Measuring Space: Perimeter and Area — Detailed Notes
6.1 Perimeter
The perimeter is the total length around a shape. Square = 4a; equilateral triangle = 3a; rectangle = 2(a + b). The ratio of perimeter to side is fixed for a given shape (4:1 for squares, 3:1 for equilateral triangles) — and this idea leads to π for circles.
6.2–6.3 Circumference & π
Circumference C = πd = 2πr (π = C/D ratio, same for every circle)
π is the constant ratio of circumference to diameter. It is irrational (proved by Lambert, 1761), so no fraction equals it exactly — we use π ≈ 22/7 ≈ 3.14 (better: 355/113). Mādhava gave the first exact formula as an infinite series: π/4 = 1 − 1/3 + 1/5 − 1/7 + …
6.4 Arc length
Arc length = 2πr × (θ° / 360°)
Semicircle = πr, quarter circle = πr/2. (This is what sets the lane staggers on a running track: outer lanes have larger-radius curves.)
6.6–6.8 Areas of basic figures
Rectangle = ab; parallelogram = base × height = bh; triangle = ½ × base × height = ½bh. Median theorem: a median divides a triangle into two triangles of equal area. Trapezium = ½(sum of parallel sides) × height = ½(a + b)h. Kite = ½ × product of diagonals.
6.8.1 Heron’s formula & triangle-circle formulas
Heron: Area = √[s(s−a)(s−b)(s−c)], s = ½(a + b + c)
Also Area = abc/(4R) (R = circumradius) and Area = r × ½(a+b+c) = rs (r = inradius). Brahmagupta’s formula for a cyclic quadrilateral with sides a, b, c, d and s = ½(a+b+c+d): Area = √[(s−a)(s−b)(s−c)(s−d)] — Heron’s formula is the special case d = 0.
6.10 Area of a circle & sector
Area of circle = πr² • Area of sector = πr² × (θ° / 360°)
Nīlakaṇṭha’s picture: slice the disc into thin sectors and rearrange into a near-parallelogram of base πr (half the circumference) and height r → area = πr². A sector is bounded by an arc and two radii; a segment is bounded by an arc and its chord. Baudhāyana’s Śulbasūtra even shows how to ‘square a rectangle’ (build a square of equal area).
Key Terms
perimeter, circumference, π, diameter, radius, arc length, sector, segment, area, parallelogram, median, trapezium, kite, Heron’s formula, semi-perimeter, circumradius, inradius, Brahmagupta’s formula, area of a circle.
End-of-Chapter Exercises — Full Solutions
Worked solutions to the end-of-chapter problems (starred * = HOTS), plus key in-text exercise answers. Using π ≈ 22/7 unless noted.
1Draw area-model figures for (a+b)(a−b)=a²−b² and (a+b+c)² = a²+b²+c²+2ab+2bc+2ca.
For (a+b)(a−b)=a²−b²: take an a×a square, cut off a b×b square from a corner; the remaining L-shape rearranges into an (a+b)×(a−b) rectangle. For (a+b+c)²: draw a square of side (a+b+c) and divide each side into parts a, b, c — the 9 pieces are three squares (a², b², c²) and six rectangles (two each of ab, bc, ca), which sum to the identity.
2Isosceles triangle: perimeter 40 cm, equal sides 15 cm each. Find its area.
Base = 40 − 15 − 15 = 10 cm. Height = √(15² − 5²) = √(225 − 25) = √200 = 10√2. Area = ½ × 10 × 10√2 = 50√2 ≈ 70.7 cm². (Heron check: s = 20, area = √(20·5·5·10) = √5000 = 50√2 ✓)
3Isosceles triangle: base 10 cm, area 60 cm². Find the equal sides.
½ × 10 × h = 60 ⇒ h = 12 cm. Each equal side = √(5² + 12²) = √(25 + 144) = √169 = 13 cm.
4A right triangle has area 54 cm² and one leg 12 cm. Find its perimeter.
½ × 12 × (other leg) = 54 ⇒ other leg = 9 cm. Hypotenuse = √(12² + 9²) = √225 = 15 cm. Perimeter = 12 + 9 + 15 = 36 cm.
5Sides of a triangle are in ratio 2 : 3 : 4, perimeter 45 cm. Find its area.
2k + 3k + 4k = 45 ⇒ k = 5, so sides 10, 15, 20 cm; s = 22.5. Area = √[22.5(12.5)(7.5)(2.5)] = √84375 / 4 = (75√15)/4 ≈ 72.6 cm².
6Find the area of a triangle with sides 7, 24, 25 cm in two different ways.
Way 1: 7² + 24² = 49 + 576 = 625 = 25², so it is right-angled; area = ½ × 7 × 24 = 84 cm². Way 2 (Heron): s = 28, area = √(28·21·4·3) = √7056 = 84 cm². Both agree.
7A bicycle wheel has diameter 60 cm. How far does the cyclist travel after 100 rotations?
One rotation = circumference = πd = (22/7)×60 = 1320/7 ≈ 188.57 cm. In 100 rotations = 100 × 1320/7 = 132000/7 ≈ 18857 cm ≈ 188.6 m.
8Find the area of a quadrant of a circle whose circumference is 66 cm.
2πr = 66 ⇒ r = 66 × 7 / (2 × 22) = 10.5 cm. Quadrant area = ¼πr² = ¼ × (22/7) × 10.5² = ¼ × (22/7) × 110.25 = 86.625 cm².
9A car wheel has outer radius 28 cm. How far does the car travel in one turn, and how many turns in 1 km?
