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📚 Class IX Maths 🧩 Worksheet Chapter 6: Measuring Space: Perimeter and Area

Perimeter & Area Worksheet Class 9 – Ganita Manjari Ch 6

Class 9 Maths Perimeter and Area worksheet with answers — 75 questions with step-by-step solutions. Ganita Manjari Ch 6. Free PDF & online practice.

This free Worksheet for CBSE Class IX Maths, Chapter 6: Measuring Space: Perimeter and Area, contains a structured worksheet with MCQs, short answer, case-based and HOTS questions in one place. It has been prepared by Sumeet Sahu at Unique Study Point, Indore, strictly following the latest NCERT syllabus for Session 2026-27.

📌 How to use this Worksheet

Measuring Space: Perimeter and Area — Class 09
UNIQUE STUDY POINT BY SUMEET SAHU

Measuring Space

Class 09 · Maths (Ganita Manjari) · Perimeter and Area · Practice Worksheet with Solutions

75 Questions 162 Marks 1 hr 30 min Time
Tap any question's "Show Answer" button to reveal the full step-by-step solution.

Section A · Very Short Answer / MCQ / Assertion–Reason (Q1–Q28)

1 Mark each
1
The diagonals of the rectangle ABCD intersect at O. If ∠OBC = 64°, then ∠OAB =MCQ
  • a) 36°
  • b) 32°
  • c) 26°
  • d) 64°
✓ Correct Answer: (c) 26°
  1. In a rectangle, all angles = 90°.
  2. ∠ABC = ∠OBC + ∠ABO
  3. 90° = 64° + ∠ABO
  4. ∠ABO = 90° − 64° = 26°
  5. Diagonals of a rectangle are equal and bisect each other, so AO = BO.
  6. In isosceles △AOB, angles opposite equal sides are equal: ∠OAB = ∠ABO = 26°
2
The length of an arc of a sector of angle θ° of a circle with radius R isMCQ
  • a) 2πRθ180
  • b) πR²θ180
  • c) πR²θ360
  • d) 2πRθ360
✓ Correct Answer: (d) 2πRθ360
  1. Standard formula: Length of arc = θ360° × 2πR = 2πRθ360
3
A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 12√2 cm, then area of the triangle isMCQ
  • a) 24√2 cm²
  • b) 64√3 cm²
  • c) 24√3 cm²
  • d) 48√3 cm²
✓ Correct Answer: (b) 64√3 cm²
  1. Diagonal of square = √2 × a
  2. 12√2 = √2 × a ⇒ a = 12 cm
  3. Perimeter of square = 4a = 4 × 12 = 48 cm
  4. Perimeter of equilateral triangle = 3x = 48 (equal perimeters)
  5. x = 16 cm
  6. Area = √34x² = √34 × 16 × 16
  7. Area = 64√3 cm²
4
The sides of a triangle are x, y and z. If x + y = 7 m, y + z = 9 m, and z + x = 8 m, then area of the triangle is:MCQ
  • a) 6 m²
  • b) 4 m²
  • c) 5 m²
  • d) 7 m²
✓ Correct Answer: (a) 6 m²
  1. Adding all three: 2(x + y + z) = 24
  2. x + y + z = 12
  3. z = 12 − 7 = 5, x = 12 − 9 = 3, y = 12 − 8 = 4
  4. s = 122 = 6 m
  5. Area = √[s(s−x)(s−y)(s−z)]
  6. = √[6(6−3)(6−4)(6−5)]
  7. = √[6 × 3 × 2 × 1] = √36
  8. = 6 m²
5
The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8 am and 8.25 am.MCQ
  • a) 125.5 cm²
  • b) 130.95 cm²
  • c) 100 cm²
  • d) 120 cm²
✓ Correct Answer: (b) 130.95 cm²
  1. Angle swept in 25 minutes: θ = 25 × 6° = 150°
  2. Area = θ360 × πr²
  3. = 150360 × 227 × 10²
  4. = 130.95 cm²
6
In the given parallelogram, find the length of the altitude from vertex A on the side DC.MCQ
A B D C 25 cm 17 cm 12 cm
Parallelogram ABCD with diagonal BD = 25 cm, BC = 17 cm, DC = 12 cm
  • a) 18 cm
  • b) 12 cm
  • c) 15 cm
  • d) 25 cm
✓ Correct Answer: (c) 15 cm
  1. In △BCD: a = 12, b = 17, c = 25 cm
  2. s = 12+17+252 = 542 = 27 cm
  3. Area = √[27(27−12)(27−17)(27−25)]
  4. = √[27 × 15 × 10 × 2] = √8100
  5. = 90 cm²
  6. Area of parallelogram = 2 × Area of △BCD = 2 × 90 = 180 cm²
  7. Area = base × height ⇒ 180 = DC × h = 12 × h
  8. h = 18012 = 15 cm
7
A circular disc of radius 6 cm is divided into three sectors with central angles 90°, 120° and 150°. The ratio of the areas of the three sectors isMCQ
  • a) 2 : 3 : 4
  • b) 1 : 5 : 6
  • c) 3 : 4 : 5
  • d) 4 : 5 : 6
✓ Correct Answer: (c) 3 : 4 : 5
  1. With the same radius, area of a sector ∝ θ.
  2. Ratio = 90 : 120 : 150
  3. Divide throughout by 30: = 3 : 4 : 5
8
Which mathematician introduced the value of π as 'asanna', meaning 'approaching' or 'approximate'?MCQ
  • a) Aryabhata
  • b) Zu Chongzhi
  • c) Nilakantha
  • d) Brahmagupta
✓ Correct Answer: (a) Aryabhata
  1. In 499 CE, Aryabhata gave π = 6283220000 = 3.1416.
  2. He described this value as 'asanna' approachingapproximate.
9
If the area of a sector of a circle of radius 36 cm is 54π cm², then the length of the corresponding arc of the sector is:MCQ
  • a) 6π cm
  • b) 5π cm
  • c) 3π cm
  • d) 4π cm
✓ Correct Answer: (c) 3π cm
  1. Area of sector = πr²θ360
  2. π(36)²θ/360 = 54π
