Ganita Manjari Class 9 Chapter 6 Solutions Measuring Space Perimeter and Area

By using Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 6 Measuring Space Perimeter and Area NCERT Solutions, students can improve their problem-solving skills.

Measuring Space Perimeter and Area Class 9 Solutions

Class 9 Ganita Manjari Chapter 6 Solutions

Class 9 Maths Ganita Manjari Chapter 6 Solutions Measuring Space Perimeter and Area

Think and Reflect (NCERT Textbook Page No. 118)

Question 1.
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?
Solution:
In a 400 m Olympic track, the stagger is necessary because the outer lanes are longer than the inner lanes. To make the race fair, runners in outer lanes start ahead so that each covers the same distance. Now, in your school’s 200 m track, the same principle applies: the outer lanes are still longer than the inner lanes. But here’s the key:

• The extra distance between lanes depends on the radius of the curves.
• Since your track is smaller (200 m vs. 400 m), the curves are tighter, and the difference in distance between lanes is less than in a 400 m track.
• Therefore, the stagger required is smaller than the Olympic stagger.
• So yes, for the same 4 × 100 m relay, your school’s track will need a smaller stagger compared to the Olympics.

Think and Reflect (NCERT Textbook Page No. 134)

Question 1.
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 a 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.
Think of various rectangles with perimeter 40 units (the sides do not have to be integers).
1. How many such rectangles are there?
2. Among them, is there one whose area is the largest? What are its dimensions?
3. Among all these rectangles, is there one whose area is the smallest?
What are its dimensions? Do either of these answers come as a surprise to you?
Solution:
Part 1: Equal-Area Polygons and Rearrangement
This is about geometric dissection (cutting and rearranging shapes).
1. Square and non-square rectangle of equal area
Yes, you can cut a square into pieces and rearrange them to form a rectangle of equal area.
Example: A 4 × 4 square (area 16) can be cut and rearranged into a 2 × 8 rectangle.
2. Two triangles of equal area but different shapes
Yes, it’s possible. Any triangle can be dissected and rearranged into another triangle of equal area, though the cuts may be more complex.
3. Triangle and square of equal area
Yes, in principle. A triangle of area 25 can be cut and rearranged into a square of side 5 (area 25). This is a classic result in geometry.
Conjecture: Any two polygons of equal area can be dissected into a finite number of straight-line pieces and rearranged to form each other.

Part 2: Rectangles with Perimeter 40 Units
Let’s analyze:
Perimeter formula: P = 2(l + w) = 40
⇒ l + w = 20
1. Infinitely many, because l and w can take infinitely many positive values that sum to 20 (not restricted to integers).
2. Area: A = l × w
With l + w = 20, the product l.w is maximized when l = w = 10.
Largest area = 10 × 10 = 100.
Dimensions: 10 × 10 (a square).
3. If one side is very small (approaching 0), the other side approaches 20.
Area approaches 0 × 20 = 0.
So the smallest area is arbitrarily close to 0, though not exactly 0 since sides must be positive.
Reflection
The maximum area rectangle for a fixed perimeter is always a square — a beautiful result showing symmetry and balance.
The minimum area rectangle tends toward a degenerate case (very long and thin), which makes intuitive sense: stretching a shape reduces its compactness.

Think and Reflect (NCERT Textbook Page No. 144)

Question 1.
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?
Solution:
Circles have captivated humans for both practical and symbolic reasons.
Practical Reasons

• Efficiency in motion: Wheels, gears, and pulleys are circular because rotation is smooth and continuous.
• Strength and stability: Circular arches and domes distribute weight evenly, making them ideal in architecture (think Roman aqueducts or the Taj Mahal’s dome).
• Space and symmetry: Circular containers (like pots or bowls) maximize volume for a given perimeter.

Symbolic & Cultural Reasons

• Unity and eternity: A circle has no beginning or end, so it often symbolizes infinity, wholeness, or the cycle of life.
• Harmony: Many cultures saw circles as representing the heavens, the sun, or cosmic balance.
• Sacred geometry: Mandalas, yantras, and other spiritual designs use circles to represent completeness and meditation focus.

