Measuring Space - Perimeter and Area

The chapter has two halves — perimeter first, then area — joined by the constant $\pi$ that shows up in both.

• Sections 6.1–6.2 define perimeter for polygons and then for the circle, introducing $\pi$ as the fixed circumference-to-diameter ratio and tracing its four-thousand-year history of approximation, from Mesopotamia and Archimedes to Mādhava's exact series.
• Section 6.3 proves nothing new but states a landmark fact: $\pi$ is irrational, so $\tfrac{22}{7}$ and $3.14$ are always approximations, never the exact value.
• Section 6.4 derives the arc-length formula $2\pi r \times \tfrac{\theta}{360}$ using rotational symmetry, and applies it to the real geometry of a 400 m athletics track.
• Section 6.5 works two classic perimeter puzzles — overlapping circles, and a "which path is longer" problem — to sharpen algebraic reasoning about lengths.
• Sections 6.6–6.8 build the area formulas for a rectangle, a parallelogram and a triangle, each by cutting one shape into another, and prove the median-bisects-area theorem.
• Section 6.8.1 introduces Heron's formula and, alongside it, Brahmagupta's formula for cyclic quadrilaterals — plus the striking fact that a quadrilateral's sides alone do not determine its area.
• Section 6.9 shows Baudhāyana's ancient construction for "squaring" a rectangle using only geometry.
• Section 6.10 proves $A = \pi r^2$ for a circle by two routes — Archimedes' polygon-limit argument and Nīlakaṇṭha's slicing-into-a-parallelogram picture — and extends it to sectors.

Worked examples below reuse NCERT's own example and figure numbers wherever the source text names them directly, so you can cross-check against your textbook.

Perimeter of basic shapes

The perimeter of any shape is the total distance around its border — imagine walking the boundary once, without doubling back. For straight-sided shapes this is pure addition, and the useful insight is that the ratio of perimeter to a side is fixed for every shape of a given kind, whatever its size.

where: $a$ is the side of the square or equilateral triangle, and $a, b$ are the length and width of the rectangle. Doubling a square's side doubles its perimeter, so the ratio $P:a$ stays fixed at $4:1$ for every square, and at $3:1$ for every equilateral triangle. This scale-invariant ratio is exactly the idea that defines $\pi$ next — a circle has its own fixed ratio too, once we compare against the right "side": the diameter.

Pi and the circumference of a circle

Ancient measurers noticed something remarkable: if you wrap a thread around any circular object and divide its length by the object's diameter, you always land close to the same number, no matter whether the circle is a coin or a cartwheel. This constant ratio of circumference to diameter is called $\pi$.

where: $C$ is the circumference, $D$ the diameter, $r$ the radius, and $D = 2r$. Pinning down $\pi$'s value spans every mathematical civilisation. Mesopotamia (c. 1900 BCE) compared a circle to an inscribed hexagon and used $3 + \tfrac{1}{8} = 3.125$. Archimedes (c. 250 BCE) trapped $\pi$ rigorously between inscribed and circumscribed 96-sided polygons, showing $3\tfrac{10}{71} < \pi < 3\tfrac{1}{7}$. Zu Chongzhi (480 CE, China) pushed the polygon method to 24,576 sides and found $\tfrac{355}{113} \approx 3.1415929$, the best rational approximation known anywhere for the next 800 years. Āryabhaṭa (499 CE) gave $\tfrac{62832}{20000} = 3.1416$ and called it āsanna — "approaching," hinting no single fraction could ever be exact. Mādhava of Sangamagrāma then found the first exact formula, an infinite sum: $\tfrac{\pi}{4} = 1 - \tfrac{1}{3} + \tfrac{1}{5} - \tfrac{1}{7} + \cdots$. This chapter mostly uses $\pi \approx \tfrac{22}{7}$ or $\pi \approx 3.14$, always remembering $\pi \neq \tfrac{22}{7}$ — the fraction only approximates it.