  3. θ = 54 × 36036 × 36 = 15°
  4. Length of arc = θ360 × 2πr
  5. = 15360 × 2π × 36
  6. = 3π cm
10
The base and hypotenuse of a right triangle are respectively 5 cm and 13 cm long. Its area is:MCQ
  • a) 40 cm²
  • b) 30 cm²
  • c) 25 cm²
  • d) 28 cm²
✓ Correct Answer: (b) 30 cm²
  1. Height = √(13² − 5²) = √(169−25) = √144
  2. Height = 12 cm
  3. Area = 12 × base × height = 12 × 5 × 12
  4. = 30 cm²
11
Area of a quadrant of a circle whose circumference is 22 cm is (π = 227):MCQ
  • a) 3.65 cm²
  • b) 17.25 cm²
  • c) 9.625 cm²
  • d) 9.625 cm³
✓ Correct Answer: (c) 9.625 cm²
  1. C = 2πr = 22 cm
  2. 2 × 227 × r = 22
  3. r = 3.5 cm
  4. Area of quadrant = πr²4
  5. = 227 × (3.5)²/4
  6. = 22 × 12.2528
  7. = 9.625 cm²
12
What is the standard unit for measuring area, and how is it defined?MCQ
  • a) A 1×1 triangle, equal to 0.5 unit².
  • b) A 1×1 square, taken to be 1 unit².
  • c) A 2×2 square, equal to 4 unit².
  • d) A circle with a radius of 1 unit, equal to π unit².
✓ Correct Answer: (b) A 1×1 square, taken to be 1 unit².
  1. The standard unit for area is a 1×1 square, whose area is defined as 1 unit².
13
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm, then find the height of the parallelogram.MCQ
  • a) 13 cm
  • b) 15 cm
  • c) 12 cm
  • d) 14 cm
✓ Correct Answer: (c) 12 cm
  1. a = 28, b = 30, c = 26 cm
  2. s = 28+30+262 = 842 = 42 cm
  3. Area = √[42(42−28)(42−30)(42−26)]
  4. = √[42 × 14 × 12 × 16] = √112896
  5. = 336 cm²
  6. Area of parallelogram = Area of triangle = 336 cm²
  7. Base × height = 336 ⇒ 28 × h = 336
  8. h = 33628 = 12 cm
14
A circular garden has a circumference of 30π meters. If the diameter of the garden is also given as x meters, choose the correct statement about x.MCQ
  • a) x is an integer but cannot be a rational number.
  • b) x is a rational number.
  • c) x is a transcendental number.
  • d) x is an irrational number.
✓ Correct Answer: (b) x is a rational number.
  1. C = πd, given C = 30π, diameter = x
  2. 30π = πx
  3. x = 30ππ = 30 meters
  4. Since 30 = 301, x is a rational number.
15
The area of an isosceles right angled triangle of equal side 30 cm, is given asMCQ
  • a) 225√3 cm²
  • b) 900 cm²
  • c) 450 cm²
  • d) 45 cm²
✓ Correct Answer: (c) 450 cm²
  1. Area = 12 × base × height
  2. = 12 × 30 × 30
  3. = 450 cm²
16
The line joining the mid-points of two chords of a circle passes through its center, then the chords are:MCQ
  • a) Equal to each other
  • b) Parallel to each other
  • c) Not equal to each other
  • d) Not parallel to each other
✓ Correct Answer: (b) Parallel to each other
  1. The line from the centre to the midpoint of any chord is the perpendicular bisector of that chord.
  2. Both chords are perpendicular to the same line through the centre.
  3. So the two chords are parallel to each other.
17
Area of an equilateral triangle of side 2 cm is:MCQ
  • a) 1 cm²
  • b) √5 cm²
  • c) √2 cm²
  • d) √3 cm²
✓ Correct Answer: (d) √3 cm²
  1. Area = √34(side)²
  2. = √34(2)²
  3. = √3 cm²
18
The area of the triangle having sides 1 m, 2 m and 2 m is:MCQ
  • a) 154 m²
  • b) √154 m²
  • c) √152 m²
  • d) 4√15 m²
✓ Correct Answer: (b) √154 m²
  1. s = 1+2+22 = 52
  2. Area = √[s(s−1)(s−2)(s−2)]
  3. = √[52321212]
  4. = √154 m²
19
The number of planks of dimensions (4m × 50cm × 20cm) that can be stored in a pit which is 16 m long, 12 m wide, and 4 m deep isMCQ
  • a) 1920
  • b) 1840
  • c) 1800
  • d) 1900
✓ Correct Answer: (a) 1920
  1. Volume of plank = 4m × 0.50m × 0.20m = 0.4 m³
  2. Volume of pit = 16m × 12m × 4m = 768 m³
  3. No. of planks = Volume of pit ÷ Volume of plank
  4. = 7680.4
  5. = 1920
20
Find the area of the sector if the radius is 12 cm and with an angle of 134°.MCQ
  • a) 167.38 cm²
  • b) 168.38 cm²
  • c) 168.00 cm²
  • d) 158.38 cm²
✓ Correct Answer: (b) 168.38 cm²
  1. Area of sector = θ360 × πr²
  2. = 134360 × 227 × 12²
  3. = 168.38 cm²
21
In the figure, the ratio of AD to DC is 3 to 2. If the area of △ABC is 40 cm², find the area of △BDC.MCQ
B A D C
△ABC with D on AC such that AD : DC = 3 : 2
  • a) 24 cm²
  • b) 30 cm²
  • c) 36 cm²
  • d) 16 cm²
✓ Correct Answer: (d) 16 cm²
  1. Let AD = 3x, DC = 2x, so AC = 5x.
  2. Area of △ABC = 12 × AC × h = 40
  3. 12 × 5x × h = 40 ⇒ 5x × h = 80
  4. x × h = 16
  5. Area of △ABD = 12 × 3x × h = 32(xh) = 32(16) = 24 cm²
  6. Area of △BDC = Area of △ABC − Area of △ABD = 40 − 24
  7. = 16 cm²
22