Uses of Circular Shapes

• Transportation: Wheels, steering systems, and gears.
• Architecture: Domes, arches, round towers.
• Astronomy: Observing celestial bodies (planets, stars, orbits).
• Everyday objects: Plates, coins, clocks, rings.
• Art & symbolism: Mandalas, circular dances, ritual spaces.

Reflection: Humans didn’t just use circles because they were practical — they also found them beautiful, meaningful, and deeply symbolic. The circle became a bridge between function and philosophy, between engineering and spirituality.

Ex 6.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 6.1 Solutions

Exercise 6.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 6.1 Solutions

Unless stated otherwise, use the approximation \(\frac {22}{7}\) for π.

Question 1.
The perimeter of a circle is 44 cm. What is its radius?
Solution:
Given perimeter of circle = 2πr = 44
⇒ 2 × \(\frac {22}{7}\) × r = 44
⇒ r = \(\frac{44 \times 7}{44}\) = 7 cm
Radius is 7 cm.

Question 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) C = 2πr
= 2 × \(\frac {22}{7}\) × 7
= 44 cm

(ii) C = 2πr
= 2 × \(\frac {22}{7}\) × 10
= \(\frac {440}{7}\)
= 62.857 cm

(iii) C = 2πr
= 2 × \(\frac {22}{7}\) × 12
= \(\frac {528}{7}\)
= 75.429 cm

Question 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) l (arc) = \(\frac{\theta}{360} \times 2 \pi r\)
= \(\frac{60}{360} \times 2 \times \frac{22}{7} \times \frac{7}{2}\)
= 3.67
∴ l (arc) is 3.67 cm

(ii) l (arc) = \(\frac{\theta}{360} \times 2 \pi r\)
= \(\frac{120}{360} \times 2 \times \frac{22}{7} \times \frac{63}{10}\)
= 13.2
∴ l (arc) is 13.2 cm

Question 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:
Perimeter of sector

Hence, the perimeter of the sector is 46.33 cm

Question 5.
Find the perimeters of the following shapes (taking the arcs to he quarter or half or three-quarters of a circle, as appropriate) (Fig. i-ix):

Solution:
(i) Here, r = \(\frac {1}{2}\) × 60 = 30 cm; L = 80 cm
P = Perimeter of (2 straight sides (80 m each) + 2 semicircles of diameter (60 m))
= 2 × (πr) + 2L
= 2πr + 2L
= (2 × \(\frac {22}{7}\) × 30 + 160) cm
= 348.57 cm

(ii) Here, R = 6 cm, r = 4 cm
R – r = 2 cm
P = Outer semicircle + inner semicircle + straight parts
= πR + πr + 2(R – r)
= π(R + r) + 2(2)
= \(\frac {22}{7}\) × (6 + 4) + 4
= \(\frac {220}{7}\) + 4
= (31.43 + 4) cm
= 35.43 cm

(iii) Here, r = \(\frac {1}{2}\) × 10 cm = 5 cm
P = 4 × semicircle length
= 4 × πr
= 4 × \(\frac {22}{7}\) × 5 cm
= 62.86 cm

(iv) r = \(\frac {1}{2}\) × 12 cm = 6 cm
P = 3 × semicircle length
= 3 × πr
= 3 × \(\frac {22}{7}\) × 6 cm
= 56.57 cm

(v) R = 14 cm, r = 7 cm
Perimeter = 4 × semicircle length + 4 × quarter circle length
= 4(\(\frac {1}{4}\)πR + 4 × (πr)
= πR + 4πr
= \(\left(\frac{22}{7} \times 14+4 \times \frac{22}{7} \times 7\right)\)
= 132 cm

(vi) For small semicircles
r = \(\frac {28}{8}\) cm = \(\frac {7}{2}\) cm
P = π(4r) + 4 × (πr)
= 8πr
= \(8 \times \frac{22}{7} \times \frac{7}{2}\) cm
= 88 cm

(vii) Here, r₁ = \(\frac {1}{2}\) × 6 cm = 3 cm
r₂ = \(\frac {1}{2}\) × 8 cm = 4 cm
r₃ = \(\frac {1}{2}\) × 10 cm = 5 cm

P = πr₁ + πr₂ + πr₃
= π(r₁ + r₂ + r₃)
= \(\frac {22}{7}\) × (3 + 4 + 5) cm
= \(\frac {22}{7}\) × 12 cm
= 37\(\frac {5}{7}\) cm

