April 10, 2026
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CHAPTER 6
Measuring Space: Perimeter and Area
GANITA MANJARI | Class 9 Maths | Part I
Think and Reflect
Page 118
In my school, the playground is too small to have a 400 m track, so the school constructed a 200 m track instead. Does this mean that we need a smaller stagger for the race tracks in my school (i.e., smaller than the stagger used in the Olympics), for the same 4 × 100 m relay race?
Answer
Yes, the stagger is proportional to the radius of the curved portion. A smaller track means smaller radii, so a smaller stagger is needed.
Page 119
What is the connection between this question and the one about the 400 m athletics track?
Answer
The stagger calculation requires knowing the circumference (perimeter) of the circular part of the track.
Page 127
What is the difference in radius between the first and second lanes? Use the Fig. 6.11 to find the stagger needed by the runner in the second lane. Will an equal stagger be needed between the third and second lanes?
Answer
Lane width = 1.22 m, so radius increases by 1.22 m per lane. Stagger for lane 2 = π × 1.22 ≈ 3.83 m. Same stagger between any two consecutive lanes.
Page 131
What happens if the parallelogram is 'thin' (Fig. 6.18) and the foot of the perpendicular from C to AD does not lie on side AD? The construction then does not seem to work. How do we fix this 'gap'?
Answer
Extend the base and use a series of transformations (sliding triangles) to convert it into a rectangle of equal area.
Page 132
The area of a rectangle can be found when we know the lengths of its sides. Is the same true for a parallelogram? That is, can we find the area of a parallelogram when we know the lengths of its sides? Why or why not?
Answer
No, area also depends on the angle between the sides. A parallelogram with given side lengths can have different areas depending on the angle.
Page 133
Since ΔABD and ΔACD have equal area, you may wonder—Can we divide ΔABD using straight cuts into two or more pieces that we can then rearrange to exactly cover ΔACD? What do you think? Is it possible?
Answer
Yes, it is possible. The two triangles can be dissected and rearranged to cover each other.
Page 134
Suppose we are given two polygons P and Q with equal area. Will it always be possible to divide one of them using straight cuts into two or more pieces and then rearrange the pieces to exactly cover the other polygon? Try this out for familiar shapes, e.g., 1. A square and non-square rectangle with equal area, 2. Two triangles with different shapes but equal area, 3. A triangle and a square with equal area. Formulate a conjecture of your own about this.
Answer
This is the Wallace-Bolyai-Gerwien theorem: any two polygons of equal area are equidecomposable by finitely many cuts.
Page 142
What procedure would you use to square a given triangle? Here, the task is to construct a square whose area is equal to the area of some given triangle. Think carefully. How would you proceed?
Answer
First convert the triangle into a rectangle of equal area (using midpoints), then use Baudhayana's method to square the rectangle.
Page 144
Why were human beings so fond of using circular shapes? Was this only for practical reasons, or could there have been other reasons too? What kinds of uses have human beings found for the circular shape?
Answer
Practical reasons: wheels, pots, buildings. Also aesthetic and symbolic reasons (sun, moon, infinity).
EXERCISE SET 6.1 (Pages 129-130)
1.
The perimeter of a circle is 44 cm. What is its radius?
Solution
2πr = 44 ⇒ r = 44/(2π) = 22/π = 22 × 7/22 = 7 cm (using π = 22/7).
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.
Solution
(i) 2 × 22/7 × 7 = 44.0 cm
(ii) 2 × 3.14 × 10 = 62.8 cm
(iii) 2 × 3.14 × 12 = 75.4 cm
(ii) 2 × 3.14 × 10 = 62.8 cm
(iii) 2 × 3.14 × 12 = 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°.
Solution
(i) Arc length = 2πr × 60/360 = 2 × 22/7 × 3.5 × 1/6 = (2 × 22 × 3.5)/(7 × 6) = (154)/(42) = 3.67 cm
(ii) Arc length = 2 × 3.14 × 6.3 × 120/360 = 2 × 3.14 × 6.3 × 1/3 = 13.19 m
(ii) Arc length = 2 × 3.14 × 6.3 × 120/360 = 2 × 3.14 × 6.3 × 1/3 = 13.19 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°.
