Measuring Space: Perimeter and Area

Contents hide

1Perimeter and Circumference

2Arc Length

3Area of Rectangles, Parallelograms and Triangles

4Heron’s Formula

5Brahmagupta’s Formula for a Cyclic Quadrilateral

6Area of a Circle and Its Sectors

Chapter mind map: how it all connects

1 · Perimeter and circumferenceThe distance around a shape; for circles, this ratio to the diameter is always π

2 · Arc lengthA fraction of the full circumference, set by the angle the arc subtends at the centre

Perimeter and Area

3 · Area of triangles and 4-gonsBase × height, Heron’s formula, and Brahmagupta’s formula for a cyclic quadrilateral

4 · Area of a circle and its sectorsπr², and the same θ/360 fraction idea applied to area instead of length

What you will learn in this chapter

How to find the perimeter and circumference of shapes, including arcs and sectors
How to find the area of a triangle in more than one way, including Heron’s formula
Why a cyclic quadrilateral’s area can be found from its four sides using Brahmagupta’s formula
How the area of a circle, πr², and the area of a sector both follow from the circle’s symmetry

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1Perimeter and Circumference

Learn by heartDefinition 1

The perimeter of a shape is the total distance around its boundary. Square: $4a$ . Equilateral triangle: $3a$ . Rectangle: $2(a+b)$ .

For any square, perimeter : side is always 4 : 1. For any equilateral triangle, it is always 3 : 1. For a circle, this fixed ratio of circumference to diameter is called π (pi), and it is the same for circles of every size.

c. 1900 BCE, MesopotamiaEstimated π ≈ 3.125 by comparing a circle to an inscribed hexagon

250 BCE, ArchimedesTrapped π between inscribed and circumscribed polygons:

3\frac{10}{71} < \pi < 3\frac{1}{7}

480 CE, Zu ChongzhiFound

\frac{355}{113} \approx 3.1415929

, unmatched in accuracy for 800 years

499 CE, ĀryabhaṭaGave π ≈ 3.1416, calling it asanna (“approaching”), hinting it has no exact fraction

14th century, MādhavaDiscovered the first exact infinite series:

\frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots

Common Mistake

π is irrational (proved by Lambert, 1761), so $\pi \ne \dfrac{22}{7}$ exactly. It is only a convenient approximation. Write $\pi \approx \dfrac{22}{7}$ , never $\pi = \dfrac{22}{7}$ .

Formula

Circumference

= \pi d = 2\pi r

2Arc Length

Figure 1: Sector OAB. Arc AB subtends angle θ° at the centre O. — **Figure 1**: Sector OAB. Arc AB subtends angle θ° at the centre O.

A semicircle (θ = 180°) is half the circumference: $2\pi r \div 2 = \pi r$ . A quarter circle (θ = 90°) is a quarter: $\pi r/2$ . In general:

Formula

Arc length

= 2\pi r \times \dfrac{\theta}{360}

\theta

= angle subtended by the arc at the centre, in degrees

Given

Radius $r=6.3$ m, sector angle $=120°$

Find

Arc length (use $\pi\approx\frac{22}{7}$ )

Solution

Arc $=2\times\frac{22}{7}\times6.3\times\frac{120}{360}=2\times\frac{22}{7}\times6.3\times\frac13=\;$ $13.2$ m

Did you know? The stagger on a running track

On a 400 m track, runners in outer lanes start further ahead because their curved sections have a larger radius, hence a longer arc. The “stagger” between lanes exactly compensates so every runner covers 400 m.

3Area of Rectangles, Parallelograms and Triangles

Shape	Area formula
Rectangle (sides $a,b$ )	$ab$
Square (side $a$ )	$a^2$ (special case of rectangle, $b=a$ )
Parallelogram	base $\times$ height
Triangle	$\tfrac12 \times$ base $\times$ height

Figure 2: Two congruent triangles fit together to make a parallelogram, so a triangle's area is half of base × height. — **Figure 2**: Two congruent triangles fit together to make a parallelogram, so a triangle’s area is half of base × height.