(viii) P = π(3r) + 3 × (πr)
= 6πr
= 6 × \(\frac {22}{7}\) × 4 cm
= 75.43 cm

(ix) Here, r = \(\frac {1}{2}\) × 10 cm = 5 cm
P = π × (2r) + 2 × πr
= 4πr
= 4 × \(\frac {22}{7}\) × 5 cm
= 62.86 cm

Question 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:
Radius of car r = \(\frac {56}{2}\) = 28 cm
Now, circumference (C) = 2 × \(\frac {22}{7}\) × 28 cm = 176 cm = 1.76 m
(i) Car travels 1.76 m in one revolution.
(ii) No. of revolutions in 10 km (10,000 m) = \(\frac {10000}{1.76}\) = 5682 revolutions.

Question 7.
Find the total perimeter of all the petals in each of the given flowers.

Solution:
(i) Here, r = \(\frac {1}{2}\) × 14 cm = 7 cm
Perimeter of all petals = 4 × πr
= 4 × \(\frac {22}{7}\) × 7 cm
= 88 cm

(ii) Here, r = 42 cm, θ = 120°
Perimeter of flower = 6 × length of arc
= \(6 \times \frac{120}{360} \times 2 \times \frac{22}{7} \times 42\) cm
= 528 cm

Question 8.
The ratio of the perimeters of two circles is 5 : 4. What is the ratio of their radii?
Solution:
\(\frac{P \text { of } A}{P \text { of } B}=\frac{2 \pi r_1}{2 \pi r_2}=\frac{5}{4}\)
⇒ \(\frac{\pi r_1}{\pi r_2}=\frac{5}{4}\)
⇒ r₁ : r₂ = 5 : 4

Ex 6.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 6.2 Solutions

Exercise 6.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 6.2 Solutions

Question 1.
Find the area of triangle ADE in Fig. (below).

Solution:
Area of ∆ADE = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 8 × 10 sq cm
= 40 sq cm

Question 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:
Here, 10² + h² = 26²
⇒ h² = 26² – 10² = 676 – 100
⇒ h² = 576 = 24²
⇒ h = 24

∴ Area of trapezium = \(\frac {1}{2}\) × (a + b) × h
= \(\frac {1}{2}\) × (40 + 20) × 24
= 60 × 12
= 720 sq. cm

Question 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:

Here, a = 8 cm, b = 11 cm
c = perimeter – sum of all sides
= 32 – (8 + 11)
= 13 cm
where c is the third side of the triangle.

Question 4.
The sides of a triangular plot are in the ratio 3 : 5 : 7; its perimeter is 300 m. Find its area.
Solution:

Question 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 d₁ = x, then d₂ = 2x
Now area of rhombus = \(\frac {1}{2}\) × d₁ × d₂ = 128
⇒ \(\frac {1}{2}\) × x × 2x = 128
⇒ x² = 128
⇒ x = √128
⇒ x = 8√2
⇒ x = 8 × 1.414
⇒ x = 11.31 cm
Hence, d₁ = 11.3 cm, d₂ = 22.6 cm

Question 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)?
Solution:
Area of triangle PCD = \(\frac {1}{2}\) × area of || gm ABCD
Area of triangle QCD = \(\frac {1}{2}\) × area of || gm ABCD

(If a triangle and a || gm have the same base and between the same parallels, the area of the triangle is half the area of the || gm)
Hence, ar ∆PCD : ar ∆QCD = 1 : 1

Question 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:

∆PLS ≅ ∆RMQ (AAS)
Then LS = MQ (CPCTC)
⇒ \(\frac {1}{2}\) × LS × PO = \(\frac {1}{2}\) × MQ × PO
⇒ area PSO = area PQO

Question 8.
If the midpoints 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.)
Solution:
Here, area of ∆PQR = \(\frac {1}{2}\) × area PBCR …..(i)
and area of ∆PSR = \(\frac {1}{2}\) × area APRD …….(ii)

adding area ∆PQR + area ∆PSR = \(\frac {1}{2}\) × area PBCR + \(\frac {1}{2}\) × area APRD
⇒ area PQRS = \(\frac {1}{2}\) × area ABCD

Question 9.
In ∆ABC, the midpoint of BC is D (Fig. below). Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).

