A frustum is a part of a right circular cone when itâ€™s cut by a plane parallel to the base of the cone. Problems on finding the volume and surface area of a frustum of a right circular cone are in this exercise. Students can access the RD Sharma Solutions Class 10 which is prepared by experts at BYJUâ€™s for the sole purpose of strengthening the concepts and problem-solving abilities. Students can also download the RD Sharma Solutions for Class 10 Maths Chapter 16 Surface Areas And Volumes Exercise 16.3 PDF provided below.

## RD Sharma Solutions for Class 10 Chapter 16 Surface Areas And Volumes Exercise 16.3 Download PDF

### Access RD Sharma Solutions for Class 10 Chapter 16 Surface Areas And Volumes Exercise 16.3

**1. A bucket has top and bottom diameters of 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 12 cm.Â Also, find the cost of tin sheet used for making the bucket at the rate of Rs 1.20 perÂ dm ^{2}.**

**Solution:**

Given,

Diameter to top of bucket = 40 cm

So, the radius (r_{1}) =Â 40/2 = 20 cm

Diameter of bottom part of the bucket = 20 cm

So, the radius (r_{2}) = 30/2 = 10cm

Depth of the bucket (h) = 12 cm

Volume of the bucket = 1/3 Ï€(r_{2}^{2}_{Â }+ r_{1}^{2} + r_{1 }r_{2Â })h

=Â Ï€/3(20^{2}Â + 10^{2}Â + 20 Ã— 10)12

=Â 8800 cm^{3}

Now,

Given that the cost of tin sheet used for making bucket perÂ dm^{2}Â = Rs 1.20

So, the total cost forÂ 17.87dm^{2}Â =Â 1.20 Ã— 17.87Â = Rs 21.40

Therefore, the cost of tin sheet used for making the bucket is Rs 21.40

**2. A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm and height 3 cm. Find the area of its whole surface and volume.Â **

**Solution:**

Given,

Base diameter of coneÂ (d_{1}) = 20 cm

So the radiusÂ (r_{1})Â =Â 20/2 cm = 10 cm

Top diameter of ConeÂ (d_{2})Â = 12 cm

So, the radiusÂ (r_{2})Â =Â 12/2 cm = 6 cm

Height of the cone (h) = 3 cm

Volume of the frustum of a right circular cone = 1/3 Ï€(r_{2}^{2}_{Â }+ r_{1}^{2} + r_{1 }r_{2Â })h

=Â Ï€/3(10^{2}Â + 6^{2Â }+ 10 Ã— 6)^{3}

= 616Â cm^{3}

Let ‘L’ be the slant height of cone, then we know that

L = âˆš(r_{1} â€“ r2_{1})^{2} + h^{2}

L = âˆš(10 â€“ 6)^{2} + 3^{2}

L = âˆš(16 + 9)

L = 5cm

So, the slant height of coneÂ (L) = 5 cm

Thus,

Total surface area of the frustum = Ï€(r_{1} + r_{2}) x L + Ï€ r_{1}^{2} + Ï€ r_{2}^{2}

=Â Ï€(10 + 6) Ã— 5 + Ï€ Ã— 10^{2}Â + Ï€ Ã— 6^{2}

=Â Ï€(80 + 100 + 36)

=Â Ï€(216)

=Â 678.85 cm^{2}

**3. The slant height of the frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm.Â Find the curved surface of the frustum.Â **