Length of an arc of a circle

An arc is a piece of a circle's circumference. Reflecting a circle in a diameter swaps its two semicircles, so they must have equal length; rotating the whole circle by a quarter-turn swaps the four quadrants, so all four quarter-arcs are equal too. This symmetry argument, generalised to any angle, is the key to the arc-length formula (proved in full below).

where: $r$ is the radius and $\theta\degree$ is the angle the arc subtends at the centre. Setting $\theta = 180$ recovers the semicircle's length $\pi r$; setting $\theta = 90$ recovers the quarter circle's length $\tfrac{\pi r}{2}$. NCERT's own case is a 400 m track: two straights of $84.39$ m plus two semicircular bends of radius $36.5$ m. A runner $0.3$ m from the inner border covers two semicircles of radius $36.8$ m — one full circle, circumference $2\pi(36.8) \approx 231.22$ m — giving $168.78 + 231.22 = 400$ m; a runner one lane out traces a larger circle, so she needs a head start, the stagger, to run a fair $400$ m too.

Area of a rectangle, a parallelogram and a triangle

Area measures the flat space a shape covers, using a $1 \times 1$ square as the unit. A rectangle with sides $a$ and $b$ is tiled exactly by $a \times b$ unit squares, giving $A = ab$. A parallelogram becomes a rectangle of the same base and height by slicing off a triangle from one end and reattaching it at the other — the areas match because the triangle simply moved. A triangle is exactly half of a parallelogram: two congruent copies, rotated and fitted together, tile a parallelogram with the same base and height.

where: $b$ is the base and $h$ is the perpendicular height from the base to the opposite side or vertex — never a slanted side. A direct consequence is the median theorem: a median joins a vertex to the midpoint of the opposite side, so it splits the triangle into two triangles with equal bases and the same height. By $A = \tfrac{1}{2}bh$, equal base and equal height force equal area — even though the two triangles are usually differently shaped. Proved in full below.

Heron's formula and two circle-radius formulas

Half base times height needs a height, which is not always convenient to measure. Heron of Alexandria found a formula that needs only the three side lengths.

where: $a, b, c$ are the three sides and $s$ is the semi-perimeter, half the perimeter. Tested on the $3$–$4$–$5$ right triangle, $s=6$ gives $\text{Area} = \sqrt{6 \times 3 \times 2 \times 1} = 6$ sq. units, matching $\tfrac{1}{2}(3)(4)=6$; for an equilateral triangle of side $a$ it reduces to the familiar $\tfrac{\sqrt3}{4}a^2$. Two more formulas link a triangle's area to circles built from it: the circumcircle (through all three vertices, radius $R$) and the incircle (touching all three sides, radius $r$).

where: $R$ is the circumradius and $r$ is the inradius; both formulas are proved fully only in Class 10, once the needed circle theorems are available.

Brahmagupta's formula and why four sides are not enough

A triangle's area is completely fixed by its three sides. A general four-sided figure (a "4-gon") is different: four rigid rods hinged at their ends, all the same length, can be flexed into a square or squashed into a thin rhombus, and the enclosed area shrinks as the shape flattens, even though every side length stays the same. So a quadrilateral's sides alone never fix its area — extra information is needed, such as a diagonal, an angle, or a special property. One such property is being cyclic: all four vertices lie on a single circle. In 628 CE, Brahmagupta found the beautiful generalisation of Heron's formula for exactly this case.

where: $a, b, c, d$ are the four sides in order and $s$ is the semi-perimeter — but the formula applies only when the quadrilateral is cyclic. Every rectangle is cyclic, and substituting $a,b,a,b$ reduces the formula to $\sqrt{ab \cdot ab} = ab$, confirming the known rectangle area. Shrinking one side to zero ($d \to 0$) collapses the quadrilateral into a triangle, and the formula collapses smoothly to $\sqrt{s(s-a)(s-b)(s-c)}$ — exactly Heron's formula, which is therefore a special case of Brahmagupta's, just as a square is a special case of a rectangle.