Assertion (A): The length of the minute hand of a clock is 7 cm. Then the area swept by the minute hand in 5 minutes is 12⅓6 cm².
Reason (R): The length of an arc of a sector of angle θ and radius r is given by l = θ360 × 2πr.Assertion-Reason
  • a) Both A and R are true and R is the correct explanation of A.
  • b) Both A and R are true but R is not the correct explanation of A.
  • c) A is true but R is false.
  • d) A is false but R is true.
✓ Correct Answer: (b) Both A and R are true but R is not the correct explanation of A.
  1. Angle swept in 5 minutes = 30° (minute hand covers 6°min)
  2. Area swept = θ360 × πr² = 30360 × 227 × 7 × 7 = 776 = 12 56 cm² — A is true.
  3. R correctly states the arc-length formula, which is true, but arc length (not area) — so it doesn't directly explain the area calculation in A.
23
Assertion (A): Area of a sector of a circle whose length of arc is 2l and length of the corresponding radius is 2r is given by 2lr.
Reason (R): Area of a sector of a circle = Length of arc × length of radius × 12Assertion-Reason
  • a) Both A and R are true and R is the correct explanation of A.
  • b) Both A and R are true but R is not the correct explanation of A.
  • c) A is true but R is false.
  • d) A is false but R is true.
✓ Correct Answer: (a) Both A and R are true and R is the correct explanation of A.
  1. Area of sector = 12 × arc length × radius (standard formula, R is true)
  2. Using arc = 2l, radius = 2r: Area = 12 × 2l × 2r = 2lr
  3. This matches A, and R directly explains it — A is true and correctly explained by R.
24
Assertion (A): Every square is a parallelogram.
Reason (R): In a square as well as in a parallelogram, the diagonals are equal in length.Assertion-Reason
  • a) Both A and R are true and R is the correct explanation of A.
  • b) Both A and R are true but R is not the correct explanation of A.
  • c) A is true but R is false.
  • d) A is false but R is true.
✓ Correct Answer: (c) A is true but R is false.
  1. A square has both pairs of opposite sides parallel, so it satisfies the definition of a parallelogram — A is true.
  2. A square's diagonals are equal, but a general parallelogram's diagonals are usually unequal — R is false.
25
Assertion (A): Sides of a triangle are 9 cm, 12 cm and 15 cm. This triangle is both a scalene triangle and a right angle triangle.
Reason (R): Area of a right triangle = 12 × base × height.Assertion-Reason
  • a) Both A and R are true and R is the correct explanation of A.
  • b) Both A and R are true but R is not the correct explanation of A.
  • c) A is true but R is false.
  • d) A is false but R is true.
✓ Correct Answer: (b) Both A and R are true but R is not the correct explanation of A.
  1. All three sides (9, 12, 15 cm) are distinct, so it's a scalene triangle.
  2. Check: 9² + 12² = 81 + 144 = 225 = 15² — Pythagoras holds, so it's also right-angled. A is true.
  3. R states a correct, general formula for the area of a right triangle, but it doesn't explain why this particular triangle is scalene and right-angled.
26
Assertion (A): If diagonals of a quadrilateral are equal, then it must be a rectangle.
Reason (R): The diagonals of a rectangle are equal.Assertion-Reason
  • a) Both A and R are true and R is the correct explanation of A.
  • b) Both A and R are true but R is not the correct explanation of A.
  • c) A is true but R is false.
  • d) A is false but R is true.
✓ Correct Answer: (d) A is false but R is true.
  1. Equal diagonals alone don't guarantee a rectangle (e.g. an isosceles trapezium also has equal diagonals) — A is false.
  2. In rectangle ABCD, AB = DC and ∠B = ∠C = 90°, with BC common.
  3. By SAS, △ABC ≅ △DCB, so AC = BD — diagonals of a rectangle are indeed equal, R is true.
27
State whether the given statement is True or False: The area of a rhombus, one side of which measures 20 cm and one whose diagonal is 24 cm, is 385 cm².TrueFalse
✓ Correct Answer: False
  1. Each half of the rhombus (split by the 24 cm diagonal) is a triangle with sides 20, 20, 24 cm.
  2. s = 20+20+242 = 32 cm
  3. Area of one triangle = √[32(32−20)(32−20)(32−24)]
  4. = √[32 × 12 × 12 × 8] = √36864 = 192 cm²
  5. Area of rhombus = 2 × 192 = 384 cm²
  6. Since 384 ≠ 385, the statement is False (correct area is 384 cm²).
28
Fill in the blank: The diagonals of a rhombus are 4 cm and 6 cm, then the area of rhombus is       cm².Fill in the Blank
  1. Area of rhombus = 12 × d₁ × d₂
  2. = 12 × 4 × 6
  3. = 12 cm²
✓ 12 cm²

