Measuring Space — Perimeter and Area Class 9 Notes and Solutions

Measuring Space — Perimeter and Area

Introduction: The Athletics Track

In a 4 × 100 m relay race, athletes stand on curved tracks. If you look closely, the starting lines are not in a straight row. Runners in outer lanes begin ahead of those in inner lanes, yet everyone finishes at the same line.

Table of Contents

Imp: The gap between starting points of adjacent lanes is called the stagger.

Why is this done? Because outer lanes are longer on the curves. The stagger makes every runner cover exactly the same total distance so the race is fair.

Imp: To calculate the stagger, we must know how to measure the distance around a circle.

6.1 Perimeter of a Shape

The perimeter of any shape is the total distance around its border. Imagine a small ant walking all the way around the edge until it returns to the start. That total walk is the perimeter.

Square

Perimeter of a square = 4a units

Rectangle

Perimeter of a rectangle = 2(a + b) units

Note: When a = b, the rectangle becomes a square. So the square formula is a special case of the rectangle formula.

Equilateral Triangle

Perimeter of equilateral triangle = 3a units

Circle — The Special Case

For a circle, the perimeter is called the circumference. How do we find it?

Imp: The ratio of circumference to side stays fixed for any regular polygon. For a circle, the ratio of circumference to diameter is also fixed.

6.2 Perimeter of a Circle — The C/D Ratio

Take any circle. Measure its circumference (C) and its diameter (D). Divide C by D. You always get the same number, no matter how big or small the circle is.

C ÷ D = constant = π (pi)

So, Circumference C = π D = 2π r, where r is the radius.

A Quick Journey Through History

Mesopotamia (c. 1900 BCE): Used π ≈ 3.125. They saw a circle is slightly bigger than a hexagon drawn inside it.
Archimedes (250 BCE): Trapped π between two fractions using 96-sided polygons: 3 10/71 < π < 3 1/7.
Zu Chongzhi (480 CE): Found two useful fractions — 22/7 (Approximate Ratio) and 355/113 (Close Ratio). The second is so accurate that no fraction with denominator below 15,000 comes closer!
Aryabhata (499 CE): Gave π ≈ 3.1416 and called it asanna (approaching), meaning it cannot be written exactly as a simple fraction.
Madhava (c. 1400 CE): Discovered the first exact formula using an infinite series:
π/4 = 1 − 1/3 + 1/5 − 1/7 + 1/9 − ...
This was the birth of calculus. He used it to find π to 11 decimal places.

6.3 π is Irrational

The digits of π go on forever without any repeating pattern:

π = 3.14159265358979...

Numbers like 1/3 = 0.333... or 1/7 = 0.142857142857... have a repeating rhythm. π does not.

Imp: π is irrational. It cannot be written as a fraction a/b where a and b are whole numbers.

We write π ≈ 22/7 or π ≈ 3.14, but remember:

π ≠ 22/7 and π ≠ 3.14
They are only close friends, not equal twins.

A much better approximation is 355/113 ≈ 3.1415929.

Fun Memory Trick
Sentence: "How I wish I could recollect pi."
Count the letters in each word: 3, 1, 4, 1, 5, 9, 2 → gives π ≈ 3.141592

Imp: March 14 (written 3/14) is Pi Day. July 22 (written 22/7) is Pi Approximation Day.

6.4 Length of an Arc of a Circle

A full circle has angle 360° at the centre and length 2πr. If we only want a part of that circle (an arc), we take the matching fraction.

Semicircle (180°)

Quarter Circle (90°)

General Arc (θ°)

Arc length = 2πr × θ/360

A Closer Look at a 400 m Track

A standard 400 m track has two straight sections (84.39 m each) and two semicircular ends. The innermost semicircle has radius 36.5 m. Each lane is 1.22 m wide.

If an athlete runs in the innermost lane at 0.3 m from the inner border, her radius on the curve is 36.5 + 0.3 = 36.8 m.

Straight distance = 2 × 84.39 = 168.78 m

Curved distance = 2π × 36.8 ≈ 231.22 m

Total = 168.78 + 231.22 = 400 m ✓

Imp: Staggers exist because runners in outer lanes have a larger radius on the curves, so they travel farther unless their start is moved forward.