Solution:
Here, AD is a median of ∆ABC
area of ∆ABD = area of ∆ACD …….(i)
AD is a median of ∆PBC
area of ∆PBD = area of ∆PCD ……(ii)
Subtracting (i) – (ii), we get
area of ABD – area of PBD = area of ACD – area of PCD
area of ABP = area of ACP
area of black region = area of red region
Area (∆ABP) = Area (∆ACP)

Question 10.
Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. below). What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the black region (∆PBC and ∆PDA)?

Solution:
Draw l || AB through P.

Area of APB = \(\frac {1}{2}\) × area of ABEF
Area of CPD = \(\frac {1}{2}\) × area of FECD
Adding area APB + area CPD = \(\frac {1}{2}\) × area ABEF + \(\frac {1}{2}\) × area FECD
⇒ area APB + area CPD = \(\frac {1}{2}\) × area ABCD
Similarly area APD + area BPC = \(\frac {1}{2}\) × area ABCD.
∴ Area of red region (∆PAB and ∆PCD) : Area of black region (∆PBC and ∆PDA) = \(\frac{\frac{1}{2} \mathrm{ABCD}}{\frac{1}{2} \mathrm{ABCD}}\) = 1 : 1

Question 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. below). Prove that Area (∆BPQ) = \(\frac {1}{2}\) Area (∆ABC).

Solution:
Join DC.
Area of QDP = area of CPD (∆s on the same base and between the same parallels are equal in area)
area QDP + area DBP = area CPD + area DBP
area BPQ = area DBC
but area DBC = \(\frac {1}{2}\) area ∆ABC
(median of a triangle divides it into two triangles of equal area).
∴ area (∆BPQ) = \(\frac {1}{2}\) area (∆ABC)

Ex 6.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 6.3 Solutions

Exercise 6.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 6.3 Solutions

Unless stated otherwise, use the approximation \(\frac {22}{7}\) for π.

Question 1.
Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Solution:
Here, θ = 60°, r = 7 cm
Area of sector = \(\frac{\theta \pi r^2}{360}\)
= \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 7 \times 7 \mathrm{~cm}^2\)
= 25.67 cm²

Question 2.
Find the area of a quadrant of a circle whose circumference is 44 cm.
Solution:
Given, 2πr = 44
⇒ 2 × \(\frac {22}{7}\) × r = 44
⇒ r = \(\frac{44 \times 7}{44}\) = 7 cm
Area of quadrant = \(\frac{1}{4} \pi r^2\)
= \(\frac{1}{4} \times \frac{22}{7} \times 7 \times 7\) sq cm
= 38.5 sq cm

Question 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:
Angle covered in one minute = \(\left(\frac{360}{60}\right)^{\circ}\) = 6°
Angle covered in 10 minutes = 10 × 6° = 60°
Length of minute hand (r) = 7 cm
Area swept = \(\frac{\theta}{360} \pi r^2\)
= \(\frac{60}{360} \times \frac{22}{7} \times 7 \times 7\) sq cm
= 25.67 sq cm

Question 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) Here, r = 10 cm, θ = 90°

Area of minor segment = \(\frac{\theta \pi r^2}{360}\)
= \(\frac {90}{360}\) × 3.14 × 10 × 10 sq cm
= 78.5 sq cm

(ii) Area of major segment = \(\frac{\theta \pi r^2}{360}\)
= \(\frac {270}{360}\) × 3.14 × 10 × 10 sq cm
= 235.5 sq cm

Question 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:
Here, r = 15 cm, θ = 60°

Area of minor segment = Area of sector AOB – Area of ∆AOB
Area of sector AOB = \(\frac{\theta \pi r^2}{360}\)
= \(\frac {60}{360}\) × 3.14 × 15 × 15 sq cm
= 117.75 sq cm
Area ∆AOB (equilateral ∆) = \(\frac{r^2 \sqrt{3}}{4}\)
= 15 × 15 × \(\frac{\sqrt{3}}{4}\) sq cm
= 97.3125 sq cm
Now area of minor segment (AB) = (117.75 – 97.3125) sq cm = 20.4375 sq cm
Area of circle = 3.14 × 15 × 15 sq cm = 706.5 sq cm
Area of major segment = Area of circle – Area of minor segment
= (706.5 – 20.4375) sq cm
= 686.0625 sq cm

