CBSE Class 10 Maths Notes & Important Questions Chapter 13 Surface Areas and Volumes

CBSE Class 10 Maths Notes & Important Questions Chapter 13 Surface Areas and Volumes

CBSE Class 10 Maths Notes Chapter 13 Surface Areas and Volumes

Here are some important points and formulas covered in this chapter:

  1. Surface area of cube = 6a²
  2. Lateral surface area of cube = 4a²
  3. Total surface area of cuboid = 2(lb + bh + lh)
  4. Lateral surface area of cuboid = 2h(l + b)
  5. Surface area of sphere = 4πr²
  6. Volume of cube = a³
  7. Volume of cuboid = lbh
  8. Volume of cylinder = πr²h
  9. Volume of cone = 1/3πr²h
  10. Volume of sphere = 4/3πr³

CBSE Class 10 Maths Important Questions Chapter 13 Surface Areas and Volumes

Some important questions from this chapter are:

  1. A rectangular tank measures 4m by 3m by 2.5m. Find the volume of water it can hold in liters.
  2. A hemispherical bowl of radius 5.6 cm is filled with water. Find the volume of water. (Take π=22/7)
  3. A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is melted and recast into a cone of height 7 cm. Determine the diameter of the base of the cone.
  4. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of canvas used for making the tent.
  5. A metallic sphere of radius 8 cm is melted and recast into a cone with radius of the base as 12 cm. Find the height of the cone so formed.

These questions cover a variety of concepts related to finding volumes and surface areas of different shapes. It is important to understand the formulas and concepts thoroughly in order to solve problems related to this chapter.

CBSE Class 10 Maths Important Questions Answers Chapter 13 Surface Areas and Volumes

  1. Find the total surface area of a cylinder with base radius 7 cm and height 10 cm. Answer: The formula for the total surface area of the cylinder is 2πrh + 2πr^2, where r is the radius of the base and h is the height of the cylinder.

Substituting the given values, we get Total surface area = 2π(7)(10) + 2π(7)^2 = 440π + 98π = 538π cm^2

  1. The diameter of a roller is 84 cm and its length is 120 cm. Find the cost of painting it on the surface at the rate of Rs 5 per 100 cm^2. (Use π= 22/7) Answer: The total surface area of the roller is the curved surface area of the cylinder plus the area of its two circular bases. We can find the curved surface area by multiplying the circumference of the base (which is the same as the perimeter of a circle with radius equal to the radius of the cylinder) by the height of the cylinder.

Radius of the cylinder = 84/2 = 42 cm Height of the cylinder = 120 cm Circumference of the base = 2πr = 2π(42) = 84π cm Curved surface area = 84π(120) = 10080π cm^2 Area of each circular base = πr^2 = π(42)^2 = 1764π cm^2 Total surface area = curved surface area + 2 × area of circular base = 10080π + 2(1764π) = 12972π cm^2

Now, we can calculate the cost of painting the surface at the given rate: Cost per 100 cm^2 = Rs 5 Cost per 1 cm^2 = Rs 5/100 = 0.05 Cost for painting 12972π cm^2 = 0.05 × 12972π = Rs 648.6 (approx)

Hence, the cost of painting the roller is approximately Rs 648.6.

  1. A solid metallic sphere of radius 6 cm is melted and recast into a cone with base radius 12 cm. Find the height of the cone. Answer: The volume of the sphere is equal to the volume of the cone, since both contain the same amount of material. We can use the formulas for the volume of a sphere and a cone to find the height of the cone.

Volume of sphere = 4/3πr^3 = 4/3π(6)^3 = 904.32π cm^3 Volume of cone = 1/3πr^2h = 1/3π(12)^2h = 144πh cm^3

Equating the two volumes, we get: 904.32π = 144πh h = 904.32π/144π = 6.27 cm (approx)

Hence, the height of the cone is approximately 6.27 cm.

  1. A toy is in the shape of a cone mounted on a hemisphere of same radius 7 cm. If the total height of the toy is 31 cm, find its volume. Answer: The toy is made up of a cone and a hemisphere. We can find the volumes of these two parts separately and then add them to get the total volume.

