Class 9 › Mathematics › Chapter
Measuring Space: Perimeter and Area
Perimeter and circumference, the number $\pi$, arc length, areas of parallelograms, triangles and circles, Heron's and Brahmagupta's formulas, sectors and segments.
1. Perimeter of a Shape
Perimeter: the total length of the boundary of a shape. It is measured in units of length (cm, m).
Imagine an insect walking once around the edge and returning to its start. The distance it walks is the perimeter.
| Shape | Perimeter |
|---|---|
| Square (side $a$) | $4a$ |
| Equilateral triangle (side $a$) | $3a$ |
| Rectangle (length $a$, width $b$) | $2(a + b)$ |
- Special case: a square is a rectangle with $a = b$, so $2(a+a) = 4a$ is the rectangle formula with $b = a$.
- Perimeter grows in the same proportion as the side: double the side and the perimeter doubles. So the ratio perimeter : side is fixed ($4:1$ for squares, $3:1$ for equilateral triangles). The question for a circle is whether such a fixed ratio exists too.
2. Circumference of a Circle and the Number $\pi$
Circumference $C$: the perimeter of a circle. The ratio $\dfrac{C}{D}$ (circumference to diameter) is the same for every circle, big or small. This constant is called $\pi$.
A bigger circle is just a scaled-up copy of a smaller one, so its circumference and diameter grow in the same proportion, and their ratio never changes.
Circumference
$$C = \pi d = 2\pi r$$
- Revolutions: in one turn a wheel covers a distance equal to its circumference, so number of turns $= \dfrac{\text{distance travelled}}{2\pi r}$.
- Finding $r$ from $C$: $r = \dfrac{C}{2\pi}$. Use $\pi \approx \dfrac{22}{7}$ when the radius is a multiple of $7$, otherwise $3.14$ (follow the question).
- Perimeters of two circles are in the same ratio as their radii (for example, ratio $5:4$ means radii are also $5:4$).
Facts about $\pi$
Home measurement: wrap a thin thread $20$ times round a reel, measure its length $L$ and the diameter $D$; then $\dfrac{L}{20D} \approx \pi$ (between $3.1$ and $3.2$).
$\pi$ is irrational (shown by Lambert, 1761): its decimal never ends and never repeats, and it is not a ratio of two integers. So there is no "best fraction": $\pi \approx \dfrac{22}{7}$ but $\pi \neq \dfrac{22}{7}$. A far better fraction is $\dfrac{355}{113}$.
The symbol $\pi$ comes from the Greek perimetros (perimeter); it was introduced by William Jones (1706) and popularised by Euler.
| Idea used | Who | Result |
|---|---|---|
| Hexagon inside the circle is shorter than the circle | Mesopotamia | $\pi \gt 3$ (used $3.125$) |
| Trap $\pi$ between inscribed and circumscribed polygons | Archimedes (up to 96 sides) | $3 \lt \pi \lt 2\sqrt{3} \approx 3.46$ with hexagons; later $3\tfrac{10}{71} \lt \pi \lt 3\tfrac17$ |
| Polygons with many more sides | Zu Chongzhi (24,576 sides) | $\dfrac{22}{7}$ and $\dfrac{355}{113} \approx 3.1415929$ |
| Approximate values | Ptolemy; Āryabhaṭa; Brahmagupta | $\dfrac{377}{120}$; $\dfrac{62832}{20000} = 3.1416$ (called āsanna, "approximate"); $\sqrt{10}$ |
| Infinite series (an exact formula) | Mādhava | $\dfrac{\pi}{4} = 1 - \dfrac13 + \dfrac15 - \dfrac17 + \cdots$ |
Early methods squeezed $\pi$ between perimeters of polygons. Mādhava changed the game: $\pi$ is the value an endless sum gets closer and closer to (a "limit"), which led to ideas of calculus.
3. Length of an Arc of a Circle
An arc that makes an angle $\theta^\circ$ at the centre is the fraction $\dfrac{\theta}{360}$ of the full circle (the full turn is $360^\circ$). So its length is that same fraction of the circumference.