Section B · Short Answer–I (Q29–Q50)

2 Marks each
29
Find the area (cm²) of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm. (Round off to the nearest integer.)
  1. a = 13 cm (equal side), b = 20 cm (base)
  2. Area = b4√(4a² − b²)
  3. = 204√[4(13)² − 20²]
  4. = 5√(676 − 400)
  5. = 5√276
  6. = 5 × 16.613
  7. = 83.07 cm²
✓ Area ≈ 83 cm²
30
The perimeter of an isosceles triangle is 42 cm and its base is 1½ times each of the equal sides. Find the height (cm) of the triangle. (Given √7 = 2.64.) (Round off to the nearest integer.)
  1. Let each equal side = a cm, so base b = 32a cm
  2. Perimeter: a + a + 32a = 42
  3. 72a = 42 ⇒ a = 12 cm
  4. b = 32(12) = 18 cm
  5. Height = √[a² − b2²]
  6. = √[12² − 9²] = √(144−81) = √63
  7. = √(9 × 7) = 3√7
  8. = 3 × 2.64 = 7.92 cm
✓ Height ≈ 8 cm
31
Find the cost of leveling the ground in the form of a triangle having its sides 40 m, 70 m and 90 m at ₹8 per square meter. (Round off to the nearest integer.)
  1. s = 40+70+902 = 2002 = 100 m
  2. Area = √[100(100−40)(100−70)(100−90)]
  3. = √[100 × 60 × 30 × 10]
  4. = √1800000
  5. = 1341.64 sq m
  6. Cost = ₹8 × 1341.64
  7. = ₹10733.12
✓ Cost ≈ ₹10,733
32
If the area of an equilateral triangle is 36√3 cm², find its perimeter (cm).
  1. Area = √34(side)²
  2. √34(side)² = 36√3
  3. (side)² = 144
  4. side = 12 cm
  5. Perimeter = 3 × side = 3 × 12
  6. = 36 cm
✓ Perimeter = 36 cm
33
Find the area (sq cm) of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.
  1. Perimeter = 3 + 4 + 5 = 12 cm
  2. s = 122 = 6 cm
  3. Area = √[6(6−3)(6−4)(6−5)]
  4. = √[6 × 3 × 2 × 1] = √36
  5. = 6 sq cm
✓ Area = 6 cm²
34
Match the following:
(a) Perimeter of a minor sector?
(b) Area of a circle?
(c) Area of minor sector?
(d) Area of minor segment?
Options: (i) πr²   (ii) 2r + 2πrθ360   (iii) (πθ360 − sinθ2)r²   (iv) πr²θ360
  1. (a) Perimeter of a minor sector = 2r + 2πrθ360, so (a) → (ii)
  2. (b) Area of a circle = πr², so (b) → (i)
  3. (c) Area of minor sector = πr²θ360, so (c) → (iv)
  4. (d) Area of minor segment = (πθ360 − sinθ2)r², so (d) → (iii)
✓ (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
35
Match the following:
(a) Angle described by minute hand in 60 seconds?
(b) Angle described by second hand in one minute?
(c) Angle described by hour hand in one hour?
(d) Angle described by hour hand in one minute?
Options: (i) 30°   (ii) ½°   (iii) 360°   (iv) 6°
  1. (a) Minute hand in 60 seconds (1 min) = 360°60 = 6°, so (a) → (iv)
  2. (b) Second hand in one minute completes a full circle = 360°, so (b) → (iii)
  3. (c) Hour hand in one hour = 360°12 = 30°, so (c) → (i)
  4. (d) Hour hand in one minute = 30°60 = ½°, so (d) → (ii)
✓ (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
36
What does it mean when we state that the circle constant π is an irrational number?
  1. It means π cannot be written exactly as a fraction pq, where p and q are integers and q ≠ 0.
  2. As a result, its decimal expansion continues forever without ending.
  3. It also never settles into a repeating pattern.
  4. Fractions like 227 are only close approximations used for calculation, not the true value of π.
37
If a chord of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle, find the area of the corresponding minor segment of the circle. (Use π = 3.14 and √3 = 1.73.)
O 60° A B
Chord AB subtends 60° at centre O, radius = 10 cm
  1. Area of minor segment = θ360πr² − 12r²sinθ
  2. = 60360(3.14)(10)² − 12(10)²√32
  3. = 3146 − (100)1.734
  4. = 3146 − 1734
  5. = 52.33 − 43.25
  6. = 9 112 or 9.08
✓ Area of minor segment = 9.08 cm²
38
A sheet is 11 cm long and 2 cm wide. Circular pieces 0.5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.
  1. Each disc fits inside a 0.5 cm × 0.5 cm square.
  2. Number of discs = Area of sheet ÷ Area of one square
  3. = 11 × 20.5 × 0.5
  4. = 220.25
  5. = 88
✓ 88 discs
39
Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.
  1. (i) Algebra: Let diagonals AC = d₁ and BD = d₂ intersect at O; in a kite, diagonals are perpendicular and one bisects the other.
