant height of the height

(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.

**Solution : **

According to the question,

**(i) **Given, Radius = 24 m

Height = 10 m

So, slant height = l = \(\displaystyle \sqrt{{{{h}^{2}}+{{r}^{2}}}}\)

⇒ l = \(\displaystyle \sqrt{{{{{10}}^{2}}+{{{24}}^{2}}}}\)

⇒ l = \(\displaystyle \sqrt{{100+576}}\)

⇒ l = \(\displaystyle \sqrt{{676}}\)

⇒ l = 26 m

∴ Slant height of the tent = 26 m.

**(ii) **Curved surface area of a cone = πrl

= \(\displaystyle \frac{{22}}{7}\) × 24 × 26

= \(\displaystyle \frac{{22}}{7}\) × 624

= \(\displaystyle \frac{{13728}}{7}\)

= 1961.14 m2

Since, the cost of 1 m2 canvas =Rs 70

∴ The cost of 1961.14 m2 canvas = Rs 70 × 1961.14

= Rs 137280

So, Total cost of canvas make is the tent = Rs 137280 .

**Q5. **What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? 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 : **

According to the question,

Given, Radius = 6 m

Height = 8 m

So, slant height = l = \(\displaystyle \sqrt{{{{h}^{2}}\,+\,{{r}^{2}}}}\)

⇒ l = \(\displaystyle \sqrt{{{{8}^{2}}\,+\,{{6}^{2}}}}\)

⇒ l = \(\displaystyle \sqrt{{64\,+\,36}}\)

⇒ l = \(\displaystyle \sqrt{{100}}\)

⇒ l = 10

Now, Curved surface area of conical = πrl

= 3.14 × 6 × 10

= 3.14 × 60

= 188.4 m2

∴ Wide of a tarpaulin = 3 m

So, Length of a tarpaulin = \(\displaystyle \frac{{188.4}}{3}\)

=

= 62.8 m

Similarly, Extra length of material that required = 20 cm

= 0.20 m [ converted **cm** to **m** ]

Total length of tarpaulin required = 62.8 + 0.20 m

= 63 m

∴ The length of required to make conical tent = 63 m .

**Q6. **The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m2 .

**Solution :**

According to the question,

Radius = \(\displaystyle \frac{d}{2}\)

Radius = \(\displaystyle \frac{{14}}{2}\)

Radius = 7 m

Given, Radius = 7 m

Slant height = 25 m

So, curved surface area of a conical tomb = πrl

= \(\displaystyle \frac{{22}}{7}\) × 7 × 25

= \(\displaystyle \frac{{22}}{7}\) × 175

= \(\displaystyle \frac{{3850}}{7}\)

= 550 m2

Now, cost of white-washing at the rate of Rs 210 per 100 m2

= Rs 550 × \(\displaystyle \frac{{210}}{{100}}\)

= Rs \(\displaystyle \frac{{115500}}{{100}}\)

= Rs 1155

∴ The cost of white-washing of a conical tomb is Rs 1155 .

**Q7. **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 :**

According to the question,

Given, Radius = 7 cm

Height = 24 cm

So, slant height = l = \(\displaystyle \sqrt{{{{h}^{2}}\,+\,{{r}^{2}}}}\)

⇒ l = \(\displaystyle \sqrt{{{{{24}}^{2}}\,+\,{{7}^{2}}}}\)

⇒ l = \(\displaystyle \sqrt{{576\,+\,49}}\)

⇒ l = \(\displaystyle \sqrt{{625}}\)

⇒ l = 25 cm

Now, Curved surface area of a right circular cone = πrl

= \(\displaystyle \frac{{22}}{7}\) × 7 × 25

= \(\displaystyle \frac{{22}}{7}\) × 175

= \(\displaystyle \frac{{3850}}{7}\)

= 550 cm2

Since, Area of sheet for 1 cap = 550 cm2

So, Area of sheet for 10 caps = 10 × 550 cm2

= 5500 cm2

∴ The area of the sheet required to make 10 caps = 5500 cm2 .

**Q8. **A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m2, what will be the cost of painting all these cones? ( Use π = 3.14 and take √1.04 = 1.02 )

**Solution :**

According to the question,

Radius = \(\displaystyle \frac{d}{2}\)

Radius = \(\displaystyle \frac{{40}}{2}\)

Radius = 20 cm

Given, Radius = 0.2 m [ converted **cm** to **m** ]

Height = 1 m

So, slant height = l = \(\displaystyle \sqrt{{{{h}^{2}}\,+\,{{r}^{2}}}}\)

⇒ l = \(\displaystyle \sqrt{{{{1}^{2}}\,+\,{{{0.2}}^{2}}}}\)

⇒ l = \(\displaystyle \sqrt{{1\,+\,0.04}}\)

⇒ l = \(\displaystyle \sqrt{{1.04}}\)

⇒ l = 1.02 m

So, curved surface area of a cone = πrl

= 3.14 × 0.2 × 1.02

= 0.64056 m2

Now, curved surface area of 50 cones = 50 × o.64056

= 32.028 m2

, cost of painting is Rs 12 per m2

= Rs 32.028 × 12

= Rs 384.34 ( approx )

∴ The cost of painting all these cones = Rs 384.34 .

Class 9 Maths Chapter 13 Surface Area And Volume :

- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.1
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.2
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.3
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.4
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.5
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.6
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.7
- NCERT Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.8