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Surface areas And Volumes

Class 10 . Mathematics . Mathematics . Surface Areas and Volumes

Chapter 12: Surface areas And Volumes

Ch 12

Mathematics

Class 10

Lecture 1

Introduction & Surface Area Of A Combination Of Solids

Reference video · Lecture 1Introduction & Surface Area Of A Combination Of Solids

Basic Solids and Composite Solids

Basic solids: cuboid, cone, cylinder and sphere.
Basic solids: cuboid, cone, cylinder and sphere.

Surface Area of Joined Solids

A cylinder joined to two hemispheres; only their curved outer surfaces are exposed.
A cylinder joined to two hemispheres; only their curved outer surfaces are exposed.
Surface-area addition rule

For joined solids, total exposed area equals the sum of the exposed areas of each part. For example, a cone joined to a hemisphere along equal circular bases has area \(\pi rl+2\pi r^2\).

Useful Curved Surface Areas
$$\begin{aligned}\text{CSA of cylinder}&=2\pi rh\\\text{CSA of cone}&=\pi rl,\quad l=\sqrt{r^2+h^2}\\\text{CSA of hemisphere}&=2\pi r^2\\\text{Area of circle}&=\pi r^2\end{aligned}$$

Use the curved areas of joined parts, plus any exposed flat faces or rings.

Finding the Exposed Area

  1. Separate the object into its basic solids and mark the dimensions.
  2. Identify the joined faces; exclude these hidden areas.
  3. Calculate each exposed curved surface, flat face or ring.
  4. Add the exposed areas and state the answer in square units.
Worked example

Find the area to colour on a cone-and-hemisphere top with total height \(5\) cm and diameter \(3.5\) cm.

  1. A cone surmounted by a hemisphere forms a top of height \(5\) cm and diameter \(3.5\) cm. Find its exposed area using \(\pi=\frac{22}{7}\).
  2. The common radius is half the diameter.
    $$r=\frac{3.5}{2}=1.75\text{ cm}$$
  3. The cone’s height is the total height minus the hemisphere’s radius.
    $$h=5-1.75=3.25\text{ cm}$$
  4. Find the cone’s slant height.
    $$l=\sqrt{r^2+h^2}=\sqrt{1.75^2+3.25^2}\approx3.69\text{ cm}$$
  5. Add the curved areas only; the joined circular faces are hidden.
    $$\begin{aligned}A&=2\pi r^2+\pi rl\\&=2\left(\frac{22}{7}\right)(1.75)^2+\left(\frac{22}{7}\right)(1.75)(3.69)\\&\approx39.6\text{ cm}^2\end{aligned}$$
Answer: The exposed area is approximately \(39.6\text{ cm}^2\).
⚠ Common mistake

Do not add the total surface areas of the separate solids without adjustment: their joined faces are hidden. Use only exposed surfaces, and remember that a hemisphere’s curved area is \(2\pi r^2\).

Practice

A cube of edge \(5\) cm has a hemisphere of diameter \(4.2\) cm fixed on its top. Find the total exposed surface area using \(\pi=\frac{22}{7}\).

Hint: Subtract the hemisphere’s circular base from the cube’s area, then add the hemisphere’s curved area.
Key step: \(A=6(5)^2-\pi(2.1)^2+2\pi(2.1)^2\).
Solution: \(A=150+\pi(2.1)^2=150+13.86=163.86\text{ cm}^2\).
Practice

A rocket consists of a cone mounted on a cylinder. The cone has height \(6\) cm and base diameter \(5\) cm; the cylinder has diameter \(3\) cm and height \(20\) cm. Find the orange area (cone and exposed ring) and yellow area (cylinder’s curved surface and bottom), using \(\pi=3.14\).

Hint: The cone’s slant height is \(6.5\) cm. Include the exposed annulus below the cone and the cylinder’s bottom face.
Key step: Orange area \(=\pi(2.5)(6.5)+\pi(2.5^2-1.5^2)\); yellow area \(=2\pi(1.5)(20)+\pi(1.5)^2\).
Solution: Orange: \(\pi(20.25)=63.585\text{ cm}^2\). Yellow: \(\pi(62.25)=195.465\text{ cm}^2\).
Practice

A bird-bath is a cylinder of height \(1.45\) m and radius \(30\) cm with a hemispherical depression of the same radius. Find its exposed area using \(\pi=\frac{22}{7}\).

Hint: Convert the cylinder’s height to centimetres. Add the cylinder’s curved area and the depression’s curved area.
Key step: \(A=2\pi rh+2\pi r^2=2\pi r(h+r)\), with \(h=145\) cm and \(r=30\) cm.
Solution: \(A=2\left(\frac{22}{7}\right)(30)(145+30)=33000\text{ cm}^2=3.3\text{ m}^2\).

Quick recap

  • A composite solid is made by joining two or more basic solids.
  • Its surface area is the sum of the areas visible from outside.
  • Exclude joined faces; include exposed flat faces and any uncovered ring-shaped areas.
  • Use \(2\pi rh\) for a cylinder, \(\pi rl\) for a cone, and \(2\pi r^2\) for a hemisphere’s curved surface.
  • For a cone, calculate slant height with \(l=\sqrt{r^2+h^2}\).
  • Keep all dimensions in consistent units before calculating.
Lecture 2

Volume Of A Combination Of Solids

Reference video · Lecture 2Volume Of A Combination Of Solids

Adding and subtracting volumes

Definition

Volume of a combination of solids — The volume of a solid made by joining basic solids is the sum of their volumes; if a part is hollow or removed, subtract its volume.