Arc length (central angle $\theta^\circ$)
$$l = 2\pi r \times \frac{\theta}{360}$$
- Semicircle ($180^\circ$): $\pi r$. Quarter circle ($90^\circ$): $\dfrac{\pi r}{2}$. (Reflection or rotation symmetry shows equal parts have equal length.)
- Perimeter of a sector = arc length $+ 2r$ (the curved part plus the two straight radii).
- Perimeter of a semicircular region = $\pi r + 2r$ (arc plus the diameter).
- Composite shapes: add only the pieces that form the outer boundary (arcs and straight edges); skip any dotted line inside the shape.
Arc, sector and segment of a circle with centre $O$.
4. Running Track and Lane Stagger
- Track shape: two straight sections plus two semicircular bends with the same centre. Together the two bends make one full circle, so one lap $= 2 \times \text{straight} + 2\pi r$, where $r$ is the radius of the lane's path.
- Why a stagger is needed: on the straights all runners cover the same distance, but on a bend an outer lane has a larger radius, so its path is longer. Outer-lane runners therefore start ahead to make the total distance equal.
Stagger between neighbouring lanes
$$\text{extra length per bend} = \pi\,\Delta r, \qquad \text{for a full lap (two bends)} = 2\pi w$$
- $\Delta r = w$, the lane width (difference in radius between adjacent lanes). A semicircle has length $\pi r$, so a bigger radius by $w$ adds $\pi w$ per bend.
- The stagger depends on the lane width and the number of bends run, not on the size of the track. It is the same between every pair of adjacent lanes.
- So a smaller (for example $200$ m) track with the same lane width needs the same stagger per bend, not a smaller one.
5. Area: Unit, Rectangle and Square
Area: the amount of space a shape covers. The unit of area is a $1 \times 1$ square, so area is measured in square units (sq. cm, $\text{m}^2$).
Rectangle and square
$$\text{Area of rectangle} = a \times b, \qquad \text{Area of square} = a^2$$
A rectangle with sides $a$ and $b$ is covered by $b$ rows of $a$ unit squares, so it holds $ab$ squares.
6. Area of a Parallelogram
Parallelogram
$$\text{Area} = \text{base} \times \text{height} = bh$$
- The height is the perpendicular distance between the two parallel sides, not the slanted side.
Moving one triangle turns the parallelogram into a rectangle with the same base and height.
Cut a right-angled triangle off one slanted end and slide it to the other end: you get a rectangle of the same base $b$ and height $h$, so their areas are equal.
- "Thin" parallelograms (foot of the perpendicular falls outside the base): shift small pieces repeatedly (move $\triangle CDD'$ to $\triangle BAA'$) until the rectangle forms; the formula $bh$ still holds.
- Sides alone do not fix the area: keeping the side lengths fixed but changing the angle between the sides changes the height and therefore the area. (A rectangle's area is fixed by its sides.)
- Parallelograms on the same base and between the same parallel lines have equal area.
7. Area of a Triangle
Triangle
$$\text{Area} = \frac12 \times \text{base} \times \text{height} = \tfrac12 bh$$
Two identical copies of a triangle fit together into a parallelogram (rotate one copy by a half-turn). The parallelogram has area $bh$, so one triangle is half of it. (Alternatively, a triangle is half of the rectangle drawn around it.)
- Any side can be the base; the height is the perpendicular from the opposite vertex (it falls outside the triangle for an obtuse triangle).
- Same base and same height $\Rightarrow$ equal areas (even if the triangles look different).
- Same height $\Rightarrow$ areas are in the ratio of the bases; same base $\Rightarrow$ ratio of heights.
Theorem: a median divides a triangle into two triangles of equal area
- Why: a median $AD$ makes $BD = DC$ and both triangles $ABD$, $ACD$ have the same height from $A$. So each has area $\tfrac12 \times BD \times h$.
- Surprising, because the two triangles are usually not congruent: equal area does not mean the same shape.
- This also gives: for any point $P$ on the median $AD$, $\text{area}(\triangle ABP) = \text{area}(\triangle ACP)$.