Exercise Set 6.1 — Solutions

Use π = 22/7 unless told otherwise.

1. The perimeter of a circle is 44 cm. What is its radius?

2πr = 44 → 2 × 22/7 × r = 44
→ r = 44 × 7/44 = 7 cm

2. Calculate the circumference (3 significant figures) for:
(i) r = 7 cm (ii) r = 10 cm (iii) r = 12 cm

(i) C = 2 × 22/7 × 7 = 44.0 cm
(ii) C = 2 × 22/7 × 10 = 440/7 ≈ 62.9 cm
(iii) C = 2 × 22/7 × 12 = 528/7 ≈ 75.4 cm

3. Arc length when:
(i) r = 3.5 cm, θ = 60° (ii) r = 6.3 m, θ = 120°

(i) l = 2πr × 60/360 = 2 × 22/7 × 3.5 × 1/6 = 11/3 ≈ 3.67 cm
(ii) l = 2 × 22/7 × 6.3 × 120/360 = 13.2 m

4. Perimeter of a sector with r = 14 cm and sector angle 75°.

Arc = 2 × 22/7 × 14 × 75/360 = 110/6 ≈ 18.33 cm
Two radii = 2 × 14 = 28 cm
Total perimeter = 18.33 + 28 = 46.3 cm

5. Find the perimeters of the shapes in Fig. 6.14 (i–ix).

Part	Shape Break-up	Perimeter
(i)	Rectangle 80×60 + two semicircles (d=60)	2×80 + π×60 = 160 + 1320/7 ≈ 348.6 m
(ii)	Outer semicircle (d=12) + inner semicircle (d=8) + two straight gaps	π×6 + π×4 + 4 = 220/7 + 4 ≈ 35.4 cm
(iii)	Four semicircles on sides of a square (side 10)	2 full circles of d=10 → 2×π×10 = 440/7 ≈ 62.9 cm
(iv)	Three semicircles on sides of equilateral triangle (side 12)	3 × ½ × 2π × 6 = 18π = 396/7 ≈ 56.6 cm
(v)	Central rectangle parts + 8 semicircles	≈ 88 cm
(vi)	Large semicircle (d=28) + three small semicircles (d=28/3 each)	π×14 + 3×π×14/3 = 28π = 88 cm
(vii)	Three semicircles on sides of right triangle (6,8,10)	π×3 + π×4 + π×5 = 12π = 264/7 ≈ 37.7 cm
(viii)	Large semicircle (d=12) + three small semicircles (d=4 each)	π×6 + 3×π×2 = 12π = 264/7 ≈ 37.7 cm
(ix)	Large semicircle (d=20) + two medium semicircles (d=10 each)	π×10 + 2×π×5 = 20π = 440/7 ≈ 62.9 cm

6. Car tyre diameter = 56 cm.
(i) Distance for one revolution?
(ii) Revolutions for 10 km?

(i) Distance = πd = 22/7 × 56 = 176 cm
(ii) 10 km = 1 000 000 cm
Revolutions = 1 000 000 ÷ 176 ≈ 5 682 revolutions

7. Total perimeter of all petals in Fig. 6.15.

(i) Square side = 14 cm. Each petal has 2 semicircular arcs of radius 7 cm.
4 petals × 2 arcs = 8 semicircles = 4 full circles.
Perimeter = 4 × 2π × 7 = 56π = 56 × 22/7 = 176 cm

(ii) Regular hexagon side = 42 cm. 6 petals, each made of two 60° arcs of radius 42 cm.
12 arcs × (1/6 × 2 × 22/7 × 42) = 12 × 44 = 528 cm

8. Ratio of perimeters of two circles is 5 : 4. What is the ratio of their radii?

C = 2πr. Since 2π is constant, C is directly proportional to r.
→ Ratio of radii = 5 : 4

6.6 Area of a Rectangle

Area tells us the amount of space a flat shape covers. We measure it in square units.

Area of rectangle = length × width = a × b square units

6.7 Area of a Parallelogram

A parallelogram can be cut and rearranged into a rectangle with the same base and same height.

Area of parallelogram = base × height = b × h square units

Imp: We must use the perpendicular height, not the slanted side length.

6.8 Area of a Triangle

From a Rectangle

Draw a diagonal in a rectangle. It splits into two equal right-angled triangles.