  2. Kite area = sum of 4 right triangles = 12(AO·BO + OC·BO + OC·DO + AO·DO)
  3. = 12(AO+OC)(BO+DO)
  4. = 12(AC)(BD)
  5. = 12d₁d₂
  6. (ii) Geometry: The diagonals divide the kite into 4 right triangles.
  7. Their combined area is exactly the area of the rectangle formed by the two diagonals as sides, halved.
  8. So area of kite = 12 × d₁ × d₂
40
If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.
  1. Let the adjacent angle = x, so the given angle = 23x
  2. Sum of adjacent angles of a parallelogram = 180°
  3. x + 23x = 180°
  4. 53x = 180°
  5. x = 108°
  6. Other angle = 23(108°) = 72°
  7. Opposite angles are equal, so the four angles are 108°, 72°, 108°, 72°
✓ Angles = 108°, 72°, 108°, 72°
41
The height of an equilateral triangle is 6 cm. Find its area.
  1. Height = √32 × side
  2. 6 = √32 × a
  3. a = 12√3 = 4√3 cm
  4. Area = √34a²
  5. = √34(4√3)²
  6. = √34(48)
  7. = 12√3 cm²
✓ Area = 12√3 cm²
42
A and B are points on a circle with center O. C is a point on the circle such that OC bisects ∠AOB. Prove that OC bisects the arc AB.
  1. Given: OC bisects ∠AOB.
  2. So ∠AOC = ∠BOC.
  3. Equal angles at the centre subtend equal arcs.
  4. So arc AC = arc BC.
  5. This means C is the mid-point of arc AB.
  6. Hence, OC bisects the arc AB. (Proved)
43
If the area of an equilateral triangle is 81√3 cm², find its height.
  1. Area = √34(side)²
  2. √34(side)² = 81√3
  3. (side)² = 324
  4. side = 18 cm
  5. Height = √32 × side
  6. = √32 × 18
  7. = 9√3 cm
✓ Height = 9√3 cm
44
The sum of the circumference and diameter of a circle is 116 cm. Find its radius.
  1. 2πr + 2r = 116
  2. 2r(π + 1) = 116
  3. 2r(227 + 1) = 116
  4. 2r297 = 116
  5. r587 = 116
  6. r = 116 × 758
  7. r = 14 cm
✓ Radius = 14 cm
45
A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.
  1. Length of arc l = 2πrθ360°
  2. 20 = 2 × π × r × 60°360°
  3. 2πr = 20 × 6
  4. r = 1202π = 60π cm
✓ Radius = 60π cm
46
The base of an isosceles triangle measures 24 cm and its area is 192 cm². Find its perimeter.
  1. 12 × BC × AL = 192
  2. 12 × 24 × h = 192
  3. h = 19212 = 16 cm
  4. BL = 12(24) = 12 cm
  5. AB² = BL² + AL² = 12² + 16² = 144 + 256 = 400
  6. AB = √400 = 20 cm
  7. Perimeter = 20 + 20 + 24
  8. = 64 cm
✓ Perimeter = 64 cm
47
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisement yields earning of Rs. 5000 per m² per year. A company hired one of its walls for 4 months. How much rent did it pay?
A B C 122 m 22 m 120 m
Triangular wall ABC with AB = 22 m, BC = 120 m, AC = 122 m
  1. AC = 122 m, AB = 22 m, BC = 120 m
  2. Check: 22² + 120² = 484 + 14400 = 14884 = 122² — right-angled at B.
  3. Area = 12 × base × height = 12 × 120 × 22
  4. = 1320 sq m
  5. Rent per year = ₹5000 × 1320
  6. Rent for 4 months = 5000 × 1320 × 412
  7. = ₹22,00,000
✓ Rent = ₹22,00,000
48
A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 msec, calculate the number of complete revolutions the wheel makes in raising the bucket.
  1. Time = 1 min 28 sec = 60 + 28 = 88 seconds
  2. Distance covered by bucket = Speed × Time = 1.1 × 88 = 96.8 m
  3. Radius of wheel = 772 = 38.5 cm
  4. Circumference = 2πr = 2 × 227 × 38.5 = 242 cm = 2.42 m
  5. Distance covered by bucket = Distance covered by wheel
  6. 96.8 = 2.42 × n
  7. n = 96.82.42 = 40
✓ 40 revolutions
49
The circumference of a circle is 88 cm. Find the area of the sector whose central angle is 72°.
  1. Circumference = 2πr = 88
  2. 2 × 227 × r = 88
  3. r = 88 × 72 × 22
  4. r = 14 cm
  5. Area of sector = πr²θ360
  6. = 227 × 14 × 14 × 72360
  7. = 123.2 cm²
✓ Area = 123.2 cm²
50
Find the area of the segment of a circle, if angle of the sector is 90° and the radius of the circle is 21 cm.
  1. Area of sector AOB = θπr²360
  2. = 90360 × 227 × 21 × 21
  3. = 11 × 3 × 212
  4. = 346.5 cm²
  5. Area of △AOB = 12 × OA × OB
  6. = 12 × 21 × 21 = 220.5 cm²
  7. Area of segment = Area of sector − Area of △AOB
  8. = 346.5 − 220.5
  9. = 126 cm²
✓ Area of segment = 126 cm²