**Solution:**

Given,

Slant height of frustum of cone (l) = 4 cm

Let ratio of the top and bottom circles beÂ r_{1}Â andÂ r_{2}

And given perimeters of its circular ends as 18 cm and 6 cm

âŸ¹Â 2Ï€r_{1}Â = 18 cm; 2Ï€r_{2}Â = 6 cm

âŸ¹Â Ï€r_{1}= 9 cm and Ï€r_{2}Â = 3 cm

We know that,

Curved surface area of frustum of a cone =Â Ï€(r_{1}Â + r_{2})l

=Â Ï€(r_{1Â }+ r_{2})l

=Â (Ï€r_{1}+Ï€r_{2})lÂ = (9 + 3)Â Ã— 4Â = (12)Â Ã— 4Â =Â 48 cm^{2}

Therefore, the curved surface area of the frustum =Â 48 cm^{2}

**4. The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.Â **

**Solution:**

Given,

Perimeter of the upper end = 44 cm

2 Ï€ r_{1} = 44

2(22/7) r_{1} = 44

r_{1 }= 7 cm

Perimeter of the lower end = 33 cm

2 Ï€ r_{2} = 33

2(22/7) r_{2} = 33

r_{2 }= 21/4 cm

Now,

Let the slant height of the frustum of a right circular cone be L

L = 16.1 cm

So, the curved surface area of the frustum cone =Â Ï€(r_{1}Â + r_{2})l

=Â Ï€(7 + 5.25)16.1

Curved surface area of the frustum cone = 619.65Â cm^{3}

Next,

The volume of the frustum cone = 1/3 Ï€(r_{2}^{2}_{Â }+ r_{1}^{2} + r_{1 }r_{2Â })h

= 1/3 Ï€(7^{2}_{Â }+ 5.25^{2} + (7) (5.25)_{Â }) x 16

= 1898.56Â cm^{3}

Thus, volume of the cone = 1898.56Â cm^{3}

Finally, the total surface area of the frustum cone

= Ï€(r_{1} + r_{2}) x L + Ï€ r_{1}^{2} + Ï€ r_{2}^{2}

= Ï€(7 + 5.25) Ã— 16.1 + Ï€7^{2Â }+ Ï€5.25^{2}

=Â Ï€(7 + 5.25) Ã— 16.1 + Ï€(7^{2}Â + 5.25^{2})Â = 860.27Â cm^{2}

Therefore, the total surface area of the frustum cone is 860.27Â cm^{2}

**5. If the radii of the circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm, find the capacity of the bucket.Â **

**Solution:**

Given,

Height of the conical bucket = 45 cm

Radii of the 2 circular ends of the conical bucket are 28 cm and 7 cm

So, r_{1}Â = 28 cmÂ r_{2}Â = 7 cm

Volume of the conical bucket = 1/3 Ï€(r_{1}^{2}_{Â }+ r_{2}^{2} + r_{1 }r_{2Â })h

=Â 1/3 Ï€(28^{2}Â + 7^{2}Â + 28 Ã— 7)45Â =Â 15435Ï€

Therefore, the volume/ capacity of the bucket is 48510Â cm^{3}.

**6. The height of a cone is 20 cm. A small cone is cut off from the top by a plane parallel to the base. If its volume be 1/125 of the volume of the original cone, determine at what height above the base the section is made.**

**Solution: **

Let the radius of the small cone be r cm

Radius if the big cone = R cm

Given, height of the big cone = 20 cm

Let the height of section made = h cm

Then, the height of small cone will be = (20 – h) cm

Now,

In â–³OAB and â–³OCD

âˆ AOB = âˆ COD [common]

âˆ OAB = âˆ OCD [each 90^{o}]

Then, OAB ~ â–³OCD [by AA similarity]

So, by C.P.S.T we have

OA/ OC = AB/ CD

(20 â€“ h)/ 20 = r/ R â€¦â€¦ (i)

Also given,

Volume of small cone = 1/125 x volume of big cone

1/3 Ï€ r^{2}(20 – h) = 1/125 x 1/3 Ï€R^{2}_{Â }x 20

r^{2}/ R^{2} = 1/125 x 20/ (20 – h) [From (i)]_{Â }

(20 – h)^{2}/ 20^{2} = 1/125 x 20/20 â€“ h

(20 – h)^{3} = 20^{3}/ 125

20 â€“ h = 20/5

20 â€“ h = 4

h = 20 – 4 = 16 cm

Therefore, itâ€™s found that the section was made at a height of 16 cm above the base.

**7. If the radii of the circular ends of a bucket 24 cm high are 5 and 15 cm respectively, find the surface area of the bucket.**

**Solution:**

Given,

Height of the bucket (h) = 24 cm

Radius of the circular ends of the bucket 5 cm and 15 cm

So, r_{1}Â = 5 cm; r_{2}Â = 15 cm

Let ‘L’ be the slant height of the bucket

Then, we know that

Now,

Curved surface area of the bucket

= Ï€(r_{1}Â + r_{2})l + Ï€r_{1}^{2}

=Â Ï€(5 + 15)26 + Ï€5^{2}Â =Â Ï€(520 + 25)Â =Â 545Ï€ cm^{2}

Therefore, the curved surface area of the bucket =Â 545Ï€ cm^{2}

**8. The radii of circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 12 cm.Â Find the total surface area and volume of frustum.Â **