Area of right-angled triangle = ½ × base × height

From a Parallelogram

Two identical (congruent) triangles fit together to make a parallelogram.

Area of any triangle = ½ × base × height = ½ b h square units

Median Theorem

A median is a line from a vertex to the midpoint of the opposite side.

Imp: A median divides a triangle into two smaller triangles of equal area.

Why? Both small triangles have equal bases (BD = DC) and the same height from A. So ½ × BD × h = ½ × DC × h.

Heron's Formula

If you know all three sides of a triangle but not the height, Heron's formula saves the day.

Let the sides be a, b, c. First find the semi-perimeter:

s = ½ (a + b + c)

Then the area is:

Area = √[ s(s − a)(s − b)(s − c) ]

Check 1 — Equilateral triangle (side a)
s = 3a/2
Area = √[ 3a/2 × a/2 × a/2 × a/2 ] = (√3/4) a²
This matches the usual formula. ✓

Check 2 — Right triangle (3, 4, 5)
s = ½ (3+4+5) = 6
Area = √[ 6 × 3 × 2 × 1 ] = √36 = 6 sq. units
Using ½ × base × height = ½ × 3 × 4 = 6. ✓

Two More Beautiful Formulas

For any triangle with sides a, b, c:

Circumradius R (radius of the circle passing through all three vertices):
Area = abc / 4R
Inradius r (radius of the circle that fits snugly inside, touching all three sides):
Area = r × s

Brahmagupta's Formula for a Cyclic Quadrilateral

A cyclic quadrilateral is a four-sided shape whose corners all lie on a single circle.

Let the sides be a, b, c, d. Semi-perimeter s = ½(a + b + c + d).

Area = √[ (s − a)(s − b)(s − c)(s − d) ]

Imp: This looks exactly like Heron's formula, but with four terms. If one side shrinks to zero (d = 0), the quadrilateral becomes a triangle and Brahmagupta's formula turns into Heron's formula. So Heron's is a special case!

Check — Rectangle (sides a, b)
All rectangles are cyclic. s = a + b.
Area = √[ (a+b−a)(a+b−b)(a+b−a)(a+b−b) ] = √[ b × a × b × a ] = √[a²b²] = ab. ✓

6.9 Squaring a Rectangle

"Squaring" a shape means constructing a square that has the same area as the given shape.

The ancient Indian mathematician Baudhayana (800 BCE) gave a clever geometric method to turn any rectangle into a square of equal area.

Imp: This construction proves geometrically that a rectangle of sides a and b can be turned into a square of side √ab.

Exercise Set 6.2 — Solutions

1. Find the area of triangle ADE in Fig. 6.31 (rectangle 10 cm × 8 cm).

Base AD = 8 cm. The horizontal distance (height) = 10 cm.
Area = ½ × 8 × 10 = 40 cm²

2. Parallel sides of an isosceles trapezium are 40 cm and 20 cm. Non-parallel sides are 26 cm each. Find the area.

Height h = √[ 26² − 10² ] = √[ 676 − 100 ] = √576 = 24 cm
(The 10 cm comes from (40−20)/2.)
Area = ½ × (40 + 20) × 24 = 30 × 24 = 720 cm²

3. Two sides are 8 cm and 11 cm. Perimeter = 32 cm. Find the area.

Third side = 32 − 8 − 11 = 13 cm
s = 32/2 = 16
Area = √[ 16 × (16−8) × (16−11) × (16−13) ] = √[ 16 × 8 × 5 × 3 ] = √1920 = 8√30 ≈ 43.8 cm²

4. Sides are in ratio 3 : 5 : 7. Perimeter = 300 m. Find area.

Sides = 60 m, 100 m, 140 m.
s = 150 m.
Area = √[ 150 × 90 × 50 × 10 ] = √6 750 000 = 1500√3 ≈ 2598 m²

5. One diagonal of a rhombus is twice the other. Area = 128 cm². Find the shorter diagonal.

Let shorter diagonal = d. Longer = 2d.
Area = ½ × d × 2d = d² = 128
d = √128 = 8√2 cm (≈ 11.3 cm)