Question 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:
Here, θ = 120°; r = 28 cm

Total area = \(\frac{\theta \pi r^2}{360}\)
= 2 × area of sector AOB
= \(2 \times \frac{120}{360} \times \frac{22}{7} \times 28 \times 28\) cm²
= 1642.66 cm²

Question 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 \(\pi r^2\left(\frac{1}{6}-\frac{\sqrt{3}}{4}\right)\).
Solution:
Here, θ = 60°

Area of minor segment = Area of sector AOB – Area of ∆AOB
= \(\frac{\theta \pi r^2}{360}-\frac{\sqrt{3} r^2}{4}\)
= \(\frac{60}{360} \pi r^2-\frac{\sqrt{3} r^2}{4}\)
= \(\frac{\pi r^2}{6}-\frac{\sqrt{3} r^2}{4}\)
= \(\pi r^2\left(\frac{1}{6}-\frac{\sqrt{3}}{4}\right)\) sq. cm

Question 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 \(\frac{3 \sqrt{3}}{4 \pi}\) ≈ 0.413.
Solution:
Here, area of equilateral triangle inscribed in a circle = \(\frac{\sqrt{3}}{4} a^2\)
= \(\frac{\sqrt{3}}{4}(\sqrt{3} r)^2\) (∵ a = √3r)
= \(\frac{3 \sqrt{3}}{4} r^2\)

Area of circle = πr²
Required ratio = \(\frac{\frac{3 \sqrt{3}}{4} r^2}{\pi r^2}=\frac{3 \sqrt{3}}{4 \pi}\) ≈ 0.413

Question 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 \(\frac{2}{\pi}\) ≈ 0.637.
Solution:

In ∆OBC, BC² = r² + r² = 2r²
BC = r√2
∴ AB = BC = r√2
Area of square ABCD = (r√2)² = 2r²
Area of circle = πr²
\(\frac{\text { Area of square }}{\text { Area of circle }}=\frac{2 r^2}{\pi r^2}=\frac{2}{\pi}\) ≈ 0.637
Area of square : Area of circle = 2 : π ≈ 0.637

Question 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 \(\frac{3 \sqrt{3}}{2 \pi}\) ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?
Solution:
Here, θ = \(\frac {360}{6}\) = 60°

∴ ∆AOB is an equilateral triangle.
Area of ∆AOB = \(\frac{r^2 \sqrt{3}}{4}\)
Area of hexagon = \(\frac{6 r^2 \sqrt{3}}{4}=\frac{3 \sqrt{3}}{2} r^2\)
Area of circle = πr²
∴ \(\frac{\text { Area of hexagon }}{\text { Area of circle }}=\frac{\frac{3 \sqrt{3}}{2} r^2}{\pi r^2}\)
= \(\frac{3 \sqrt{3}}{2} r^2 \times \frac{1}{\pi r^2}\)
= \(\frac{3 \sqrt{3}}{2 \pi}\)
≈ 0.827
An equilateral triangle (Question 8) is made of 3 such smaller triangles meeting at the centre, whereas a hexagon (Question 10) is made of 6.
Since both are inscribed in the same circle of radius r, the hexagon simply covers exactly double the area that the inscribed equilateral triangle does.

Ganita Manjari Class 9 Maths Chapter 6 End of Chapter Exercise Solutions

Measuring Space Perimeter and Area End of Chapter Exercise Solutions

In the problems below, unless stated otherwise, use the approximation \(\frac {22}{7}\) for π.

Question 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:
Here, a² – b² = (a + b)(a – b)

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Question 2.
An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Solution:
Here, AB = AC = 15 cm

AC = (40 – 2 × 15) cm = 10 cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{10+15+15}{2}\)
= \(\frac {40}{2}\)
= 20

Question 3.
An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?
Solution:
ar ∆ABC = 60 cm²
Now, \(\frac {1}{2}\) × 10 × h = 60
where h is the height of ∆ABC.