One turn = 2πr = 2 × (22/7) × 28 = 176 cm = 1.76 m. Turns in 1 km (100000 cm) = 100000 ÷ 176 ≈ 568 turns.
10(*) Two rectangles have the same area and the same perimeter. Must they be congruent?
Yes. If area ab and perimeter 2(a+b) are both fixed, then a + b and a·b are fixed, so a and b are the two roots of the same quadratic x² − (a+b)x + ab = 0 — i.e. the side lengths are determined. Hence the two rectangles have identical dimensions and are congruent.
11Using area of a parallelogram, show the area of a trapezium is ½(a + b)h.
Place the trapezium (parallel sides a, b, height h) and a copy of it rotated 180° together — they form a parallelogram with base (a + b) and height h, area (a + b)h. The trapezium is half of it, so its area = ½(a + b)h.
12By dividing a trapezium into two triangles, show its area is ½(a + b)h.
A diagonal splits the trapezium into two triangles, both of height h: one with base a and one with base b. Area = ½·a·h + ½·b·h = ½(a + b)h.
13Show how two identical copies of a trapezium make a parallelogram, giving the area formula.
Rotate one copy of the trapezium 180° and join it to the other along a slant side: the two parallel sides a and b line up end-to-end to give a parallelogram of base (a + b) and height h (area (a + b)h). One trapezium is half this, so area = ½(a + b)h.
14Show that the area of a kite is half the product of its diagonals, (i) by algebra, (ii) by geometry.
A kite’s diagonals are perpendicular; let them be p and q, meeting so that one diagonal bisects the other. (i) Algebra: the kite = 2 triangles each with base q and heights p₁, p₂ (p₁ + p₂ = p): area = ½·q·p₁ + ½·q·p₂ = ½·q·p = ½ pq. (ii) Geometry: enclose the kite in a rectangle p × q (sides parallel to the diagonals); the kite covers exactly half of each of the four cells, i.e. half the rectangle = ½ pq.
15Scaling: (i) rectangle sides doubled, (ii) triangle sides doubled, (iii) triangle sides tripled. Compare areas and whether copies fit.
(i) 2a × 2b = 4ab = 4× the area; yes, 4 copies of the a×b rectangle tile it. (ii) A triangle with sides 2a, 2b, 2c is similar with scale factor 2, so area = 4×; yes, 4 copies fit (divide it with the midpoint triangle). (iii) Scale factor 3 → area = 9×; yes, 9 copies fit. (Area scales as the square of the length factor.)
17–18Fraction of a rectangle covered by equal circles packed in rows (conjecture and proof).
Each circle of diameter d sits in a d × d square cell and covers π(d/2)² = πd²/4 of that cell’s d² area, i.e. a fraction π/4 ≈ 0.785. Since the whole rectangle is just a grid of such cells, the circles always cover π/4 of it — for 10, 20, 50 circles alike. Proof: n circles cover n·πd²/4 and the rectangle is n·d², ratio = π/4 regardless of n.
19(*) Nine identical rectangles form a large rectangle of area 72 cm². Find the perimeter of each small rectangle.
Each small rectangle has area 72 ÷ 9 = 8 cm². In the standard stacking the long side is twice the short side (l = 2w), so 2w² = 8 ⇒ w = 2 cm, l = 4 cm. Perimeter of each = 2(4 + 2) = 12 cm.
23(*) Concentric circles; a chord BC of the outer circle of length l touches the inner circle. Show the ring’s area is ¼πl².
The point of contact A is the foot of the perpendicular from the centre O to BC, so OA = r (inner radius) and A is the midpoint of BC. In right triangle OAB: (l/2)² + r² = R² ⇒ R² − r² = l²/4. Area of ring = πR² − πr² = π(R² − r²) = ¼πl².
24(*) Semicircles are drawn on the three sides of a right triangle. Show Area(A) + Area(B) = Area(C).
A semicircle on a side of length x has area ½π(x/2)² = πx²/8 — proportional to the square of the side. With legs a, b and hypotenuse c, a² + b² = c² (Baudhāyana–Pythagoras), so πa²/8 + πb²/8 = πc²/8, i.e. Area(A) + Area(B) = Area(C).
25(*) Two circles of radius r pass through each other’s centres. Find the area enclosed by the two circles.
The centres are r apart. Area of the overlapping lens = 2[½ r²(θ − sinθ)] with θ = 120° = 2π/3 → lens = (2π/3)r² − (√3/2)r². Union = 2πr² − lens = 2πr² − (2π/3)r² + (√3/2)r² = (4π/3 + √3/2)·r² ≈ 4.19r² + 0.87r² = 5.05r².
IXKey in-text exercise answers (Sets 6.1 & 6.3).
Set 6.1 Q1 — 2πr = 44 ⇒ r = 44×7/(2×22) = 7 cm. Set 6.1 Q6 — tyre diameter 56 cm: one revolution = πd = (22/7)×56 = 176 cm; in 10 km (10⁵ cm) = 1000000/176 ≈ 5682 revolutions. Set 6.3 Q1 — sector r = 7, 60°: area = πr²×60/360 = (22/7)×49/6 = 77/3 ≈ 25.67 cm². Set 6.3 Q3 — minute hand 7 cm, 10 min = 60°: swept area = (22/7)×49×60/360 = 77/3 ≈ 25.67 cm². Set 6.3 Q6 — wiper blade 28 cm, 120°, two wipers: 2 × (22/7)×28²×120/360 = 2 × (22/7)×784/3 = ≈ 2051.8 cm². Figure-construction/proof problems 1, 16, 20, 21, 22, 26, 27 are done by the area-model and symmetry methods described in the Notes.