Theorem: a median splits a triangle into two equal areas

If AD is a median of △ABC (D is the midpoint of BC), then △ABD and △ACD have equal bases (BD = DC) and the same height from A, so both have area $\tfrac12 \times BD \times h$ : exactly equal, even though the two triangles are usually different shapes.

4Heron’s Formula

Formula

s=\dfrac{a+b+c}{2}

, Area

=\sqrt{s(s-a)(s-b)(s-c)}

a, b, c

are the three side lengths;

s

is the semi-perimeter. Works even when the height is not known.

Given

A triangle with sides 3, 4, 5 units

Find

Its area, and verify against $\tfrac12\times$ base $\times$ height

Solution

$s=\dfrac{3+4+5}{2}=6$ . Area $=\sqrt{6(6-3)(6-4)(6-5)}=\sqrt{6\times3\times2\times1}=\sqrt{36}=\;$ $6$ sq. units

Check: $3^2+4^2=5^2$ , so this is right-angled at the vertex between sides 3, 4. Area $=\tfrac12(3)(4)=6$ sq. units. Matches!

Two more triangle-area formulas

Area $=\dfrac{abc}{4R}$ ( $R$ = circumradius) and Area $=rs$ ( $r$ = inradius, $s$ = semi-perimeter). Both connect a triangle’s area to circles associated with it: the circumcircle and the incircle.

5Brahmagupta’s Formula for a Cyclic Quadrilateral

Common Mistake

Unlike a triangle, a quadrilateral’s area is not determined by its four side lengths alone: the same four sides 3, 3, 3, 3 can enclose very different areas depending on the angles. One more piece of information (an angle, a diagonal, or a special property like being cyclic) is always needed.

Formula

s=\dfrac{a+b+c+d}{2}

, Area

=\sqrt{(s-a)(s-b)(s-c)(s-d)}

Only valid for a cyclic quadrilateral (Brahmagupta, 628 CE). Reduces to Heron’s formula when one side

d\to0

(the quadrilateral collapses to a triangle).

Given

A cyclic quadrilateral with sides 5, 5, 12, 12 units

Find

Its area

Solution

$s=\dfrac{5+5+12+12}{2}=17$ . Area $=\sqrt{(17-5)(17-5)(17-12)(17-12)}=\sqrt{12\times12\times5\times5}=12\times5=\;$ $60$ sq. units

Did you know?

Baudhāyana’s Śhulbasūtra (800 BCE) describes how to construct, using only a compass and straightedge, a square with exactly the same area as a given rectangle, an ancient technique called “squaring”.

6Area of a Circle and Its Sectors

Figure 3: A circle sliced into sectors. As the slices get thinner, laying them side by side approaches a parallelogram, giving Area = base × height = πr × r = πr². — **Figure 3**: A circle sliced into sectors. As the slices get thinner, laying them side by side approaches a parallelogram, giving Area = base × height = πr × r = πr².

Formula

Area of a circle

= \pi r^2

Archimedes (250 BCE) proved this using the limit of regular polygons; Nīlakaṇṭha (c. 1500) gave the slice-rearrangement argument shown above.

Learn by heartDefinition 2

A sector is the region bounded by an arc and the two radii to its endpoints. A segment is the region bounded by an arc and the chord joining its endpoints.

Formula

Area of a sector

= \pi r^2 \times \dfrac{\theta}{360}

Semicircular disc

=\tfrac12\pi r^2

; quarter disc

=\tfrac14\pi r^2

, by the same reasoning as arc length

Given

Radius $=14$ cm, sector angle $=75°$

Find

Area of the sector (use $\pi\approx\frac{22}{7}$ )

Solution

Area $=\dfrac{22}{7}\times14^2\times\dfrac{75}{360}=\dfrac{22}{7}\times196\times\dfrac{75}{360}=\;$ $128.33$ sq. cm

A circle’s circumference and area both come from the same idea: cut it into a fraction θ/360 of the whole, or slice it thin enough that curves become straight lines.

the single idea to carry out of this chapter

Check yourself before the exam

Can I find the circumference and arc length of a circle, and the perimeter of a sector?
Can I apply Heron’s formula, and check it against ½ × base × height when possible?
Can I use Brahmagupta’s formula for a cyclic quadrilateral’s area?
Can I find the area of a circle and of any sector or semicircular/quarter disc?