⇒ h = \(\frac{60 \times 2}{10}\) = 12 cm
In ∆ADC, ∠ADC = 90°
DC = \(\frac {BC}{2}\)
= \(\frac {1}{2}\) × 10 cm
= 5 cm
AC² = 5² + 12² (by Baudhayana theorem)
⇒ AC² = 25 + 144 = 169 = 13²
⇒ AC = 13 cm but AB = AC
∴ AB = AC = 13 cm

Question 4.
The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.
Solution:

Here, ar ∆ABC = 54 cm2
⇒ \(\frac {1}{2}\) × base × height = 54
⇒ \(\frac {1}{2}\) × 12 × AB = 54
⇒ AB = \(\frac{54 \times 2}{12}\) = 9 cm
AC = \(\sqrt{A B^2+B C^2}\) (by Baudhayana theorem)
= \(\sqrt{9^2+12^2}\)
= \(\sqrt{81+144}\)
= √225
= 15 cm
Perimeter ∆ABC = AB + BC + AC
= (9 + 12 + 15) cm
= 36 cm

Question 5.
The sides of a triangle are in the ratio 2 : 3 : 4, and its perimeter is 45 cm. Find its area.
Solution:
Let a = 2x, b = 3x, c = 4x
then 2x + 3x + 4x = 45 (given)
⇒ 9x = 45
⇒ x = 5
∴ a = 10 cm, b = 15 cm, c = 20 cm

Question 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:
Here, a = 7, b = 24, c = 25


Now, (7, 24, 25) is a Baudhayana triplet.
Hence, such a triangle is a right triangle.
ar. of Δ = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 7 × 24
= 84 sq cm

Question 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:
Here, r = \(\frac {1}{2}\) × 60 cm = 30 cm
Distance in 100 revolutions = 100 × 2πr
= 200 × 3.14 × 30 cm
= 18840 m
= 18.84 km

Question 8.
Find the area of a quadrant of a circle whose circumference is 66 cm.
Solution:
Given circumference, C = 66

Question 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:
Here, r = 28 cm
C = 2πr
= 2 × \(\frac {22}{7}\) × 28 cm
= 176 cm
= 1.76 m
Distance travelled in one revolution = 1.76 m
∴ Number of turns in 1 km \(=\frac{\text { Total distance }}{\text { Circumference }}\)
= \(\frac{1000}{1.76}\)
≈ 568

Question 10.
Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?
Solution:
2(a + b) = 2(c + d) and ab = cd

Both equations are true only if a = c and b = d.
Hence, both rectangles are congruent.

Question 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., \(\frac {1}{2}\)(a + b)h.

Solution:

Area of trapezium ABCD = Area of || gm AECD + ar of ∆CEB
= AE × h + \(\frac {1}{2}\) × BE × h
= a × h + \(\frac {1}{2}\) × (b – a) × h
= \(\frac{2 a h+b h-a h}{2}\)
= \(\frac{a h+b h}{2}\)
= \(\frac {1}{2}\)(a + b)h

Question 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:

Here, Area trapezium ABCD = Area ∆ABD + Area ∆BCD
= \(\frac {1}{2}\) × a × h + \(\frac {1}{2}\) × b × h
= \(\frac {1}{2}\)(a + b) × h

Question 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:

I and II are two identical copies of trapezium ABCD; they are placed as shown.
AEFD is a parallelogram = 2 × area of trapezium ABCD
= ar. of ||gm AEFD
= 2 × area of trapezium ABCD
= (a + b) × h
area of trapezium ABCD = \(\frac {1}{2}\)(a + b) × h

Question 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:

Area of kite ABCD = ar ΔABC + ar ΔADC
= \(\frac {1}{2}\) × AC × BO + \(\frac {1}{2}\) × AC × DO
= \(\frac {1}{2}\) × AC × (BO + DO)
= \(\frac {1}{2}\) × AC × BD
Using Geometry (Rectangular Bound)
If we draw a rectangle around the kite, its area is d₁ × d₂.
We can see that the kite occupies exactly half of the rectangle’s space because the four outer triangles are congruent to the four inner triangles of the kite.

A = \(\frac{1}{2} d_2 d_1\) = \(\frac {1}{2}\) × AC × BD

Question 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 of ABCD = a × b
Area of PQRS = 4ab
Area of PQRS = 4 ar ABCD
4 copies of ABCD will get into PQRS.

(ii) Here, s = \(\frac{a+b+c}{2}\)

4 copies of ΔABC will get into ΔPQR.
(iii) Similarly, 9 copies of ΔABC will get into ΔPQR.

