Class 9 Maths Chapter 14: Math of Space: Surface Area and Volume — Notes, Examples & Practice

A cuboid with length $l$, breadth $b$ and height $h$ has 6 rectangular faces in 3 equal pairs. Unfold it into a net and add the areas: $2(lb + bh + hl)$. The four side faces alone (walls of a room) give the lateral surface area $2h(l + b)$.

Its volume counts unit cubes: $l \times b \times h$. A cube is a cuboid with $l = b = h = a$, so its surface area is $6a^2$ and its volume $a^3$.

Cutting a solid into pieces keeps the total volume but increases the total surface area — new faces appear at every cut. That is why crushed ice melts faster than a block.

Cut the curved surface of a cylinder (radius $r$, height $h$) straight down and unroll it: you get a rectangle with length equal to the base circumference $2\pi r$ and width $h$. So the curved surface area is $2\pi rh$.

Add the two circular ends for the total surface area: $2\pi rh + 2\pi r^2 = 2\pi r(h + r)$. An open cylinder (a glass or a tank with no lid) has only one end.

Volume = base area × height $= \pi r^2 h$ — the same idea as a cuboid. A hollow pipe's volume of material is $\pi(R^2 - r^2)h$.

A cone has radius $r$, vertical height $h$ (from the vertex straight down to the centre of the base) and slant height $l$ (from the vertex to the edge of the base). These form a right triangle, so $l^2 = r^2 + h^2$.

Unroll the curved surface and you get a sector of a circle of radius $l$ whose arc is $2\pi r$. Its area is $\pi rl$ — the curved surface area. Add the base for the total: $\pi r(l + r)$.

Fill a cone with water and pour it into a cylinder with the same base and height: it takes exactly three cones to fill it. So $V = \dfrac{1}{3}\pi r^2 h$.

A pyramid has a polygon base and triangular faces meeting at a vertex (apex). The Egyptian pyramids have square bases.

Just as a cone is one-third of a cylinder, a pyramid is one-third of a prism with the same base and height: $V = \dfrac{1}{3} \times \text{base area} \times h$, where $h$ is the perpendicular height.

Surface area has no single formula — add the base and the areas of all the triangular faces. For a square pyramid with base side $a$ and slant height $s$ of each face (apex to the midpoint of a base edge), lateral area $= 4 \times \dfrac{1}{2} a s = 2as$.

A sphere of radius $r$ has surface area $4\pi r^2$ — exactly four times the area of its largest circle. (Archimedes showed it equals the curved surface of the cylinder that just encloses it: $2\pi r \times 2r$.)

Its volume is $\dfrac{4}{3}\pi r^3$ — two-thirds of that enclosing cylinder ($\pi r^2 \times 2r$). Archimedes was so proud of this that he asked for a sphere inside a cylinder on his tomb.

A hemisphere is half a sphere: curved surface $2\pi r^2$, but its total surface also includes the flat circular face: $3\pi r^2$. Volume $\dfrac{2}{3}\pi r^3$.

Melting and recasting: when a solid is melted and reshaped, its volume stays the same (ignoring waste). Set old volume = new volume, or old volume = (number of pieces) × (volume of one piece).

Water displacement: an object sunk in water raises the level by (object's volume) ÷ (base area of the container).

Guesstimates: to estimate, say, the water a school uses in a day, choose a simple model, state assumptions clearly (number of students, litres per person), and calculate. Different assumptions give different but reasonable answers — the reasoning is what counts.