Area of a circle and of a sector

Just as a square's ratio $P^2 : A$ is fixed at $16:1$ and an equilateral triangle's is fixed at $36 : 3\sqrt3$, the ratio $C^2 : A$ for circles is a fixed constant too — and Archimedes proved that constant is exactly $\pi$, giving the area formula.

where: $r$ is the radius. Archimedes' argument treats the circle as the limit of regular polygons: any regular polygon's area equals half its perimeter times the radius of its inscribed circle (the apothem). As the number of sides grows without bound, the perimeter approaches the circumference $2\pi r$ and the apothem approaches $r$, so the area approaches $\tfrac{1}{2}(2\pi r)(r) = \pi r^2$ (proved step by step below). Nīlakaṇṭha Somayājī (c. 1500 CE) gave an equally convincing visual argument: slice the disc into thin pie wedges and lay them out alternately up and down; as the wedges thin, the zigzag top straightens into a parallelogram of base $\pi r$ (half the circumference) and height $r$, so area $= \pi r \times r$. The same reasoning, applied to a slice instead of the whole, gives the sector area: a region bounded by two radii and an arc subtending $\theta\degree$ has area $\pi r^2 \times \tfrac{\theta}{360}$, mirroring the arc-length formula. A segment is instead bounded by an arc and the straight chord joining its endpoints.

Board level - area of a triangle by Heron's formula (reusing NCERT Example 5)

Find the area of a triangle whose sides are $3$ units, $4$ units and $5$ units, using Heron's formula, and check the answer against the half-base-times-height formula.

Step 1. Compute the semi-perimeter: $s = \tfrac{1}{2}(3+4+5) = \tfrac{1}{2}(12) = 6$ units.

Step 2. Apply Heron's formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{6(6-3)(6-4)(6-5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6$ sq. units.

Step 3. Cross-check using the half-base-times-height formula. Since $3^2 + 4^2 = 9 + 16 = 25 = 5^2$, the converse of the Baudhāyana–Pythagoras theorem tells us the triangle is right-angled, with the right angle between the sides of length $3$ and $4$. So base $= 3$, height $= 4$, and $\text{Area} = \tfrac{1}{2}(3)(4) = 6$ sq. units — the same answer. The area of the $3$–$4$–$5$ triangle is $\mathbf{6}$ sq. units.

Board level - perimeter of a sector of a circle (reusing NCERT Exercise 6.1)

Find the perimeter of a sector of a circle of radius $14$ cm whose sector angle is $75\degree$. Use $\pi \approx \tfrac{22}{7}$.

Step 1. A sector's perimeter is the two straight radii plus the curved arc between them, not the arc alone.

Step 2. Arc length $= 2\pi r \times \tfrac{\theta}{360} = 2 \times \tfrac{22}{7} \times 14 \times \tfrac{75}{360}$. First $2 \times \tfrac{22}{7} \times 14 = 88$ cm (full circumference); then $88 \times \tfrac{75}{360} = 88 \times \tfrac{5}{24} = \tfrac{55}{3} \approx 18.33$ cm.

Step 3. Add the two radii: perimeter $= 14 + 14 + 18.33 = 46.33$ cm. The perimeter of the sector is approximately $\mathbf{46.33}$ cm (exactly $\tfrac{139}{3}$ cm).

Higher-order thinking (HOTS / olympiad-style) - perimeter of two overlapping circles (reusing NCERT Example 1)

Two circles of equal radius $r$ are placed so that each passes through the centre of the other. They intersect at points $C$ and $D$. Find the perimeter of the lens-and-outer shape formed — that is, the total length of the two outer (major) arcs, ignoring the two arcs that lie inside the overlap.

Step 1. Let the circles be centred at $A$ and $B$, each radius $r$, with $AB = r$ since each passes through the other's centre. Intersection point $C$ lies on both circles, so $AC = BC = r$ too — $\triangle ABC$ is equilateral, giving $\angle CAB = \angle CBA = 60\degree$.