Quick Revision: read this the night before the exam

Circumference $=2\pi r$ ; arc length $=2\pi r\times\frac{\theta}{360}$
π is irrational; $\pi\approx\frac{22}{7}$ or $3.14$ , never exactly equal
Area of triangle $=\frac12$ base × height $=\sqrt{s(s-a)(s-b)(s-c)}$ (Heron’s)
A median splits a triangle into two equal-area triangles
Cyclic quadrilateral area (Brahmagupta): $\sqrt{(s-a)(s-b)(s-c)(s-d)}$
Area of circle $=\pi r^2$ ; area of sector $=\pi r^2\times\frac{\theta}{360}$

Practice Questions

1 markFind the circumference of a circle of radius 21 cm (use $\pi\approx\frac{22}{7}$ ).
2 marksFind the length of an arc of radius 10.5 cm subtending 60° at the centre.
3 marksFind the area of a triangle with sides 8 cm, 15 cm, 17 cm using Heron’s formula, and verify using base × height.
2 marksFind the area of a cyclic quadrilateral with sides 9, 9, 16, 16 units.
2 marksFind the area of a sector of radius 21 cm with a central angle of 60°.
2 marksThe diameter of a wheel is 63 cm. How far does it travel in 500 revolutions?
3 marksAD is a median of △ABC with area 48 sq. cm. Find the area of △ABD.
2 marksFind the perimeter of a sector of radius 7 cm with a central angle of 90°.
3 marksA park is shaped as a rhombus with all sides 13 m and one diagonal 24 m. Find its area using Heron’s formula (split into two triangles).
2 marksThe ratio of the circumferences of two circles is 3:2. Find the ratio of their areas.

Multiple Choice

1 markπ is:
(a) exactly 22/7(b) a rational number(c) an irrational number(d) an integer
1 markThe area of a sector with θ = 180° is:
(a) πr²(b) πr²/2(c) πr²/4(d) 2πr²
1 markBrahmagupta’s formula applies to:
(a) any quadrilateral(b) any parallelogram(c) only a cyclic quadrilateral(d) only a square

Assertion (A): A median of a triangle always divides it into two triangles of equal area.
Reason (R): The two triangles formed have equal bases and share the same height from the opposite vertex.

Case Study

A circular flower bed has radius 14 m. A gardener wants to plant flowers in a sector of the bed with a central angle of 90°, and build a path along the sector’s curved and straight edges.

1 markWhat is the arc length of the sector (use $\pi\approx\frac{22}{7}$ )?
1 markWhat is the perimeter of the sector (including the two straight edges)?
2 marksWhat is the area to be planted with flowers?

Answer Key

2\times\frac{22}{7}\times21=132

2\times\frac{22}{7}\times10.5\times\frac{60}{360}=11

s=20

; Area

=\sqrt{20\times12\times5\times3}=\sqrt{3600}=60

sq. cm; also right-angled (

8^2+15^2=17^2

\tfrac12(8)(15)=60

sq. cm

s=25

; Area

=\sqrt{16\times16\times9\times9}=16\times9=144

sq. units

\frac{22}{7}\times21^2\times\frac{60}{360}=231

sq. cm

Q6Circumference

=\frac{22}{7}\times63=198

cm; distance

=198\times500=99000

=990

Q724 sq. cm (half of 48)

Q8Arc

=2\times\frac{22}{7}\times7\times\frac{90}{360}=11

cm; perimeter

=11+7+7=25

Q9Each half-diagonal splits the rhombus into 2 triangles with sides 13,13,24:

s=25

, Area

=\sqrt{25\times12\times12\times1}=60

; total rhombus area

=2\times60=120

sq. m

Q10Circumference ratio 3:2 means radius ratio 3:2, so area ratio

=3^2:2^2=9:4

MCQ 1–3(c) · (b) · (c)

A/RBoth A and R are true, and R is the correct explanation of A

Case Study22 m · 50 m · 154 sq. m