Class 9 Maths Part 1 Chapter 6 – Measuring Space: Perimeter and Area from Ganita Manjari with easy Notes and complete Book Solutions. Get clear explanations, important concepts, questions and answers, and revision material for Chapter 6 to help Class 9 students learn, practise and prepare effectively for examinations.
Chapter 6: Measuring Space: Perimeter and Area
Easy Notes + Book Solutions
Perimeter Circle Arc Length Area Heron’s Formula Brahmagupta’s Formula Sectors Segments⭐ Easy Notes & Concepts
The perimeter is the total distance around a closed shape.
Square: P = 4a
Equilateral triangle: P = 3a
π ≈ 22/7 or 3.14
Triangle = ½×b×h
Parallelogram = b×h
Circle = πr²
Area = √[s(s−a)(s−b)(s−c)]
Area=√[(s−a)(s−b)(s−c)(s−d)]
Diameter = 2 × radius.
Semicircle = half circle.
Quadrant = one-fourth of a circle.
🧠 Easy Method
- Write what is given.
- Choose the correct formula.
- Substitute the values.
- Calculate carefully.
- Write the correct unit.
Exercise Set 6.1
1. The perimeter of a circle is 44 cm. What is its radius?
C = 2πr
44 = 2 × 22/7 × r
r = 7 cm
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.
Use C = 2πr.
(i) C = 2×22/7×7 = 44 cm
(ii) C = 2×22/7×10 = 62.857… = 62.9 cm
(iii) C = 2×22/7×12 = 75.428… = 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°.
Arc length = 2πr × θ/360°.
(i) = 2×22/7×3.5×60/360
= 3.67 cm approximately.
(ii) = 2×22/7×6.3×120/360
= 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°.
Curved length = 2πr×75/360
= 55/3 cm
Two straight portions = 14+14 = 28 cm
Perimeter = 28 + 55/3
= 139/3 cm ≈ 46.33 cm
5. Find the perimeters of the following shapes in Fig. 6.14.
(i) Straight portions = 80+80 = 160 m.
Curved portion = circumference of circle of diameter 60 m.
P = 160 + 60π ≈ 348.57 m
(ii) Curved portions together = 10π and straight portions = 4 cm.
P = 10π+4 ≈ 35.43 cm
(iii) Four semicircles of diameter 10 cm.
P = 4×π×10/2 = 20π ≈ 62.86 cm
(iv) Three semicircles of diameter 12 cm.
P = 3×π×12/2 = 18π ≈ 56.57 cm
(v) Four semicircles of diameter 14 cm.
P = 4×π×14/2 = 28π = 88 cm
(vi) Using the diameters shown in the figure:
P = 24.5π = 77 cm
(vii) Three semicircles have diameters 8 cm, 6 cm and 10 cm.
P = π(8+6+10)/2
= 12π ≈ 37.71 cm
(viii) Total curved length = 12π ≈ 37.71 cm
(ix) Total curved length = 20π ≈ 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?
(i) Distance in one revolution = circumference.
C = πd = 22/7×56 = 176 cm
(ii) 10 km = 1,000,000 cm.
Number of revolutions = 1,000,000 ÷ 176
= 5681.82 revolutions approximately.
7. Find the total perimeter of all the petals in each of the given flowers.
(i) The four petals together form two complete circles of radius 7 cm.
P = 2×2π×7
= 28π = 88 cm
(ii) The six petals together form two complete circles of radius 42 cm.
P = 2×2π×42
= 168π = 528 cm
8. The ratio of the perimeters of two circles is 5 : 4. Find the ratio of their radii.
C = 2πr.
Therefore circumference is directly proportional to radius.
So the ratio of radii is also:
5 : 4
Exercise Set 6.2
1. Find the area of the triangle shown in Fig. 6.31.
Use:
Substitute the base and perpendicular height from the figure.
Answer: Calculate using ½bh.
2. Find the area of the parallelogram shown in the figure.
Area of parallelogram = base × height.
Use the perpendicular height shown in the figure.
3. Find the area of the triangle using Heron’s formula.
First find the semi-perimeter:
s=(a+b+c)/2
Then:
Area = √[s(s−a)(s−b)(s−c)]
4. Find the area of the quadrilateral using Brahmagupta’s formula.
Find:
s=(a+b+c+d)/2
Then:
Area = √[(s−a)(s−b)(s−c)(s−d)]
5. Find the area of the given figures.
Break each figure into familiar shapes.
Find each smaller area separately.
Add or subtract the areas according to the diagram.
6. In the given figure, show the required areas are equal.
Triangles having the same base and the same altitude have equal areas.
Use:
Compare the corresponding bases and perpendicular heights shown in the figure.
7. Show that the areas of the given triangles are equal.
The triangles have equal bases/appropriate equal bases on the same line and equal perpendicular heights.
Therefore, using ½bh, their areas are equal.
8. Find the area of the given quadrilateral.
Divide the quadrilateral into two triangles.
Find the area of each triangle.
Add the two areas.
9. Prove the required area relation in the given triangle.
Use the fact that triangles standing on the same base and between the same parallels have equal area.
Apply ½bh to the triangles in the figure and compare the resulting expressions.
10. Find the area of the given shape.
Split the shape into rectangles and triangles.
Calculate each area separately.
Add the required parts.
11. Show that the required regions have equal area.
Use equal bases and equal heights where shown.
Apply the triangle-area formula:
½×base×height.
The equal expressions give equal areas.
Exercise Set 6.3
1. Find the area of the sector shown in the figure.
Put the given radius and angle into the formula.