6. ABCD is a parallelogram. P and Q are any two points on side AB. What is area(△PCD) : area(△QCD)?

Both triangles share base CD and have the same height (the perpendicular distance between the parallel lines AB and CD).
→ Ratio = 1 : 1

7. O is any point on diagonal PR of parallelogram PQRS. Prove that area(△PSO) = area(△PQO).

Triangles PSO and PQO share vertex O and have equal sub-areas because the diagonal PR divides the parallelogram into two equal triangles, and any point O on PR creates sub-triangles with equal area by altitude reasoning.
→ area(△PSO) = area(△PQO)

8. Mid-points of the sides of any quadrilateral are joined. Prove the area of the formed parallelogram is half the area of the original quadrilateral.

The inner shape formed (Varignon's parallelogram) has each side parallel to a diagonal and half its length. Its area = ½ × area of original quadrilateral. Proved.

9. In △ABC, D is the midpoint of BC. P is any point on AD. Show that area(△ABP) = area(△ACP).

Median AD: area(△ABD) = area(△ACD).
PD is also a median for △PBC: area(△PBD) = area(△PCD).
Subtracting: area(△ABD) − area(△PBD) = area(△ACD) − area(△PCD).
→ area(△ABP) = area(△ACP)

10. P is any point inside square ABCD. Find the ratio of (area △PAB + area △PCD) to (area △PBC + area △PDA).

Let side = s. Perpendicular distances from P to AB and CD be h₁ and h₂ with h₁ + h₂ = s.
Red area = ½s·h₁ + ½s·h₂ = ½s². Similarly green area = ½s².
→ Ratio = 1 : 1

11. In △ABC, D is the midpoint of AB. P is on BC, Q is on AB such that CQ ∥ PD. Prove that area(△BPQ) = ½ area(△ABC).

By the midpoint and parallel-line properties, Q is positioned so that the base BP = ½BC.
→ area(△BPQ) = ½ area(△ABC)

6.10 Area of a Circle

Early civilizations knew that the area inside a circle is proportional to the square of its radius.

Babylonians (before 1500 BCE): C² : A ≈ 12 : 1
Egyptians & Baudhayana (800 BCE): A ≈ 256/81 r² ≈ 3.16 r²

In 250 BCE, Archimedes proved the constant is exactly π.

Area of a circle = π r²

Visual Proof (Nilakantha, c. 1500)

Cut the circle into many equal pizza-like slices. Rearrange them alternately to form a shape very close to a parallelogram.

As the slices become thinner:

Base = half the circumference = πr
Height = radius = r

Area = base × height = πr × r = π r²

6.10.1 Area of a Sector of a Circle

A sector is the "pie-slice" region bounded by two radii and an arc.

Area of sector = π r² × θ/360

A segment is the region bounded by an arc and its chord. To find its area, subtract the triangle area from the sector area.

Exercise Set 6.3 — Solutions

Use π = 22/7 unless stated otherwise.

1. Sector radius 7 cm, angle 60°. Find area.

Area = (60/360) × (22/7) × 7² = (1/6) × 22 × 7 = 77/3 ≈ 25.7 cm²

2. Find the area of a quadrant of a circle whose circumference is 44 cm.

2πr = 44 → r = 7 cm.
Quadrant area = (1/4) × (22/7) × 7² = 77/2 = 38.5 cm²

3. Minute hand = 7 cm. Find area swept in 10 minutes.

In 60 min → 360°. In 10 min → 60°.
Area = (60/360) × (22/7) × 7² = 77/3 ≈ 25.7 cm²

4. Chord of radius 10 cm subtends 90° at centre. Find minor and major sector areas (π ≈ 3.14).

(i) Minor sector = (90/360) × 3.14 × 100 = 78.5 cm²
(ii) Major sector = (270/360) × 3.14 × 100 = 235.5 cm²

5. Chord radius 15 cm, angle 60°. Find minor and major segment areas (π ≈ 3.14, √3 ≈ 1.73).

Minor sector = (60/360) × 3.14 × 225 = 117.75 cm²
Triangle area = ½ × 15 × 15 × sin 60° ≈ 97.31 cm²
Minor segment = 117.75 − 97.31 = 20.44 cm²
Major segment = 3.14 × 225 − 20.44 = 686.06 cm²

6. Two wipers, blade 28 cm, sweep 120° each. Find total cleaned area.

One wiper = (120/360) × (22/7) × 784 = 2464/3 cm²
Two wipers = 4928/3 ≈ 1642.7 cm²

7. Prove that the minor segment for 60° in a circle of radius r has area πr²(1/6 − √3/4).

Sector area = (1/6)πr².
Triangle area = ½r² sin 60° = (√3/4)r².
Segment = Sector − Triangle = πr²(1/6 − √3/4).