- Cutting and rearranging: two shapes of equal area can often be cut into pieces and rearranged to cover each other (for example the two halves made by a median).
- Rectangles with a fixed perimeter: infinitely many exist; the square has the largest area, and the area can be made as small as we like (but not zero) by making the rectangle very thin.
8. Trapezium, Rhombus and Kite
| Shape | Area | Why (in simple words) |
|---|---|---|
| Trapezium (parallel sides $a, b$, height $h$) | $\tfrac12 (a+b)\,h$ | Two copies (one turned upside down) make a parallelogram with base $a + b$ and height $h$; take half. Or split into two triangles with bases $a$ and $b$ and the same height $h$. |
| Rhombus or kite (diagonals $d_1, d_2$) | $\tfrac12 d_1 d_2$ | The diagonals are perpendicular; one diagonal splits the figure into two triangles whose heights add up to the other diagonal. |
9. Heron's Formula
When we know the three sides of a triangle but not its height, Heron's formula gives the area directly from the sides.
Heron's formula (sides $a, b, c$)
$$s = \frac{a+b+c}{2}, \qquad \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$
- $s$ is the semi-perimeter (half the perimeter).
- Steps: find $s$, find the three differences $s-a$, $s-b$, $s-c$, multiply all four numbers, take the square root.
- Check each side is less than $s$ (a triangle must exist).
Checking it against $\tfrac12 bh$ (special cases)
Equilateral (side $a$): $s = \tfrac{3a}{2}$, area $= \dfrac{\sqrt{3}}{4}a^2$. (Height $h = \tfrac{\sqrt3}{2}a$ from Pythagoras gives the same.)
Isosceles (equal sides $a$, base $2b$): $s = a + b$, area $= b\sqrt{a^2 - b^2}$ (height $\sqrt{a^2 - b^2}$).
Right-angled: if $a^2 + b^2 = c^2$ the area is also $\tfrac12 \times$ (product of the two shorter sides).
Two more area formulas (circles of the triangle)
$$\text{Area} = \frac{abc}{4R} \qquad\qquad \text{Area} = r \cdot s = \frac{r(a+b+c)}{2}$$
- $R$ = radius of the circumcircle (the circle through the three vertices); $r$ = radius of the incircle (the circle that touches all three sides from inside).
- Proofs of these (and of Heron's formula) come in Grade 10.
10. Brahmagupta's Formula for a Cyclic Quadrilateral
For a triangle, the three sides fix the area. For a 4-sided figure the four sides alone do not: a four-rod frame can be pushed into different shapes with different areas (a rhombus of side $3$ can have area $9$, $8.01$ or $5.41$). So we need extra information, such as the shape being cyclic (all four vertices on a circle).
Brahmagupta's formula (cyclic 4-gon with sides $a, b, c, d$)
$$s = \frac{a+b+c+d}{2}, \qquad \text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}$$
- Check with a rectangle (always cyclic, sides $a, b$): $s = a + b$, so area $= \sqrt{b \cdot a \cdot b \cdot a} = ab$. Isosceles trapeziums are also cyclic and the formula agrees with $\tfrac12(a+b)h$.
- Heron's formula is a special case: take a triangle as a 4-gon whose fourth side is $d = 0$ (two vertices coincide). A triangle is always cyclic, so the formula applies and becomes $\sqrt{s(s-a)(s-b)(s-c)}$.
Special case and generalisation: a special case comes from a general result by adding an extra condition (a square is a rectangle with $a = b$; $(a+b)^2$ is $(a+b+c)^2$ with $c = 0$). Going the other way, from a special case to a more general result, is called generalisation. Brahmagupta's formula generalises Heron's.
11. Squaring a Rectangle (Baudhāyana's Construction)
To "square" a shape: construct a square whose area equals the area of the shape. For a rectangle with sides $a \gt b$, the target square has area $ab$ (side $\sqrt{ab}$).