Section C · Short Answer–II (Q51–Q65)

3 Marks each
51
A rectangular plot is 24 m long and 20 m wide. A cubical pit of edge 4 m is dug at each of the four corners of the field and the soil removed is evenly spread over the remaining part of the plot. By what height does the remaining plot get raised?
  1. Area of plot = 24 × 20 = 480 m²
  2. Volume of one pit = 4³ = 64 m³
  3. Volume of 4 pits = 4 × 64 = 256 m³
  4. Base area of one pit = 4² = 16 m²; area of 4 pit bases = 4 × 16 = 64 m²
  5. Remaining area = 480 − 64 = 416 m²
  6. Let height raised = h; Volume of soil raised = Volume of soil dug
  7. 416 × h = 256
  8. h = 256416 = 813 m
✓ Height raised = 813 m
52
In the trapezium ABCD, AB || DC and ∠ADC = 90°, AB = 10 cm, DC = 40 cm and the diagonal AC = 41 cm. Then, find the area of trapezium.
  1. In △ADC, AC² = AD² + DC² (Pythagoras)
  2. 41² = AD² + 40²
  3. AD² = 41² − 40² = (41+40)(41−40)
  4. = 81 × 1 = 81
  5. AD = 9 cm
  6. Area of trapezium = 12(AB + DC) × AD
  7. = 12(10 + 40) × 9
  8. = 12(50)(9) = 25 × 9
  9. = 225 cm²
✓ Area of trapezium = 225 cm²
53
A conical tent is made by stitching 12 triangular pieces of cloth of two different colours, red and white alternately, each piece measuring 10 cm, 20 cm and 20 cm.
  • How much cloth of red colour is required to make a conical tent?
  • Find the total length of triangular pieces of white colour.
  1. a = 10, b = 20, c = 20 cm
  2. s = 10+20+202 = 502 = 25 cm
  3. Area of one piece = √[25(25−10)(25−20)(25−20)]
  4. = √[25 × 15 × 5 × 5] = 5 × 5√15 = 25√15 cm²
  5. (i) There are 6 red pieces (alternating out of 12).
  6. Red cloth = 6 × 25√15 = 150√15 cm²
  7. (ii) Perimeter of one triangular piece = 10 + 20 + 20 = 50 cm
  8. Total length of 6 white pieces = 6 × 50 = 300 cm
✓ (i) 150√15 cm²   (ii) 300 cm
54
Construct a rectangle whose adjacent sides are 4.7 cm and 3.2 cm.Construction
  1. Draw AB = 4.7 cm.
  2. Construct ∠ABX = 90°.
  3. With B as centre and radius 3.2 cm, draw an arc cutting BX at C.
  4. With A as centre and radius 3.2 cm, draw an arc.
  5. With C as centre and radius 4.7 cm, draw an arc cutting the previous arc at D.
  6. Join DC and DA.
  7. ABCD is the required rectangle.
55
A chord PQ of length 12 cm subtends an angle 120° at the center of a circle. Find the area of the minor segment cut off by the chord PQ.
  1. ∠POQ = 120°, PQ = 12 cm
  2. PL = PQ × 0.5 = 6 cm (L = foot of perpendicular from O)
  3. ∠POL = ∠QOL = 60°
  4. In △OPL: sin60° = PLOP
  5. OP = 6sin60° = 6√3/2 = 12√3 = 4√3 cm (radius r)
  6. Area of sector = 120360πr² = 13π(48) = 16π cm²
  7. Area of △POQ = 12r²sin120° = 12(48)√32 = 12√3 cm²
  8. Area of minor segment = 16π − 12√3
✓ Area = 4(4π − 3√3) cm²
56
A certain quantity of wood costs ₹25,000 per m³. A solid cubical block of such wood is bought for ₹18,225. Calculate the volume of the block and use the method of factor to find the length of one edge of the block.
  1. Volume of block = Cost of block ÷ Cost per m³
  2. = 1822525000
  3. = 0.729 m³
  4. (side)³ = 0.729 = 7291000
  5. Factorising: 7291000 = 3×3×3×3×3×32×2×2×5×5×5
  6. side = 3×32×5 = 910
  7. side = 0.9 m
✓ Volume = 0.729 m³; edge = 0.9 m
57
A rectangular plot of land measures 45 m × 30 m. A boundary wall of height 2.4 m is built all around the plot at a distance of 1 m from the plot. Find the area of the inner surface of the boundary wall.
  1. Length of the wall's inner boundary = 45 + 2 = 47 m (1 m clearance on each side)
  2. Breadth of the wall's inner boundary = 30 + 2 = 32 m
  3. Area of inner surface = 2(47 × 2.4) + 2(32 × 2.4)
  4. = 225.6 + 153.6
  5. = 379.2 m²
✓ Area = 379.2 m²
58
Area of a sector of central angle 200° of a circle is 770 cm². Find the length of the corresponding arc of this sector.
  1. Area of sector = πr²360 × θ
  2. 770 = πr²360 × 200
  3. r² = 770 × 18π = 770 × 18 × 722
  4. r² = 9 × 49
  5. r = 21 cm
  6. Length of arc = θ360 × 2πr
  7. = 200360 × 2 × 227 × 21
  8. = 2018 × 21 × 227
  9. = 2203 cm = 73 13 cm
✓ Arc length = 73 13 cm
59
Calculate the area of a triangle, whose sides are in the ratio 7 : 5 : 4 and perimeter equal to 32 cm.
  1. Let sides = 7x, 5x, 4x
  2. Perimeter: 7x + 5x + 4x = 32
  3. 16x = 32 ⇒ x = 2 cm
  4. Sides: 14 cm, 10 cm, 8 cm
  5. s = 322 = 16 cm
  6. Area = √[16(16−14)(16−10)(16−8)]
  7. = √[16 × 2 × 6 × 8] = √1536
  8. = 16√6 cm²
✓ Area = 16√6 cm²
60
The sides of a triangle are in the ratio of 13 : 14 : 15 and its perimeter is 84 cm. Find the area of the triangle.
  1. Sum of ratio parts = 13 + 14 + 15 = 42
  2. Side a = 1342 × 84 = 26 cm
  3. Side b = 1442 × 84 = 28 cm
  4. Side c = 1542 × 84 = 30 cm
  5. s = 842 = 42 cm
  6. Area = √[42(42−26)(42−28)(42−30)]
  7. = √[42 × 16 × 14 × 12]
  8. = (42)(4)(2)
  9. = 336 cm²
✓ Area = 336 cm²
61
Find the area of an isosceles triangle, by using Heron's formula, having base 4 cm and length of one of the equal sides as 6 cm.
  1. a = 4, b = 6, c = 6
  2. s = 4+6+62 = 8
  3. Area = √[8(8−4)(8−6)(8−6)]
  4. = √[8 × 4 × 2 × 2] = √128
  5. = 8√2 cm²
✓ Area = 8√2 cm²
62
You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., 12(a+b)h.
a h b
Trapezium split into a parallelogram and a triangle sharing height h
  1. Area of the parallelogram part = base × height = ah
  2. The triangle sharing height h has base = (b − a)
  3. Area of triangle = 12(b−a)h
  4. Total area = ah + 12(b−a)h
  5. Factor out h: = h[a + 12b − 12a]
  6. = h[12a + 12b]
  7. = 12(a+b)h
63
Area of a sector of a circle of radius 36 cm is 54π cm². Find the length of the corresponding arc of the sector.
  1. r = 36 cm, Area of sector = 54π cm²
  2. πr²θ360 = 54π
  3. θ = 15°
  4. Length of arc = πrθ180
  5. = 36π × 15180
  6. = 3π cm
✓ Arc length = 3π cm
64
Find the radius of a circle if a 90° arc has a length of 3.5π cm. Hence, find the area of the sector formed by this arc.
  1. Circumference of a quadrant (90° arc) = 2πr4 = πr2
  2. πr2 = 3.5π
  3. r2 = 3.5 ⇒ r = 7 cm
  4. Area of sector = πr² × 90360
  5. = 227 × 7 × 7 × 14
  6. = 22 × 7 × 14
  7. = 38.5 cm²
✓ Radius = 7 cm; Area of sector = 38.5 cm²
65
The radii of two circles are in the ratio 3 : 8. If the difference between their areas is 385π cm², then find the area of the bigger circle. [Take π = 227]
  1. Let radii = 3x (smaller) and 8x (bigger)
  2. π[(8x)² − (3x)²] = 385π
  3. 64x² − 9x² = 385
  4. 55x² = 385
  5. x² = 7
  6. Area of bigger circle = π(8x)² = 64πx²
  7. = 64 × 227 × 7
  8. = 64 × 22
  9. = 1408 cm²
✓ Area of bigger circle = 1408 cm²

Section D · Case–Based / Data–Based (Q66–Q70)

