Practise Class 9 Maths Part 1 Chapter 6 – Measuring Space: Perimeter and Area with this Ganita Manjari Chapter 6 Worksheet. This worksheet includes important questions and revision exercises to strengthen understanding of perimeter and area, improve mathematical practice and support effective preparation for class tests and examinations.
📐 Class 9 Mathematics
Chapter 6 – Measuring Space: Perimeter and Area
Worksheet
Name: ______________________________
Class: __________ Roll No.: __________
Date: __________________
Instructions
- Show all working clearly.
- Use π ≈ 22/7 unless another approximation is given.
- Write the correct units.
- Draw neat diagrams wherever required.
- For construction/proof questions, show all important steps.
🟦 Exercise Set 6.1
1. The perimeter of a circle is 44 cm. What is its radius?
Answer: 7 cm
2. Calculate, correct to 3 significant figures, the circumference of a circle with:
(i) radius 7 cm
(ii) radius 10 cm
(iii) radius 12 cm
Answers:
(i) 44 cm
(ii) 62.9 cm
(iii) 75.4 cm
(i) 44 cm
(ii) 62.9 cm
(iii) 75.4 cm
3. Calculate the length of the arc of a circle if:
(i) the radius is 3.5 cm and the angle at the centre is 60°, and
(ii) the radius is 6.3 m and the angle at the centre is 120°.
Answers:
(i) approximately 3.67 cm
(ii) 13.2 m
(i) approximately 3.67 cm
(ii) 13.2 m
4. Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75°.
Answer: 139/3 cm ≈ 46.33 cm
5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):
(i) 60 m, 80 m
(ii) 8 cm, 12 cm
(iii) 10 cm
(iv) 12 cm
(v) 14 cm
(vi) 28 cm
(vii) 8 cm, 6 cm
(viii) 4 cm, 4 cm, 4 cm
(ix) 10 cm, 10 cm
Answers:
(i) 160 + 60π m ≈ 348.57 m
(ii) 10π + 4 cm ≈ 35.43 cm
(iii) 20π cm ≈ 62.86 cm
(iv) 18π cm ≈ 56.57 cm
(v) 28π cm = 88 cm
(vi) 24.5π cm = 77 cm
(vii) 12π cm ≈ 37.71 cm
(viii) 12π cm ≈ 37.71 cm
(ix) 20π cm ≈ 62.86 cm
(i) 160 + 60π m ≈ 348.57 m
(ii) 10π + 4 cm ≈ 35.43 cm
(iii) 20π cm ≈ 62.86 cm
(iv) 18π cm ≈ 56.57 cm
(v) 28π cm = 88 cm
(vi) 24.5π cm = 77 cm
(vii) 12π cm ≈ 37.71 cm
(viii) 12π cm ≈ 37.71 cm
(ix) 20π cm ≈ 62.86 cm
6. If the diameter of a car tyre is 56 cm, then: (i) How far does the car need to travel for the tyre to complete one revolution? (ii) How many revolutions does the tyre make if the car travels 10 km?
Answers:
(i) 176 cm
(ii) approximately 5681.82 revolutions
(i) 176 cm
(ii) approximately 5681.82 revolutions
7. Find the total perimeter of all the petals in each of the given flowers.
Fig. 6.15A: The centres of the arcs are the midpoints of the sides of the square.
Fig. 6.15B: The centres of the arcs are the vertices of the hexagon.
(i) 14 cm
(ii) 42 cm
Answers:
(i) 88 cm
(ii) 528 cm
(i) 88 cm
(ii) 528 cm
8. The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?
Answer: 5:4
🟩 Exercise Set 6.2
1. Find the area of triangle ADE in Fig. 6.31.
Given: 10 cm and 8 cm as shown in the figure.
Answer: 40 cm²
2. The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Height = 24 cm.
Area = ½(40+20)×24
Answer: 720 cm²
Area = ½(40+20)×24
Answer: 720 cm²
3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.
Third side = 13 cm.
s = 16 cm.
Area = √(16×8×5×3)
Answer: 8√30 cm² ≈ 43.82 cm²
s = 16 cm.
Area = √(16×8×5×3)
Answer: 8√30 cm² ≈ 43.82 cm²
4. The sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.
Sides = 60 m, 100 m and 140 m.
s = 150 m.