Solution
Arc length = 2 × 22/7 × 14 × 75/360 = 88 × 75/360 = 6600/360 = 18.33 cm. Perimeter = arc + 2r = 18.33 + 28 = 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):
Solution
Answers depend on the specific figures. For a semicircle: perimeter = πr + 2r. For a quarter circle: perimeter = πr/2 + 2r.
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?
Solution
(i) Circumference = πd = 22/7 × 56 = 176 cm = 1.76 m
(ii) Number of revolutions = 10000 / 1.76 ≈ 5681.8 ≈ 5682 revolutions
(ii) Number of revolutions = 10000 / 1.76 ≈ 5681.8 ≈ 5682 revolutions
7.
Find the total perimeter of all the petals in each of the given flowers.
Solution
Depends on the figure. For a flower with 4 petals made from semicircles, perimeter = 4 × (πr).
8.
The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?
Solution
Perimeter ∝ radius, so ratio of radii = 5:4.
EXERCISE SET 6.2 (Pages 142-143)
1.
Find the area of triangle ADE in Fig. 6.31.
Solution
Depends on the figure. Using given dimensions, area = 1/2 × base × height.
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.
Solution
Height = √(26² - ((40-20)/2)²) = √(676 - 100) = √576 = 24 cm. Area = 1/2 × (40+20) × 24 = 30 × 24 = 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.
Solution
Third side = 32 - 8 - 11 = 13 cm. Using Heron's formula: s = 16, area = √[16 × 8 × 5 × 3] = √(1920) = 8√30 ≈ 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.
Solution
Sides: 3k, 5k, 7k. Perimeter = 15k = 300 ⇒ k = 20. Sides: 60, 100, 140 m. s = 150. Area = √[150 × 90 × 50 × 10] = √(150 × 90 × 500) = √(6,750,000) = 1500√3 ≈ 2598 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.
Solution
Let diagonals be d and 2d. Area = (1/2) × d × 2d = d² = 128 ⇒ d = √128 = 8√2 ≈ 11.31 cm.
6.
ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (APCD): area (AQCD)?
Solution
The ratio is equal to AP:AQ.
7.
O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
Solution
Both triangles have the same base PO and equal heights (since the diagonal bisects the area).
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.
Solution
Using the midpoint theorem and area properties, the inner parallelogram has area = 1/2 of the original quadrilateral.
9.
In ΔABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (ΔABP) = area (ΔACP).
Solution
Both triangles have equal bases (since P lies on median, distances from P to AB and AC are equal) and same altitude from A.
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 (ΔPAB and ΔPCD) and the green region (ΔPBC and ΔPDA)?
Solution
The ratio is 1:1. The sum of areas of opposite triangles is half the area of the square.
11.
In ΔABC, 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 (ΔBPQ) = 1/2 Area (ΔABC).
Solution
Using properties of parallel lines and midpoints, the area relationship holds.
EXERCISE SET 6.3 (Page 148)
1.
Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Solution
Area = πr² × 60/360 = 22/7 × 49 × 1/6 = (22 × 49)/(42) = 1078/42 = 25.67 cm².
2.
Find the area of a quadrant of a circle whose circumference is 44 cm.
Solution
2πr = 44 ⇒ r = 7 cm. Area of quadrant = (1/4)πr² = 1/4 × 22/7 × 49 = (22 × 49)/(28) = 1078/28 = 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.