4 Marks each
66
Case Study — History of π: Maths teacher Ms. Kavitha is teaching her class about the history of the constant π. She explains that in ancient Mesopotamia (around 1900 BCE), mathematicians set π = 3 + 18 = 3.125 after observing that a circle's perimeter was larger than the perimeter of a hexagon inscribed within it. Archimedes of Syracuse (around 250 BCE) used polygons with up to 96 sides to show that 31071 < π < 317. Indian mathematician Āryabhaṭa (499 CE) gave the value 6283220000 = 3.1416, describing it as asanna (approximate). Zu Chongzhi (480 CE) discovered the fraction 355113 ≈ 3.1415929. Finally, Mādhava of Sangamagrāma discovered the first exact formula: π4 = 1 − 13 + 15 − 17 + ...
  • What does the ratio CD represent for any circle? (1)
  • Who introduced the symbol π, and what does the letter π stand for? (1)
  • Why is π called an irrational number? How is this different from rational numbers like 13 or 711? (2)
    OR — Āryabhaṭa described his value of π as asanna (approximate). Why was this a profound insight for the time? (2)
  1. (i) For any circle, CD is always the same constant value — this constant is π itself.
  2. (ii) Welsh mathematician William Jones introduced the symbol π in 1706.
  3. The letter π is the first letter of the Greek word "perimetros", meaning perimeter; it was later popularised by Leonhard Euler.
  4. (iii) A rational number can be written as a fraction ab (a, b integers, b ≠ 0); its decimal either terminates (1.4) or repeats in a pattern (13 = 0.333..., 711 = 0.636363...).
  5. π cannot be written as any such fraction; its decimal expansion (3.14159265358...) goes on forever with no repeating pattern — this is why it's called irrational.
  6. OR: By calling his value asanna approachingapproximate, Aryabhata was signalling that π could not be expressed exactly as a simple fraction.
  7. This effectively recognised π's irrational nature centuries before Lambert formally proved it in 1761 — a remarkable piece of mathematical intuition for the time.
67
Case Study — Earth Day Badges: For the inauguration of 'Earth day' week in a school, badges were given to volunteers. Organisers purchased these badges from an NGO, who made these badges in the form of a circle inscribed in a square of side 8 cm. O is the centre of the circle and ∠AOB = 90°.
D C A B O P Q 90° 8 cm
Square ABCD (side 8 cm) with inscribed circle; sector OPQ spans 90°
  • What is the area of square ABCD? (1)
  • What is the length of diagonal AC of square ABCD? (1)
  • Find the area of sector OPRQO. (2)
    OR — Find the area of the remaining part of square ABCD when the area of the circle is excluded. (2)
  1. (i) Area of square ABCD = (side)² = 8²
  2. = 64 cm²
  3. (ii) In △ABC, ∠B = 90°, so AC² = AB² + BC² = 2(AB)²
  4. AC = AB√2 = 8√2 cm
  5. (iii) Radius of inscribed circle = half the side = 4 cm
  6. Area of sector OPRQO = θ360 × πr² = 90360 × 227 × 4²
  7. = 14 × 227 × 16
  8. = 887 cm² = 12 47 cm²
  9. OR: Area of full circle = πr² = 227 × 16 = 3527 cm²
  10. Remaining area = Area of square − Area of circle = 64 − 3527
  11. = 448 − 3527
  12. = 967 cm² = 13 57 cm²
✓ (i) 64 cm²   (ii) 8√2 cm   (iii) 887 cm²   OR 967 cm²
68
Case Study — NSS Wheel Symbol: NSS (National Service Scheme) aims to connect students to the community. The NSS symbol is based on the Rath wheel of the Konark Sun Temple, Odisha, signifying the progress cycle of life. The diameters of the inner circle are equally placed (8 equal spokes). Given that OP = 21 cm (outer radius), OS = 10 cm (inner radius).
O P Q R S
Wheel with 8 equal spokes; outer radius OP = 21 cm, inner radius OS = 10 cm
  • Find m∠ROS.
  • Find the perimeter of sector OPQ.
  • Find the area of shaded region PQRS.
    OR — Find the area of shaded region ACB, i.e. the segment ACB.
  1. (i) The 8 equally-spaced spokes divide the circle into 8 equal sectors.
  2. m∠ROS = 360°8 = 45°
  3. (ii) Arc PQ = 45360 × 2π(21) = 18 × 2 × 227 × 21 = 16.5 cm
  4. Perimeter of sector OPQ = OP + OQ + arc PQ = 21 + 21 + 16.5
  5. = 58.5 cm
  6. (iii) Area of shaded PQRS = θ360 × π × (OP² − OS²)
  7. = 45360 × 227 × (21² − 10²)
  8. = 18 × 227 × (441 − 100) = 18 × 227 × 341
  9. = 375128 cm² ≈ 133.96 cm²
  10. OR: Taking ∠AOB = 90°, r = OP = 21 cm:
  11. Area of sector AOB = 90360 × 227 × 21² = 346.5 cm²
  12. Area of △AOB = 12 × 21 × 21 = 220.5 cm²
  13. Area of segment ACB = 346.5 − 220.5 = 126 cm²
✓ (i) 45°   (ii) 58.5 cm   (iii) ≈133.96 cm²   OR 126 cm²
69
Case Study — Surveying Land: A civil engineer is surveying three triangular plots of land for a housing project. Plot A: 13 m, 14 m, and 15 m. Plot B: sides in the ratio 5 : 12 : 13 with perimeter 60 m. Plot C: an equilateral triangle with each side 10 m. The land will be levelled at a cost of ₹80 per m².
  • What is the semi-perimeter of Plot A? (1)
  • Using Heron's formula, find the area of Plot A. (1)
  • For Plot B, the sides are in the ratio 5 : 12 : 13 and the perimeter is 60 m. What are the actual side lengths? (2)
    OR — Show that Plot B is a right-angled triangle. Then find its area without using Heron's formula. (2)
  1. (i) s = 13+14+152 = 422
  2. = 21 m
  3. (ii) Area = √[21(21−13)(21−14)(21−15)]
  4. = √[21 × 8 × 7 × 6] = √7056
  5. = 84 m²
  6. (iii) Let sides = 5k, 12k, 13k
  7. 5k + 12k + 13k = 60 ⇒ 30k = 60 ⇒ k = 2
  8. Sides = 10 m, 24 m, 26 m
  9. OR: Check: 10² + 24² = 100 + 576 = 676 = 26² — Plot B is right-angled (between the 10 m and 24 m sides).
  10. Area = 12 × base × height = 12 × 10 × 24
  11. = 120 m²
✓ (i) 21 m   (ii) 84 m²   (iii) 10, 24, 26 m   OR 120 m²
70
Case Study: Renu made a picture of an aeroplane with coloured paper, made up of 5 labelled portions (I–V) with the dimensions shown.
I II III IV V 5 cm 6 cm 6.5 cm 1 cm 1 cm 2 cm
Renu's aeroplane craft — portions I (nose), II (body), IIIIV/V (wings and tail)
  • What is the area of portion I? (1)
  • What is the area of portion II? (1)
  • What is the area of portion III? (2)
    OR — What is the area of portion IV? (2)
  1. (i) Portion I is an isosceles triangle with sides 1, 5, 5 cm (base 1 cm).
  2. 2s = 1 + 5 + 5 = 11 ⇒ s = 5.5 cm
  3. Area = √[5.5(5.5−1)(5.5−5)(5.5−5)] = √[5.5 × 4.5 × 0.5 × 0.5]
  4. = √6.1875 ≈ 2.5 cm²
  5. (ii) Portion II is a rectangle with sides 6.5 cm and 1 cm.
  6. Area = length × breadth = 6.5 × 1
  7. = 6.5 cm²
  8. (iii) Portion III is a trapezium with parallel sides 2 cm and 1 cm, slant side 1 cm.
  9. Height = √[1² − 0.5²] = √0.75 = √32 ≈ 0.866 cm
  10. Area = 12(2+1) × height = 32(0.866)
  11. ≈ 1.3 cm²
  12. OR: Portion IV is a triangle with base 1.5 cm and height 6 cm.
  13. Area = 12 × base × height = 12(1.5)(6)
  14. = 4.5 cm²
✓ (i) ≈2.5 cm²   (ii) 6.5 cm²   (iii) ≈1.3 cm²   OR 4.5 cm²