Step 2. By the same reasoning at the other intersection point $D$, $\triangle ABD$ is also equilateral, so $\angle BAD = 60\degree$. Adding the two angles at $A$: $\angle CAD = 60\degree + 60\degree = 120\degree$; by symmetry $\angle CBD = 120\degree$ too.

Step 3. The $120\degree$ angle at $A$ subtends the minor arc $CD$ (inside the overlapping lens, excluded by the problem). The major arc — the outer boundary we want — subtends the remaining $360\degree - 120\degree = 240\degree$, which is $\tfrac{2}{3}$ of the full circle.

Step 4. Each major arc has length $\tfrac{2}{3} \times 2\pi r = \tfrac{4\pi r}{3}$; with two such arcs, one per circle, the total perimeter is $2 \times \tfrac{4\pi r}{3} = \dfrac{8\pi r}{3}$. The perimeter of the shape formed by the two circles is $\mathbf{\dfrac{8\pi r}{3}}$ units. Timing note: allow four to five minutes. The problem hinges on spotting the two equilateral triangles — once $\angle CAD = 120\degree$ is found, the rest is bookkeeping. The classic trap is computing the excluded minor arc instead of the major, outer arc that forms the visible boundary — re-read which portion is wanted before reaching for the arc-length formula.

• Writing $\pi = \tfrac{22}{7}$ instead of $\pi \approx \tfrac{22}{7}$. The fraction only approximates an irrational number; treating it as exact silently introduces error downstream.
• Confusing diameter and radius inside a formula. $C = \pi D = 2\pi r$ — plugging a diameter into $2\pi r$ (or vice versa) doubles or halves every answer.
• Forgetting to convert the angle to a fraction of $360\degree$ before using the arc-length or sector-area formulas.
• Assuming a quadrilateral's area is fixed once its four sides are known. Side lengths alone leave it free to flex into many different areas; Brahmagupta's formula needs the extra condition that it is cyclic.
• Forgetting a sector's perimeter includes the two straight radii, not just the curved arc. A segment's perimeter, by contrast, is arc plus one chord.
• Using a slant side as the "height" in $A = \tfrac12 bh$ or $A = bh$. The height must be perpendicular to the base, never a slanted length.
• Sign errors inside Heron's or Brahmagupta's formula. Every bracket like $(s-a)$ must come out positive for a real shape; a negative value signals an arithmetic slip, not a valid shape.

• State and apply the perimeter formulas for a square, an equilateral triangle, a rectangle and a circle.
• Explain, in your own words, why the ratio of a shape's perimeter to its side (or of a circle's circumference to its diameter) stays fixed regardless of the shape's size.
• Recall that $\pi$ is irrational, name at least two historical approximations to it, and use $\pi \approx \tfrac{22}{7}$ or $3.14$ correctly with the sign $\approx$, never $=$.
• Derive and apply the arc-length formula $2\pi r \times \tfrac{\theta}{360}$ and the sector-area formula $\pi r^2 \times \tfrac{\theta}{360}$ for a given central angle.
• Build the area formulas for a rectangle, a parallelogram and a triangle from cutting-and-rearranging arguments, and state and prove the median-bisects-area theorem.
• Apply Heron's formula to find a triangle's area from its three sides, including sanity-checking it against $\tfrac12 bh$ for a right triangle.
• State the condition under which Brahmagupta's formula applies (a cyclic quadrilateral), and explain why side lengths alone never fix a general quadrilateral's area.
• Derive $A = \pi r^2$ for a circle using the regular-polygon limiting argument or the wedge-to-parallelogram picture.
• Distinguish a sector's perimeter (arc plus two radii) from a segment's perimeter (arc plus one chord), and compute both.

Ask the tutor: if a quadrilateral is not cyclic, what extra piece of information would let you pin down its exact area?