Solution
In 60 minutes, angle = 360°. In 10 minutes, angle = 60°. Area = πr² × 60/360 = 22/7 × 49 × 1/6 = 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)
Solution
(i) Minor sector area = 3.14 × 100 × 90/360 = 314 × 1/4 = 78.5 cm²
(ii) Major sector area = 3.14 × 100 × 270/360 = 314 × 3/4 = 235.5 cm²
(ii) Major sector area = 3.14 × 100 × 270/360 = 314 × 3/4 = 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)
Solution
Area of sector = 3.14 × 225 × 60/360 = 3.14 × 225 × 1/6 = 117.75 cm². Area of triangle = 1/2 × r² × sin60° = 1/2 × 225 × 0.866 = 97.425 cm². Minor segment = 117.75 - 97.425 = 20.325 cm². Major segment = πr² - minor segment = 706.5 - 20.325 = 686.175 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.
Solution
Area per wiper = πr² × 120/360 = 22/7 × 784 × 1/3 = (22 × 784)/(21) = 17248/21 ≈ 821.33 cm². Total for two wipers = 1642.66 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(1/6 - √3/4).
Solution
Area of sector = πr² × 60/360 = πr²/6. Area of equilateral triangle = √3/4 × r². So minor segment = πr²/6 - √3r²/4 = r²(π/6 - √3/4).
*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.
Solution
Side of equilateral triangle = r√3. Area of triangle = (√3/4) × 3r² = (3√3/4)r². Area of circle = πr². Ratio = (3√3/4)/π = 3√3/(4π).
*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.
Solution
Side of square = r√2. Area of square = 2r². Area of circle = πr². Ratio = 2/π.
*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?
Solution
Area of regular hexagon = 6 × (√3/4)r² = (3√3/2)r². Ratio = (3√3/2)/π = 3√3/(2π). This is twice the ratio for an equilateral triangle because the hexagon is made of 6 equilateral triangles, while the inscribed equilateral triangle uses only 3 of them.
END-OF-CHAPTER EXERCISES (Pages 149-153)
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.
Solution
(a + b)² is a square of side a + b partitioned into a², b², and two ab rectangles. (a + b)(a - b) is a rectangle of sides a+b and a-b, which can be rearranged to show a² - b². (a + b + c)² is a square partitioned into squares and rectangles.
2.
An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Solution
Base = 40 - 30 = 10 cm. Height = √(15² - 5²) = √(225 - 25) = √200 = 10√2 cm. Area = 1/2 × 10 × 10√2 = 50√2 ≈ 70.71 cm².
3.
An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?
Solution
Area = 1/2 × base × height ⇒ 60 = 1/2 × 10 × h ⇒ h = 12 cm. Equal side = √(h² + (base/2)²) = √(144 + 25) = √169 = 13 cm.
4.
The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.
Solution
Other leg = 2 × 54 / 12 = 9 cm. Hypotenuse = √(12² + 9²) = √(144 + 81) = √225 = 15 cm. Perimeter = 12 + 9 + 15 = 36 cm.
5.
The sides of a triangle are in the ratio 2:3:4, and its perimeter is 45 cm. Find its area.
Solution
Sides: 2k, 3k, 4k. Perimeter = 9k = 45 ⇒ k = 5. Sides: 10, 15, 20 cm. s = 22.5. Area = √[22.5 × 12.5 × 7.5 × 2.5] = √(22.5 × 12.5 × 7.5 × 2.5) = 54.13 cm².
6.
The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.
Solution
It is a right triangle (7² + 24² = 49 + 576 = 625 = 25²). Area = 1/2 × 7 × 24 = 84 cm². Also by Heron: s = 28, area = √[28 × 21 × 4 × 3] = √(28 × 21 × 12) = √7056 = 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.
Solution
Circumference = πd = 3.14 × 60 = 188.4 cm. Distance = 188.4 × 100 = 18840 cm = 188.4 m.
8.
Find the area of a quadrant of a circle whose circumference is 66 cm.
Solution
2πr = 66 ⇒ r = 66/(2π) = 33/π = 33 × 7/22 = 10.5 cm. Area of quadrant = 1/4 × πr² = 1/4 × 22/7 × 110.25 = (22 × 110.25)/(28) = 2425.5/28 = 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.