Construction (rectangle $ABCD$, $AD = a$, $AB = b$)
1. Mark $E$ on $AD$ with $AE = b$ (a square of side $b$ is cut off).
2. Let $F$ be the midpoint of $ED$, so $AF = \dfrac{a+b}{2}$.
3. Draw square $AFGH$ with $H$ on $AB$ extended.
4. With centre $H$ and radius $HG$ (so $HK = \tfrac{a+b}{2}$), draw an arc cutting $BC$ at $K$.
5. Draw $KP \parallel AH$ meeting $GH$ at $P$. Then the square on $HP$ has the same area as the rectangle.
Why it works: in right triangle $HKP$, $HK = \tfrac{a+b}{2}$ and $KP = BH = \tfrac{a-b}{2}$. By the Baudhāyana–Pythagoras theorem, $HP^2 = \left(\tfrac{a+b}{2}\right)^2 - \left(\tfrac{a-b}{2}\right)^2 = ab$. It is the identity $\left(\tfrac{a+b}{2}\right)^2 - \left(\tfrac{a-b}{2}\right)^2 = ab$ drawn as a picture.
- Squaring a triangle: first turn it into a rectangle of the same area (base $\times$ half the height), then square that rectangle.
12. Area of a Circle
Area of a circle
$$A = \pi r^2 \;=\; \tfrac12 \times (\text{circumference}) \times r$$
Slices of a circle rearranged to look like a parallelogram (Nīlakaṇṭha's picture).
Three ways to see $A = \pi r^2$
1. Proportion idea: for any shape, $\dfrac{\text{perimeter}^2}{\text{area}}$ does not change when the shape is scaled (it is $16:1$ for every square and $36:\sqrt3$ for every equilateral triangle). For circles $\dfrac{C^2}{A}$ is therefore a fixed constant; it equals $\dfrac{(2\pi r)^2}{\pi r^2} = 4\pi \approx 12.6$. (The Babylonians used about $12$, giving $A \approx C^2/12$; the Egyptians used $A \approx \left(\tfrac{8d}{9}\right)^2 = \tfrac{256}{81}r^2$.)
2. Archimedes: a regular polygon has area $\tfrac12 \times$ perimeter $\times$ (radius of its inscribed circle). With more and more sides the polygon becomes the circle: area $= \tfrac12 \times 2\pi r \times r = \pi r^2$. In words: area of circle = area of a right triangle with legs $r$ and $C$.
3. Slicing (Nīlakaṇṭha): cut the circle into many thin sectors and place them alternately up and down. The shape tends to a parallelogram with base half the circumference ($\pi r$) and height $r$, so area $= \pi r \times r = \pi r^2$.
- Area of a semicircle $= \tfrac12 \pi r^2$, of a quarter circle $= \tfrac14 \pi r^2$.
- Ring between two concentric circles (radii $R \gt r$): $\pi(R^2 - r^2)$. If a chord of the larger circle of length $l$ touches the smaller circle, then $(l/2)^2 = R^2 - r^2$, so the ring has area $\dfrac{\pi l^2}{4}$.
- Doubling the radius multiplies the area by $4$.
13. Sector and Segment of a Circle
Sector: the region bounded by an arc and the two radii to its end points.
Segment: the region bounded by an arc and the chord joining its end points.
Segment: the region bounded by an arc and the chord joining its end points.
A sector with central angle $\theta^\circ$ is the fraction $\dfrac{\theta}{360}$ of the whole circle, so its area is that fraction of $\pi r^2$ (the same idea as arc length).
Area of a sector
$$\text{Area} = \pi r^2 \times \frac{\theta}{360} \;=\; \tfrac12 \times l \times r \quad (l = \text{arc length})$$
- Minor segment = area of sector $-$ area of triangle $OAB$ (the triangle formed by the two radii and the chord). Major segment = area of circle $-$ area of minor segment.
- For $\theta = 60^\circ$, triangle $OAB$ is equilateral, so minor segment $= \dfrac{\pi r^2}{6} - \dfrac{\sqrt3}{4}r^2 = r^2\left(\dfrac{\pi}{6} - \dfrac{\sqrt3}{4}\right)$.