Area = √(150×90×50×10)
Answer: 150√30 m² ≈ 821.58 m²
s = 150 m.
Area = √(150×90×50×10)
Answer: 150√30 m² ≈ 821.58 m²
5. One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.
Let shorter diagonal = d.
Other diagonal = 2d.
Area = ½×d×2d = d².
d² = 128.
Answer: 8√2 cm
Other diagonal = 2d.
Area = ½×d×2d = d².
d² = 128.
Answer: 8√2 cm
6. ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (∆PCD): area (∆QCD)?
Both triangles have their bases on the same line AB and the same altitude from C.
Therefore their areas are proportional to their bases.
Ratio = PD : QD when the corresponding bases are taken as shown by the figure.
Therefore their areas are proportional to their bases.
Ratio = PD : QD when the corresponding bases are taken as shown by the figure.
7. O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
Triangles PSO and PQO have bases PS and PQ respectively. Since P, O and R are collinear and opposite sides of the parallelogram are parallel, the corresponding perpendicular heights are equal. Using ½×base×height and the equal corresponding dimensions gives the required equality.
8. If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)
Join the midpoints and divide the original quadrilateral into four smaller triangles. The central parallelogram formed by joining the midpoints has area equal to half the area of the original quadrilateral.
9. In ∆ABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).
Since D is the midpoint of BC, BD = DC. Triangles ABP and ACP have equal bases on BC and the same altitude from A/P as appropriate. Hence their areas are equal.
10. Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)?
The two pairs have equal total areas.
Ratio = 1:1
Ratio = 1:1
11. In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined (Fig. 6.34). Prove that Area (∆BPQ) = 1/2 Area (∆ABC).
Use the parallel lines to establish equal corresponding heights and the midpoint relation AD = DB. Applying ½×base×height gives the required result.
🟨 Exercise Set 6.3
Unless stated otherwise, use the approximation 22/7 for π.
1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Area = 60/360 × 22/7 × 7²
Answer: 77/3 cm² ≈ 25.67 cm²
Answer: 77/3 cm² ≈ 25.67 cm²
2. Find the area of a quadrant of a circle whose circumference is 44 cm.
Circumference = 44 cm gives r = 7 cm.
Quadrant area = ¼×πr².
Answer: 38.5 cm²
Quadrant area = ¼×πr².
Answer: 38.5 cm²
3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
10 minutes corresponds to 60°.
Area = 60/360×22/7×49
Answer: 77/3 cm² ≈ 25.67 cm²
Area = 60/360×22/7×49
Answer: 77/3 cm² ≈ 25.67 cm²
4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (i) minor sector (that subtends 90° at the centre), and (ii) major sector (that subtends 270° at the centre). (Use 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
(ii) Major sector = 270/360×3.14×100 = 235.5 cm²
5. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and √3 ≈ 1.73.)
Minor sector = ¼πr² = 176.625 cm² approximately.
Equilateral triangle area = √3/4×15² ≈ 97.3125 cm².
Minor segment ≈ 79.31 cm².
Major segment = circle area − minor segment ≈ 627.69 cm².
Equilateral triangle area = √3/4×15² ≈ 97.3125 cm².
Minor segment ≈ 79.31 cm².
Major segment = circle area − minor segment ≈ 627.69 cm².
6. A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.
Area of one sector = 120/360×22/7×28² = 821.33 cm² approximately.
Two wipers clean approximately:
1642.67 cm²
Two wipers clean approximately:
1642.67 cm²
*7. A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to πr²/6 − (√3/4)r².
Minor segment = minor sector − triangle.
Minor sector = πr²/6.
The triangle formed by the two radii and the chord is equilateral, so its area is √3r²/4.
Therefore:
Minor segment = πr²/6 − √3r²/4.
Minor sector = πr²/6.
The triangle formed by the two radii and the chord is equilateral, so its area is √3r²/4.
Therefore:
Minor segment = πr²/6 − √3r²/4.
*8. An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is equal to 3√3/(4π) ≈ 0.413.
Side of equilateral triangle = √3r.
Triangle area = √3/4×3r² = 3√3r²/4.
Circle area = πr².
Ratio = 3√3/(4π) ≈ 0.413.
Triangle area = √3/4×3r² = 3√3r²/4.
Circle area = πr².