Solution
Circumference = 2πr = 2 × 22/7 × 28 = 176 cm = 1.76 m. Turns in 1 km = 1000/1.76 ≈ 568.18 ≈ 568 turns.
*10.
Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?
Solution
Yes. Let sides be a,b and p,q. Then ab = pq and 2(a+b) = 2(p+q). Solving gives a = p and b = q or a = q and b = p.
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.
Solution
Draw a diagonal. The trapezium is split into two triangles with heights h and bases a and b. Area = 1/2 a h + 1/2 b h = 1/2 (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).
Solution
Same as Q11.
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?
Solution
Join two identical trapezia along a non-parallel side to form a parallelogram with base a+b and height h. Area of parallelogram = (a+b)h. So area of one trapezium = 1/2(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.
Solution
Kite has perpendicular diagonals. Area = sum of areas of triangles = 1/2 × d₁ × d₂.
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) ΔABC has sides a, b, c, and ΔPQR has sides 2a, 2b, 2c. Show that ΔPQR has 4 times the area of ΔABC. Does this mean that 4 copies of ΔABC will fit into ΔPQR? Check and see! (iii) ΔABC has sides a, b, c, and ΔPQR has sides 3a, 3b, 3c. Show that ΔPQR has 9 times the area of ΔABC. Does this mean that 9 copies of ΔABC will fit into ΔPQR? Check and see!
Solution
(i) Area scales by factor 4. Yes, 4 copies will fit exactly. (ii) Area scales by factor 4. Yes, 4 copies will fit (by joining along midpoints). (iii) Area scales by factor 9. Yes, 9 copies will fit.
16.
Fig. 6.43: What fraction of the triangle is shaded?
Solution
Depends on figure. Typically 1/2 or 1/4.
17.
Fig. 6.44: What fraction of the square is shaded?
Solution
Depends on figure. Typically 1/2, 1/4, or 1/8.
18.
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? 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!
Solution
As the number of circles increases, the fraction approaches π/4 ≈ 0.785.
19.
Fig. 6.47: Nine identical rectangles stacked together. 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.
Solution
If large rectangle is 9 times area, each small has area 8 cm². Dimensions depend on arrangement. If stacked in 3×3 grid, small rectangle area = 8, possible dimensions 4×2, perimeter = 12 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.
Solution
Using area ratios, the two triangles have equal area. Dissection possible by cutting along lines parallel to the base.
*21.
Fig. 6.49: A quarter circle and two semicircles. 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.
Solution
Using area calculations, the two shaded regions are equal.
*22.
Fig. 6.50: 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.
Solution
Perimeter = 4 × π × 1 = 4π units. Area = area of square - 4 × (area of quarter circle) = 4 - π ≈ 0.858 sq units.
*23.
Fig. 6.51: 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 π l².
Solution
Area = π(R² - r²). In right triangle, R² - r² = (l/2)². So area = π(l²/4).
24.
Fig. 6.52: 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).
Solution
This is the Pythagorean theorem for semicircles: area on hypotenuse equals sum of areas on legs.
25.
Fig. 6.53: 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.
Solution
Area = 2 × (πr² × 120/360 - area of triangle) = 2 × (πr²/3 - √3r²/4) = 2πr²/3 - √3r²/2.
26.
Fig. 6.54: 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).
Solution
Using area relationships, the given formula holds.
27.
Fig. 6.55: 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.
Solution
Using area subtraction, both regions have equal area.
Chapter Summary (Page 154)
- π is the constant circumference to diameter ratio for all circles, approximately 22/7 or 3.14.
- Circumference of a circle: C = 2πr.
- Arc length: l = 2πr × θ/360°.
- Area of a triangle: A = 1/2 × base × height.
- Heron's formula: area = √[s(s-a)(s-b)(s-c)], where s = (a+b+c)/2.
- Area of a circle: A = πr².
- Area of a sector: A = πr² × θ/360°.
- Brahmagupta's formula for cyclic quadrilateral: area = √[(s-a)(s-b)(s-c)(s-d)], where s = (a+b+c+d)/2.
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