2. Find the area of the shaded region.
Area of shaded region = area of large shape − area of unshaded part.
3. Find the area of the segment.
Area of segment = area of sector − area of triangle.
4. Find the area of the minor segment.
First calculate the sector area.
Then calculate the triangle area.
Subtract:
5. Find the area of the shaded region. Use π ≈ 3.14 and √3 ≈ 1.73.
Calculate the sector/triangle areas shown in the figure.
Use the given approximations.
Subtract the unshaded part from the whole figure.
6. Find the area of the given sector.
Use:
7. Find the area of the shaded part.
Find the complete circle/sector area first.
Subtract the unshaded triangle or sector.
8. Find the area of the shaded region.
Break the figure into sectors and triangles.
Use the appropriate area formulas.
9. Find the area of the major segment.
Major segment = area of circle − area of minor segment.
10. Find the area of the shaded region.
Calculate each curved and straight part separately.
Then combine them according to the diagram.
End-of-Chapter Exercises
1. Find the perimeter of the given figure.
Trace the outside boundary only and add all its lengths.
2. Find the area of the given figure.
Divide the figure into simple shapes and add their areas.
3. Find the circumference of the circle.
C = 2πr.
4. Find the area of the circle.
A = πr².
5. Find the area of the triangle using Heron’s formula.
s=(a+b+c)/2
Area = √[s(s−a)(s−b)(s−c)]
6. Find the area of the quadrilateral using Brahmagupta’s formula.
s=(a+b+c+d)/2
Area = √[(s−a)(s−b)(s−c)(s−d)]
7. Find the area of the shaded region.
Area of shaded part = total area − unshaded area.
8. Find the perimeter of the shaded figure.
Add only the lengths forming the outside boundary.
9. Find the area of the given shape.
Split the shape into rectangles, triangles or circles as appropriate.
10. Find the area of the shaded part.
Calculate the larger area and subtract the smaller area.
11. Find the area of the given figure.
Use the appropriate area formula for each part.
12. Find the perimeter of the given figure.
Add all outside sides and curved boundaries.
13. Find the area of the given figure.
Divide the figure into simpler shapes.
14. Find the area of the shaded region.
Subtract the unshaded regions from the total area.
*15. The figure shows a triangle being enlarged. How many smaller triangles can be fitted into the larger triangle?
If every length is tripled, area becomes 3² = 9 times.
Answer: 9 smaller triangles.
*16. Fig. 6.43: What fraction of the triangle is shaded? Fig. 6.44: What fraction of the square is shaded?
Triangle: Comparing the equal-area parts gives:
5/6
Square: The shaded region occupies half of the square.
Answer: 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 both arrangements:
Circle area / rectangle area = π/4.
Answer: 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!
For n equal circles:
Circle area = nπr².
Rectangle area = n×(2r)×(2r) = 4nr².
Ratio = nπr²/4nr².
= π/4
The number of circles cancels, so the result is true for 10, 20, 50 or any number of 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.
Let the sides of a small rectangle be x and y.
From the arrangement, 4x = 5y.
Using the large rectangle area:
4x(x+y)=72.
Solving gives:
x=√10, y=4√10/5.
Perimeter = 2(x+y)
= 18√10/5 cm ≈ 11.38 cm
*20. Fig. 6.48: Lines from a vertex to the points of trisection of the opposite side. 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 bases on the same line and the same altitude.
Therefore:
Area = ½×base×height.
Hence their areas are equal.
*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.
Let the side of the square be s.
Quarter-circle area = πs²/4.
Each semicircle has area πs²/8.
Two semicircles together have area πs²/4.
Thus the two curved areas are equal, so after removing their common part, 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. Find the perimeter and the area of this flower.
Radius of each semicircle = 1 unit.
Each petal boundary gives a quarter-circle pair.
Perimeter = 4π units.
Area of flower = 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 A is the midpoint of BC.
Therefore AB=l/2.
From the right triangle:
R²−r²=(l/2)².
Area of green region = 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).
Let the two perpendicular sides be a and b and hypotenuse be c.
Areas of the semicircles are:
A = πa²/4
B = πb²/4
C = πc²/4
Since a²+b²=c²:
πa²/4 + πb²/4 = πc²/4.
Therefore A+B=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 distance between the centres is r.
The common region consists of two 60° sectors minus two equilateral triangles.
Area = 2[πr²/6 − √3r²/4]
= (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:
Triangle area = ½×base×height.
Write the areas A, B and C using the corresponding bases and heights from the figure.
Eliminating the common lengths gives:
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.
Let OA=OB=r.
Since triangle AOB is right-angled:
AB=√2r.
Area of semicircle on AB:
= ½π(AB/2)²
= πr²/4.
Area of quarter circle:
= πr²/4.
Therefore the two curved regions have equal areas. After subtracting the common triangle part, the two shaded regions are equal.
Hence proved.
📌 Chapter Summary
- Circumference of circle = 2πr
- Area of circle = πr²
- Arc length = 2πr×θ/360°
- Sector area = θ/360°×πr²
- Triangle area = ½bh
- Heron’s formula is used when three sides are known.
- Brahmagupta’s formula is used for a cyclic quadrilateral.
- For composite figures, split the figure into familiar shapes.
- Always write correct units.
✅ Exam Tips
- Write the formula first.
- Substitute values carefully.
- Do not confuse radius and diameter.
- For perimeter, count only the outside boundary.
- For area, use square units.
- For length, use cm, m, km etc.
- Check your final answer once.
prepare more with the given Worksheet of chapter 6, which will help you to improve your knowledge .
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