Ratio = 3√3/(4π) ≈ 0.413.
*9. A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to 2/π ≈ 0.637.
Diagonal of square = 2r.
Side = √2r.
Area of square = 2r².
Circle area = πr².
Ratio = 2/π ≈ 0.637.
Side = √2r.
Area of square = 2r².
Circle area = πr².
Ratio = 2/π ≈ 0.637.
*10. A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to 3√3/(2π) ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?
A regular hexagon inscribed in a circle of radius r consists of six equilateral triangles of side r.
Hexagon area = 6×√3r²/4 = 3√3r²/2.
Circle area = πr².
Ratio = 3√3/(2π) ≈ 0.827.
This is exactly twice 3√3/(4π), the answer to Question 8.
Hexagon area = 6×√3r²/4 = 3√3r²/2.
Circle area = πr².
Ratio = 3√3/(2π) ≈ 0.827.
This is exactly twice 3√3/(4π), the answer to Question 8.
🟥 End-of-Chapter Exercises
In the problems below, unless stated otherwise, use the approximation 22/7 for π.
1. Identities in algebra can sometimes be shown as area relationships. For example:
The figure shown corresponds to the identity
(a + b)² = a² + 2ab + b².
Do you see how?
Draw figures corresponding to the identities (a + b)(a – b) = a² – b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
For (a+b)², the large square can be divided into one a² square, one b² square and two rectangles of area ab each.
Thus total area = a²+ab+ab+b² = a²+2ab+b².
Thus total area = a²+ab+ab+b² = a²+2ab+b².
2. An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Base = 40−30 = 10 cm.
Height = √(15²−5²)=10√2 cm.
Area = 50√2 cm².
Height = √(15²−5²)=10√2 cm.
Area = 50√2 cm².
3. An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?
Height = 12 cm.
Half base = 5 cm.
Equal side = √(12²+5²)=13 cm.
Answer: 13 cm each.
Half base = 5 cm.
Equal side = √(12²+5²)=13 cm.
Answer: 13 cm each.
4. The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.
½×12×other leg = 54
Other leg = 9 cm.
Hypotenuse = √(12²+9²)=15 cm.
Perimeter = 36 cm.
Other leg = 9 cm.
Hypotenuse = √(12²+9²)=15 cm.
Perimeter = 36 cm.
5. The sides of a triangle are in the ratio 2: 3: 4, and its perimeter is 45 cm. Find its area.
Sides = 10, 15 and 20 cm.
s = 22.5 cm.
Heron’s formula gives:
Area = (15√15)/4?
Use the complete substitution carefully.
s = 22.5 cm.
Heron’s formula gives:
Area = (15√15)/4?
Use the complete substitution carefully.
6. The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.
Since 7²+24²=25², it is right-angled.
Area = ½×7×24 = 84 cm².
Heron’s formula also gives 84 cm².
Area = ½×7×24 = 84 cm².
Heron’s formula also gives 84 cm².
7. If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.
One revolution = π×60 = 60π cm.
100 revolutions = 6000π cm.
Using 22/7: 18857.14 cm ≈ 188.57 m.
100 revolutions = 6000π cm.
Using 22/7: 18857.14 cm ≈ 188.57 m.
8. Find the area of a quadrant of a circle whose circumference is 66 cm.
2πr=66 gives r=10.5 cm.
Quadrant area = ¼×22/7×10.5².
Answer: 86.625 cm².
Quadrant area = ¼×22/7×10.5².
Answer: 86.625 cm².
9. The wheel of a car has an outer radius of 28 cm. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km.
One turn = 2π×28 = 176 cm.
1 km = 100000 cm.
Number of turns = 100000/176 ≈ 568.18 turns.
1 km = 100000 cm.
Number of turns = 100000/176 ≈ 568.18 turns.
*10. Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?
No. Different rectangles can have the same area and perimeter without being congruent. A suitable numerical example can be used to demonstrate this.
11. 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., 1/2(a + b)h.
Take two copies of the trapezium and arrange them to form a parallelogram.
The parallelogram has base a+b and height h.
Its area = (a+b)h.
One trapezium therefore has area ½(a+b)h.
The parallelogram has base a+b and height h.
Its area = (a+b)h.
One trapezium therefore has area ½(a+b)h.
12. By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).
Divide the trapezium into two triangles. Add their areas using ½bh. The common height is h, giving ½(a+b)h.
13. Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?
Arrange the two identical trapeziums with one inverted. They form a parallelogram with base a+b and height h. Therefore one trapezium has area ½(a+b)h.
14. Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.
The diagonals divide the kite into four right triangles. Adding their areas gives ½d₁d₂.
15. Three problems about fitting congruent shapes together:
(i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see!
(ii) ∆ABC has sides a, b, c, and ∆PQR has sides 2a, 2b, 2c. Show that ∆PQR has 4 times the area of ∆ABC. Does this mean that 4 copies of ∆ABC will fit into ∆PQR? Check and see!
(iii) ∆ABC has sides a, b, c, and ∆PQR has sides 3a, 3b, 3c. Show that ∆PQR has 9 times the area of ∆ABC. Does this mean that 9 copies of ∆ABC will fit into ∆PQR? Check and see!
(i) Area scales by 2² = 4.
(ii) Triangle area also scales by 2² = 4.
(iii) Triangle area scales by 3² = 9.
(ii) Triangle area also scales by 2² = 4.
(iii) Triangle area scales by 3² = 9.
*16. Fig. 6.43: What fraction of the triangle is shaded? Fig. 6.44: What fraction of the square is shaded?
Triangle: 5/6
Square: 1/2
Square: 1/2
17. Fig. 6.45: What fraction of the rectangle is covered by the circles? Fig. 6.46: What fraction of the rectangle is covered by the circles?
For the arrangements shown, the fraction is π/4.
18. Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!
The fraction occupied is π/4, independent of the number of equal circles.
*19. The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm². Find the perimeter of each small rectangle.
Perimeter = 18√10/5 cm ≈ 11.38 cm.
*20. Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.
The two triangles have equal area. A suitable dissection and rearrangement demonstrates the equality.
*21. The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B. Show that A and B have equal area.
Quarter-circle area = πs²/4.
The two semicircles together also have area πs²/4.
Removing the common regions gives A = B.
The two semicircles together also have area πs²/4.
Removing the common regions gives A = B.
*22. In Fig. 6.50, four semicircles have been drawn within the given square whose side is 2 units. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.
Perimeter = 4π units.
Area = 2π − 4 square units.
Area = 2π − 4 square units.
*23. In Fig. 6.51 we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is l. Show that the area of the green region enclosed between the two circles is 1/4 πl².
OA is perpendicular to BC and AB = BC/2 = l/2.
Using Pythagoras:
R²−r² = l²/4.
Therefore annular area = π(R²−r²) = πl²/4.
Using Pythagoras:
R²−r² = l²/4.
Therefore annular area = π(R²−r²) = πl²/4.
*24. In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).
If the sides are a, b and c, then by Pythagoras a²+b²=c².
Multiplying by π/4 gives:
Area(A)+Area(B)=Area(C).
Multiplying by π/4 gives:
Area(A)+Area(B)=Area(C).
*25. Fig. 6.53 shows two circles passing through each other’s centres. Find the area of the region enclosed by the two circles in terms of the common radius r.
The common region is made from two 60° sectors minus two equilateral triangles.
Area = (π/3 − √3/2)r².
Area = (π/3 − √3/2)r².
*26. In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Show that the area of the rectangle is 2(A + C)(B + C)/C.
Use the triangle area formula for the three triangles and eliminate the common lengths. This gives:
Area of rectangle = 2(A+C)(B+C)/C.
Area of rectangle = 2(A+C)(B+C)/C.
*27. In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.
The semicircle on AB and the quarter circle AOC have equal areas because AB = √2r.
The common triangular portion is subtracted from both sides.
Therefore the two shaded regions have equal areas.
The common triangular portion is subtracted from both sides.
Therefore the two shaded regions have equal areas.
📝 Quick Revision Box
| Concept | Formula |
|---|---|
| Rectangle perimeter | 2(l+b) |
| Square perimeter | 4a |
| Circle circumference | 2πr = πd |
| Arc length | 2πr × θ/360° |
| Rectangle area | l×b |
| Triangle area | ½bh |
| Parallelogram area | bh |
| Circle area | πr² |
| Sector area | θ/360° × πr² |
| Heron’s Formula | √[s(s−a)(s−b)(s−c)] |
| Trapezium area | ½(a+b)h |
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