Radius of the hemisphere = 7 cm Total height of the toy = 31 cm Height of the cone = Total height – Radius of hemisphere = 31 – 7 = 24 cm

CBSE Class 10 Maths Important Questions Answers MCQs Chapter 13 Surface Areas and Volumes

  1. A cylinder has a height of 8 cm and a radius of 2 cm. What is its curved surface area? a) 12π cm^2 b) 16π cm^2 c) 32π cm^2 d) 64π cm^2

Answer: c) 32π cm^2

  1. The diameter of a sphere is 28 cm. What is its surface area? a) 1232 cm^2 b) 2464 cm^2 c) 615 cm^2 d) 307.5 cm^2

Answer: b) 2464 cm^2

  1. A right circular cylinder has a radius of 5 cm and a height of 12 cm. What is its volume? a) 60π cm^3 b) 300π cm^3 c) 600π cm^3 d) 900π cm^3

Answer: b) 300π cm^3

  1. The dimensions of a rectangular box are 6 cm, 8 cm, and 10 cm. What is its total surface area? a) 120 cm^2 b) 240 cm^2 c) 360 cm^2 d) 480 cm^2

Answer: d) 480 cm^2

  1. The base of a pyramid is a square with side length 12 cm, and the height of the pyramid is 9 cm. What is the volume of the pyramid? a) 432 cm^3 b) 648 cm^3 c) 864 cm^3 d) 1296 cm^3

Answer: b) 648 cm^3

CBSE Class 10 Maths Notes Chapter 13 Surface Areas and Volumes

Surface Areas and Volumes Class 10 Notes Maths Chapter 13 1

SURFACE AREA AND VOLUME OF COMBINATIONS

Cone on a Cylinder.
Surface Areas and Volumes Class 10 Notes Maths Chapter 13 2
r = radius of cone & cylinder;
h1 = height of cone
h2 = height of cylinder
Total Surface area = Curved surface area of cone + Curved surface area of cylinder + area of circular base
= πrl + 2πrh2 +πr2;
Slant height, l = r2+h21−−−−−−√
Total Volume = Volume of cone + Volume of cylinder
= 13πr2h1+πr2h2

Cone on a Hemisphere:
Surface Areas and Volumes Class 10 Notes Maths Chapter 13 3
h = height of cone;
l = slant height of cone = r2+h2−−−−−−√
r = radius of cone and hemisphere
Total Surface area = Curved surface area of cone + Curved surface area of hemisphere = πrl + 2πr2
Volume = Volume of cone + Volume of hemisphere = 13πr2h+23πr3

Conical Cavity in a Cylinder
Surface Areas and Volumes Class 10 Notes Maths Chapter 13 4
r = radius of cone and cylinder;
h = height of cylinder and conical cavity;
l = Slant height
Total Surface area = Curved surface area of cylinder + Area of the bottom face of cylinder + Curved surface area of cone = 2πrh + πr2 + πrl
Volume = Volume of cylinder – Volume of cone = πr2h−13πr2h=23πr2h

Cones on Either Side of the Cylinder.
Surface Areas and Volumes Class 10 Notes Maths Chapter 13 5
r = radius of cylinder and cone;
h1 = height of the cylinder
h2 = height of cones
Slant height of cone, l = h22+r2−−−−−−√
Surface area = Curved surface area of 2 cones + Curved surface area of cylinder = 2πrl + 2πrh1
Volume = 2(Volume of cone) + Volume of cylinder = 23πr2h2+πr2h1

Cylinder with Hemispherical Ends.
Surface Areas and Volumes Class 10 Notes Maths Chapter 13 6
r = radius of cylinder and hemispherical ends;
h = height of cylinder
Total surface area= Curved surface area of cylinder + Curved surface area of 2 hemispheres = 2πrh + 4πr2
Volume = Volume of cylinder + Volume of 2 hemispheres = πr2h+43πr3

Hemisphere on Cube or Hemispherical Cavity on Cube
Surface Areas and Volumes Class 10 Notes Maths Chapter 13 7
a = side of cube;
r = radius of hemisphere.
Surface area = Surface area of cube – Area of hemisphere face + Curved surface area of hemisphere
= 6a2 – πr2 + 2πr2 = 6a2 + πr2
Volume = Volume of cube + Volume of hemisphere = a3+43πr3

Hemispherical Cavity in a Cylinder
Surface Areas and Volumes Class 10 Notes Maths Chapter 13 8
r = radius of hemisphere;
h = height of cylinder
Total surface area = Curved surface area of cylinder + Surface area of base + Curved surface area of hemisphere
= 2πrh + πr2 + 2πr2 = 2πrh + 3πr2
Volume = Volume of cylinder – Volume of hemisphere = πr2h−23πr3

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