Volume rule

Joined solids: Add the volumes of the parts. Space removed or occupied: Subtract that volume from the total capacity. Joining surfaces do not reduce volume.

Volumes used in this lecture
$$V_{\text{cuboid}}=lbh,\quad V_{\text{cylinder}}=\pi r^2h,\quad V_{\text{hemisphere}}=\frac{2}{3}\pi r^3,\quad V_{\text{cone}}=\frac{1}{3}\pi r^2h$$

Use consistent units; volume is measured in cubic units.

Finding the volume or capacity

  1. Identify the basic solids and note their dimensions.
  2. Find each part's volume using the appropriate formula.
  3. Add volumes for joined parts; subtract cavities, raised parts, or occupied space when required.
  4. Check that all dimensions use the same unit and state the answer in cubic units.
A shed formed by a cuboid surmounted by a half-cylinder.
A shed formed by a cuboid surmounted by a half-cylinder.
Worked example

Find the shed's total capacity and the air remaining after the machinery and workers are inside.

  1. A shed has a cuboidal part measuring 15 m by 7 m by 8 m and a half-cylinder roof of length 15 m and diameter 7 m. Find its air capacity. Then find the air left after machinery occupies 300 m³ and 20 workers occupy 0.08 m³ each. Use π = 22/7.
  2. Find the cuboid volume and the half-cylinder radius.
    $$V_{\text{cuboid}}=15\times7\times8=840\text{ m}^3,\qquad r=\frac{7}{2}=3.5\text{ m}$$
  3. Add the volume of the half-cylinder roof to the cuboid volume.
    $$V_{\text{roof}}=\frac{1}{2}\times\frac{22}{7}\times(3.5)^2\times15=288.75\text{ m}^3,\qquad V_{\text{shed}}=840+288.75=1128.75\text{ m}^3$$
  4. Subtract the space occupied by the machinery and workers.
    $$V_{\text{workers}}=20\times0.08=1.6\text{ m}^3,\qquad V_{\text{air}}=1128.75-(300+1.6)=827.15\text{ m}^3$$
Answer: The shed holds 1128.75 m³ of air when empty. With the machinery and workers inside, 827.15 m³ of air remains.
A cylindrical glass with a raised hemispherical base that reduces its capacity.
A cylindrical glass with a raised hemispherical base that reduces its capacity.
Practice

A cylindrical glass has inner diameter 5 cm and height 10 cm. A raised hemisphere of radius 2.5 cm is at its base. Find its apparent and actual capacities using π = 3.14.

Hint: Apparent capacity is the full cylinder volume; subtract the raised hemisphere's volume for actual capacity.
Key step: Use r = 2.5 cm for both shapes.
Solution: Apparent capacity = 3.14 × (2.5)² × 10 = 196.25 cm³. Raised hemisphere volume = (2/3) × 3.14 × (2.5)³ ≈ 32.71 cm³. Actual capacity = 196.25 − 32.71 = 163.54 cm³.
A toy made of a hemisphere and a cone, with a cylinder circumscribing it.
A toy made of a hemisphere and a cone, with a cylinder circumscribing it.
Practice

A toy consists of a hemisphere surmounted by a cone. Both have radius 2 cm; the cone height is 2 cm. Find the toy's volume and the difference between the volume of its circumscribing cylinder and the toy. Use π = 3.14.

Hint: The circumscribing cylinder has radius 2 cm and height equal to the hemisphere radius plus the cone height.
Key step: The cylinder height is 2 + 2 = 4 cm; add the hemisphere and cone volumes for the toy.
Solution: Toy volume = (2/3) × 3.14 × 2³ + (1/3) × 3.14 × 2² × 2 = 25.12 cm³. Cylinder volume = 3.14 × 2² × 4 = 50.24 cm³. Difference = 50.24 − 25.12 = 25.12 cm³.
⚠ Common mistake

Do not add every volume automatically: subtract a raised part or hollow space from the container's volume, and subtract machinery or people when finding air actually remaining.

Quick recap

  • Joined solids: add the volumes of their constituent parts.
  • Cavities and raised internal parts reduce a container's actual capacity.
  • Air remaining = total capacity − space occupied.
  • For a hemisphere, use half the volume of a sphere: (2/3)πr³.
  • A circumscribing cylinder's height includes the vertical heights of all parts of the solid.

Formula sheet

Useful Curved Surface Areas
$$\begin{aligned}\text{CSA of cylinder}&=2\pi rh\\\text{CSA of cone}&=\pi rl,\quad l=\sqrt{r^2+h^2}\\\text{CSA of hemisphere}&=2\pi r^2\\\text{Area of circle}&=\pi r^2\end{aligned}$$
Volumes used in this lecture
$$V_{\text{cuboid}}=lbh,\quad V_{\text{cylinder}}=\pi r^2h,\quad V_{\text{hemisphere}}=\frac{2}{3}\pi r^3,\quad V_{\text{cone}}=\frac{1}{3}\pi r^2h$$

Mathematics — ALL CHAPTERS

1

Real Numbers

2

Polynomials

3

Pair Of Linear Equations In Two Variables

4

Quadratic Equations

5

Arithmetic Progressions

6

Triangles

7

Coordinate Geometry

8

Introduction To Trigonometry

9

Some Applications Of Trigonometry

10

Circles

11

Areas Related To Circles

12

Surface areas And Volumes

13

Statistics 

14

Probability