- Area swept by a rotating hand or wiper of length $r$ through $\theta^\circ$ is a sector of radius $r$ and angle $\theta$ (for example a minute hand moves $6^\circ$ per minute).
| Regular shape inscribed in a circle of radius $r$ | Area of shape | Ratio to circle |
|---|---|---|
| Equilateral triangle | $\dfrac{3\sqrt3}{4}r^2$ | $\dfrac{3\sqrt3}{4\pi} \approx 0.413$ |
| Square | $2r^2$ | $\dfrac{2}{\pi} \approx 0.637$ |
| Hexagon | $\dfrac{3\sqrt3}{2}r^2$ | $\dfrac{3\sqrt3}{2\pi} \approx 0.827$ |
The hexagon is made of $6$ equilateral triangles of side $r$, and the triangle is half of that hexagon, which is why the hexagon's ratio is exactly twice the triangle's.
14. How Perimeter and Area Change with Size
Scaling a shape by a factor $k$
$$\text{lengths and perimeter} \times k, \qquad \text{area} \times k^2$$
Both dimensions (length and width, or base and height) get multiplied by $k$, so area, which uses two lengths, gets multiplied by $k\times k$. Doubling every side makes the area $4$ times; tripling makes it $9$ times (and $4$ or $9$ copies of the original fit exactly in these cases).
- Equal perimeter and equal area for two rectangles forces equal $a + b$ and $ab$, which fix the sides, so such rectangles are congruent.
- Algebra from areas: identities such as $(a+b)^2 = a^2 + 2ab + b^2$ can be drawn as area pictures (a square of side $a+b$ split into $a^2$, $b^2$ and two $ab$ rectangles).
- Semicircles on a diameter: a path of one big semicircle equals in length a path of several smaller semicircles along the same diameter, because the small diameters add up to the big diameter (so the radii add up, and the lengths are $\pi\times$ radius).
15. Formula Sheet and Common Mistakes
| Quantity | Formula |
|---|---|
| Rectangle / Square | Perimeter $2(a+b)$ / $4a$; area $ab$ / $a^2$ |
| Parallelogram | Area $bh$ |
| Triangle | Area $\tfrac12 bh$; Heron: $\sqrt{s(s-a)(s-b)(s-c)}$; also $\dfrac{abc}{4R}$ and $rs$ |
| Equilateral triangle | Perimeter $3a$; area $\dfrac{\sqrt3}{4}a^2$ |
| Trapezium | $\tfrac12(a+b)h$ |
| Rhombus / kite | $\tfrac12 d_1 d_2$ |
| Cyclic 4-gon | $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ |
| Circle | $C = 2\pi r$; $A = \pi r^2$ |
| Arc length | $2\pi r\,\dfrac{\theta}{360}$ |
| Sector | Area $\pi r^2 \dfrac{\theta}{360}$; perimeter $=$ arc $+ 2r$ |
| Segment (minor) | Sector $-$ triangle |
| Lane stagger | $\pi w$ per bend; $2\pi w$ per lap |
Common mistakes:
• Mixing up perimeter (length, units) and area (square units).
• Using the slanted side instead of the perpendicular height.
• Forgetting to halve in $\tfrac12 bh$, or using the full perimeter instead of $s$ in Heron's formula.
• Taking the radius for the diameter (or the other way round) in circle formulas.
• Forgetting the two radii in the perimeter of a sector (or the diameter in a semicircular region).
• Using $\theta$ in radians or forgetting to divide by $360$.
• Writing $\pi = \tfrac{22}{7}$ as exact; it is only an approximation.
• Applying Brahmagupta's formula to a 4-gon that is not cyclic.
• Mixing up perimeter (length, units) and area (square units).
• Using the slanted side instead of the perpendicular height.
• Forgetting to halve in $\tfrac12 bh$, or using the full perimeter instead of $s$ in Heron's formula.
• Taking the radius for the diameter (or the other way round) in circle formulas.
• Forgetting the two radii in the perimeter of a sector (or the diameter in a semicircular region).
• Using $\theta$ in radians or forgetting to divide by $360$.
• Writing $\pi = \tfrac{22}{7}$ as exact; it is only an approximation.
• Applying Brahmagupta's formula to a 4-gon that is not cyclic.