**Solution:**

Given the height of frustum cone = 12 cm

Radii of a frustum cone are 12 cm and 3 cm

So, r_{1}Â = 12 cm; r_{2}Â = 3 cm

Let slant height of the frustum cone be ‘L’

Then, we know that

L = 15 cm

Now, the total surface area of frustum of a cone = Ï€ (r_{1} + r_{2}) x L + Ï€ r_{1}^{2} + Ï€ r_{2}^{2}

=Â Ï€ (12 + 3)15 + Ï€12^{2}Â + Ï€3^{2}

Thus, total surface area of the frustum = 378Â Ï€ cm^{2}

Next,

Volume of frustum cone = 1/3 Ï€(r_{2}^{2}_{Â }+ r_{1}^{2} + r_{1 }r_{2Â })h

=Â 1/3 Ï€(12^{2}Â + 3^{2}Â + 12 Ã— 3) Ã— 12Â =Â 756Ï€ cm^{3}

Therefore, the volume of the frustum cone =Â 756 Ï€ cm^{3}

**9. A tent consists of a frustum of a cone capped by a cone. If radii of the ends of the frustum be 13 m and 7 m, the height of frustum be 8 m and the slant height of the conical cap be 12 m, find the canvas required for the tent.Â **

**Solution:**

Given,

Height of frustum (h) = 8 m

Radii of the frustum cone are 13 cm and 7 cm

So, r_{1}Â = 13 cm and r_{2}Â = 7Â cm

Let â€˜Lâ€™ be slant height of the frustum cone

Then, we know that

Curved surface area of the frustumÂ (s_{1}) = Ï€(r_{1}Â + r_{2}) Ã— L =Â Ï€(13 + 7) Ã— 10Â =Â 200 Ï€ m^{2}

Then, given slant height of conical cap = 12 m

Base radius of upper cap cone = 7 m

So, the curved surface area of upper cap coneÂ (s_{2}) = Ï€rlÂ =Â Ï€ Ã— 7 Ã— 12Â =Â 264Â m^{2}

Thus, the total canvas required for tentÂ (S) = s_{1}Â + s_{2}

S =Â 200Ï€ + 264 = 892.57 m^{2}

Therefore, the canvas required for the tent is 892.57 m^{2}.

**10. A milk container of height 16 cm is made of metal sheet in the form of frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate ofÂ Rs.44 perÂ litreÂ which the container can hold.**

**Solution: **

Given,

A milk container of the form of frustum of a cone

The radius of the lower end (r_{1}) = 8 cm

The radius of the upper end (r_{2}) = 20 cm

Let h be its height, h = 16 cm

Then, the capacity of the container = Volume of frustum of the cone

= 1/3 Ï€(r_{2}^{2}_{Â }+ r_{1}^{2} + r_{1 }r_{2Â })h

= 1/3 Ï€(20^{2}_{Â }+ 8^{2} + (20) (8)_{Â }) x 16

= 10459.42 cm^{3}

= 10.46 litres

Now, given that cost of 1 litre of milk = Rs 44

Then the cost of 10.46 litres of milk = Rs (44 x 10.46) = Rs 460.24

**11. A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the container, at the rate ofÂ Rs.25 perÂ litre.**

**Solution: **

Let R and r be the radii of the top and base of the bucket respectively,

Let h be its height.

Then, we have R = 20 cm, r = 10 cm, h = 30 cm

Capacity of the bucket = Volume of the frustum of the cone

= 1/3 Ï€(R^{2}_{Â }+ r^{2} + R r_{Â })h

= 1/3 Ï€(20^{2}_{Â }+ 10^{2} + 20 x 10_{Â }) x 30

= 3.14 x 10 (400 + 100 + 200)

= 21980 cm^{3} = 21.98 litres

Now,

Surface area of the bucket = CSA of the bucket + Surface area of the bottom

= Ï€ l (R + r) + Ï€r^{2}

We know that,

l = âˆšh^{2} + (R – r)^{2}

= âˆš[30^{2} + (20 – 10)^{2}] = âˆš(900 + 100)

= âˆš1000 = 31.62 cm

So,

The Surface area of the bucket = (3.14) x 31.62 x (20 + 10) + (3.14) x 10^{2}

= 2978.60 + 314

= 3292.60 cm^{2}

Next, given that the cost of 1 litre milk = Rs 25

Thus, the cost of 21.98 litres of milk = Rs (25 x 21.98) = Rs 549.50