8. Equilateral triangle inscribed in circle radius r. Show area ratio = 3√3/4π ≈ 0.413.

Side = r√3. Area of triangle = (√3/4)(r√3)² = (3√3/4)r².
Ratio = (3√3/4)r² ÷ πr² = 3√3/4π.

9. Square inscribed in circle radius r. Show area ratio = 2/π ≈ 0.637.

Diagonal = 2r → side = r√2. Area = 2r².
Ratio = 2r² ÷ πr² = 2/π.

10. Regular hexagon inscribed in circle radius r. Show area ratio = 3√3/2π ≈ 0.827.

Area of hexagon = 6 × (√3/4)r² = (3√3/2)r².
Ratio = (3√3/2)r² ÷ πr² = 3√3/2π.
It is exactly twice Q8 because the hexagon contains 6 equilateral triangles of side r, while the inscribed equilateral triangle in the same circle covers only 3 of them.

End-of-Chapter Exercises — Solutions

Use π = 22/7 unless stated otherwise.

1. Draw area models for the identities:
(a+b)(a−b) = a² − b² and (a+b+c)² = a²+b²+c²+2ab+2bc+2ca

The large square of side (a+b+c) is divided into 9 rectangles whose areas add up to give the identity.

2. Isosceles triangle perimeter 40 cm, equal sides 15 cm each. Find area.

Base = 40 − 15 − 15 = 10 cm
Height = √[15² − 5²] = √200 = 10√2 cm
Area = ½ × 10 × 10√2 = 50√2 ≈ 70.7 cm²

3. Isosceles triangle base 10 cm, area 60 cm². Find equal sides.

½ × 10 × h = 60 → h = 12 cm
Equal side = √[5² + 12²] = √169 = 13 cm

4. Right-angled triangle area 54 cm², one leg 12 cm. Find perimeter.

Other leg = (2 × 54)/12 = 9 cm
Hypotenuse = √[144 + 81] = 15 cm
Perimeter = 12 + 9 + 15 = 36 cm

5. Sides in ratio 2 : 3 : 4, perimeter 45 cm. Find area.

Sides = 10, 15, 20 cm. s = 22.5 cm
Area = √[22.5 × 12.5 × 7.5 × 2.5] = 75√15/4 ≈ 72.6 cm²

6. Sides 7 cm, 24 cm, 25 cm. Find area in two ways.

Method 1: 7² + 24² = 625 = 25² → right-angled.
Area = ½ × 7 × 24 = 84 cm²

Method 2 — Heron's: s = 28.
Area = √[28 × 21 × 4 × 3] = √7056 = 84 cm²

7. Bicycle wheel diameter 60 cm. How far after 100 rotations?

Circumference = (22/7) × 60 = 1320/7 cm
Distance = 100 × 1320/7 ≈ 18 857 cm ≈ 188.6 m

8. Find area of a quadrant of a circle whose circumference is 66 cm.

2πr = 66 → r = 10.5 cm
Quadrant area = ¼ × (22/7) × 110.25 = 86.625 cm²

9. Car wheel outer radius 28 cm. Distance after one turn? Turns in 1 km?

Circumference = 2 × (22/7) × 28 = 176 cm
Turns in 1 km = 100 000 ÷ 176 ≈ 568 full turns

10. Two rectangles have the same area and the same perimeter. Are they congruent?

a + b = c + d and ab = cd.
Both pairs are roots of the same quadratic x² − (a+b)x + ab = 0, so {a, b} = {c, d}.
→ Yes, they are congruent.

11. Using the figure, show that area of trapezium = ½(a+b)h.

Trapezium = parallelogram (area ah) + triangle (area ½(b−a)h).
Total = ah + ½(b−a)h = ½(2a + b − a)h = ½(a+b)h.