Section E · Long Answer (Q71–Q75)

5 Marks each
71
Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 90°.
  1. In △BAD, ∠A = 90°, so BD² = AB² + AD²
  2. BD² = 40² + 9² = 1600 + 81 = 1681
  3. BD = √1681 = 41 m
  4. Area of △BAD = 12 × AB × AD = 12(40)(9)
  5. = 180 m²
  6. For △BDC: a = BD = 41, b = BC = 28, c = CD = 15 m
  7. s = 41+28+152 = 842 = 42 m
  8. Area = √[42(42−41)(42−28)(42−15)]
  9. = √[42 × 1 × 14 × 27] = √15876
  10. = 126 m²
  11. Area of ABCD = Area of △BAD + Area of △BDC = 180 + 126
  12. = 306 m²
✓ Area of ABCD = 306 m²
72
Find the perimeter of the shape (taking the arcs to be quarter, half or three-quarters of a circle, as appropriate):
14 cm
A 3×3 grid (14 cm wide) with 4 semicircular bumps on alternating outer edges
  1. The central grid is 3×3 squares, total width 14 cm.
  2. Side of one small square = 143 cm.
  3. The 4 semicircles each have diameter D = 143 cm (equal to one small square's side).
  4. 4 semicircles together = 2 full circles.
  5. Curved length = 2 × πD = 2 × 227 × 143
  6. = 2 × 22 × 23 = 883 ≈ 29.33 cm
  7. Straight segments: 4 sides × 2 segments each = 8 segments, each 143 cm.
  8. Straight length = 8 × 143 = 1123 ≈ 37.33 cm
  9. Total perimeter = 883 + 1123 = 2003
  10. ≈ 66.67 cm
✓ Perimeter ≈ 66.67 cm
73
If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.
  1. Let length = x, breadth = y.
  2. Perimeter: 2(x+y) = 68 ⇒ x + y = 34 …(1)
  3. Diagonal: x² + y² = 26² = 676 …(2)
  4. From (1): (x+y)² = 34² = 1156
  5. x² + 2xy + y² = 1156
  6. 676 + 2xy = 1156 (using (2))
  7. 2xy = 480
  8. xy = 240
  9. Area of rectangle = x × y
✓ Area = 240 m²
74
Since △ABD and △ACD have equal area, you may wonder — can we divide △ABD using straight cuts into two or more pieces that we can then rearrange to exactly cover △ACD? What do you think? Is it possible?
  1. Yes, this is possible — it follows from the Bolyai–Gerwien theorem.
  2. The theorem states that any two simple polygons of equal area are "scissors-congruent" (one can be cut into finitely many polygonal pieces and rearranged into the other).
  3. Since Area(△ABD) = Area(△ACD), such a dissection must exist.
  4. In practice, both triangles can be transformed into a common intermediate shape, usually a rectangle of the same area, via a specific set of straight cuts.
  5. Reversing a different set of cuts on that rectangle then rebuilds it into the second triangle.
  6. This principle underlies geometric puzzles like tangrams, and shows that while a shape's form is flexible, area is conserved under such dissections.
75
In the given figure, lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, then what is the distance of the other chord from the centre?
O A B C D
Parallel chords AB = 6 cm and CD = 8 cm; AB is 4 cm from centre O
  1. AB = 6 cm (smaller chord), CD = 8 cm, distance OP (to AB) = 4 cm
  2. The perpendicular from the centre bisects a chord.
  3. AP = PB = AB2 = 3 cm
  4. CQ = QD = CD2 = 4 cm
  5. In △OPA: OA² = OP² + AP² = 4² + 3² = 25
  6. OA = 5 cm (this is the radius)
  7. OA = OC = radius = 5 cm
  8. In △OQC: OC² = OQ² + CQ²
  9. 25 = OQ² + 4² = OQ² + 16
  10. OQ² = 9 ⇒ OQ = 3 cm
✓ Distance of the 8 cm chord from centre = 3 cm
Prepared by Sumeet Sahu · Mob: 8103405051 · Unique Study Point
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📋 Details

ClassClass IX (CBSE / NCERT)
SubjectMaths
ChapterChapter 6: Measuring Space: Perimeter and Area
Resource TypeWorksheet
Last Updated05 September 2026
Session2026-27 (Latest NCERT Syllabus)
Downloads0+
Prepared bySumeet Sahu, Unique Study Point, Indore
CostFree
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