12. By dividing a trapezium into two triangles, show the same formula.

Draw a diagonal. Two triangles with same height h and bases a and b.
Total area = ½ah + ½bh = ½(a+b)h.

13. Show how two identical trapeziums make a parallelogram and derive the formula.

Rotate one trapezium 180° and join. New parallelogram: base = (a+b), height = h.
Area of parallelogram = (a+b)h → one trapezium = ½(a+b)h.

14. Show that area of a kite = ½ × d₁ × d₂.

The diagonals are perpendicular. Two triangles: each has base d₂ and combined heights = d₁.
Total area = ½ × d₂ × d₁ = ½d₁d₂.

15. Fitting congruent shapes together.

(i) Four copies of a×b fit into 2a×2b (2×2 grid). Yes.
(ii) Four copies of a triangle fit into a triangle with doubled sides (join at midpoints). Yes.
(iii) Nine copies fit into a triangle with tripled sides. Yes.

16. What fraction of the triangle is shaded? (Fig. 6.43)

After removing three corner triangles (each 1/9) and three side triangles (each 2/9):
Central region = 1 − 3(1/9) − 3(2/9) = 1/3 of the triangle.

17. What fraction of the square is shaded? (Fig. 6.44)

Using coordinate geometry with trisection lines:
Shaded parallelogram = 1/5 of the square.

18. What fraction of the rectangle is covered by the circles? (Fig. 6.45 and 6.46)

For n circles of radius r in a row: Circle area/Rectangle area = nπr²/(4nr²) = π/4 ≈ 0.785

Imp: This fraction is the same no matter how many circles are packed in the row!

19. Nine identical rectangles make a large rectangle of area 72 cm². Find the perimeter of each small rectangle.

From the tiling: 4L = 5W → L = 5W/4.
Area equation gives W² = 32/5, so W = 4√5/5 cm, L = √10 cm.
Perimeter = 2(L+W) = 18√10/5 ≈ 11.38 cm.

20. Show that the blue and red triangles in Fig. 6.48 have equal area.

Both share the same vertex and have equal base lengths (each 1/3 of the large base) with the same height. → Equal areas.

21. In Fig. 6.49, show that shaded regions A and B have equal area.

Let square side = r. Quarter circle area = πr²/4. Semicircle area = πr²/8.
By subtraction with overlapping regions and symmetry, A = B = πr²/8 − r²/4. → A = B.

22. Four semicircles drawn inside a square of side 2 units. Find perimeter and area of the flower.

Radius of each semicircle = 1 unit.
Perimeter: 4 semicircular arcs = 4 × π × 1 = 4π ≈ 12.57 units.
Area: Flower area = 2π − 4 = 2(π − 2) ≈ 2.28 sq. units.

23. Two concentric circles. Chord BC of larger circle touches smaller at A. BC = l. Show green area = ¼πl².

OA ⊥ BC, A is midpoint, BA = l/2.
In △OAB: R² = r² + (l/2)² → R² − r² = l²/4.
Green area = π(R² − r²) = πl²/4.

24. Semicircles on all sides of a right triangle. Show A + B = C.

Semicircle on side x has area πx²/8.
A + B = π(a²+b²)/8 = πc²/8 = C (by Pythagoras). → A + B = C.

25. Two circles pass through each other's centres (radius r). Find area of enclosed lens.

Each sector angle = 120°. Sector area = πr²/3. Equilateral triangle area = (√3/4)r².
Lens = 2 segments = 2(πr²/3 − (√3/4)r²) = r²(2π/3 − √3/2).

26. Three triangles inside a rectangle have areas A, B, C. Show rectangle area = 2(A+C)(B+C)/C.

Using distance-from-sides coordinates (x, y), with w × h = rectangle area:
A = hx/2, C = h(w−x)/2, B = wy/2. After eliminating x, y and combining, rectangle area wh = 2(A+C)(B+C)/C.

27. Show the two shaded regions in Fig. 6.55 have equal area.

Quarter circle (radius r) area = πr²/4. Semicircle on radius OA (diameter r, radius r/2) area = πr²/8.
Triangle OAB area = r²/2.
By careful subtraction of overlapping regions, both shaded parts equal πr²/8 − r²/4. → Both shaded areas are equal.