Question 16.

Solution:
(i) Here,

BC is the median of ΔACE
∴ ar ΔABC = ar ΔBCE ……(1)
ED is the median of ΔCEF
∴ ar ΔFED = ar ΔCDE ……(2)
Adding (1) and (2)
ar ABC + ar FED = ar BCE + ar CDE
Area of unshaded part = area of shaded part = \(\frac {1}{2}\) area of triangle.
∴ area of shaded part = \(\frac {1}{2}\) ar of triangle.

(ii) Here,

EBGD is a parallelogram
⇒ DE || GB
⇒ DS || GR
In ΔCDS, G is the midpoint of DC and DS || GR
⇒ R is the midpoint of SC.
GR is the line segment joining the midpoints of sides CD and CS of ΔCDS.
Let area CGR = x
Then area CDS = 4x
In the same way, area DPA = area ABQ = area BRC = 4x
Shaded area = Area of square ABCD – 16x = a² – 16x
area CDH = 5x
⇒ \(\frac{1}{2} \times a \times \frac{a}{2}\) = 5x
⇒ 5x = \(\frac{a^2}{4}\)
⇒ x = \(\frac{a^2}{20}\)
∴ Shaded area = \(a^2-16 \times \frac{a^2}{20}=\frac{a^2}{5}\)

Question 17.

Solution:
(i)

Let L = 6r, B = 2r
Area of rectangle = 6r × 2r = 12r²
Area of circles = 3πr²
Required fraction = \(\frac{3 \pi r^2}{12 r^2}=\frac{\pi}{4}\)
(ii)

Let L = 8r, B = 2r
Area of rectangle = 8r × 2r = 16r²
Area of circles = 4πr²
Required fraction = \(\frac{4 \pi r^2}{16 r^2}=\frac{\pi}{4}\)

Question 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!
Solution:
n circles are filled in a rectangle as shown.

Then \(\frac{\text { area of circles }}{\text { area of rectangle }}=\frac{\pi}{4}\)
For 10 circles, fraction = \(\frac{\pi}{4}\)
For 20 circles, fraction = \(\frac{\pi}{4}\)
For 50 circles, fraction = \(\frac{\pi}{4}\)

Question 19.
The figure below shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm². Find the perimeter of each small rectangle.

Solution:

Area of 9 rectangles = 72 sq cm
Area of 1 rectangle = \(\frac {72}{9}\) = 8 sq cm
Now, 4l = 5b
l = \(\frac {5}{4}\)b
∴ l × b = \(\frac {5}{4}\)b × b
but lb = 8
∴ \(\frac{5}{4} b^2\) = 8
⇒ b² = \(\frac {32}{5}\)
⇒ b = \(\sqrt{\frac{32}{5}}=\frac{4 \sqrt{10}}{5}\)
Then l = \(\frac{5}{4} \times \frac{4 \sqrt{10}}{5}=\sqrt{10}\)
Perimeter of each rectangle = 2(l + b)
= \(2\left(\sqrt{10}+\frac{4}{5} \sqrt{10}\right)\)
= \(\frac{18}{5} \sqrt{10}\) cm

Question 20.
Show that the areas of the shaded black triangle and the shaded red triangle are equal.

Find a way of cutting up the black triangle into some number of pieces and rearranging the pieces to cover the red triangle.
Solution:
I Part:

Here, BD = DE = EC = \(\frac {1}{3}\) BC
ar ΔABD = \(\frac{1}{2} \times \frac{1}{3} \mathrm{BC} \times h=\frac{\mathrm{BC} \times h}{6}\)
ar ΔAEC = \(\frac{1}{2} \times \frac{1}{3} \mathrm{BC} \times h=\frac{\mathrm{BC} \times h}{6}\)
Hence, ar ΔABD = ar ΔAEC
II Part: Do it yourself.

Question 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.

Solution:
Here, area of quarter circle ABC = \(\frac{1}{4} \pi r^2\)
= \(\frac{1}{4} \pi(2 r)^2\)
= πr²

Area of semicircle AOB = \(\frac{1}{2} \pi r^2\)
Area of semicircle BOC = \(\frac{1}{2} \pi r^2\)
Now, area = a + b + c + d = πr²
⇒ \(\left(\frac{1}{2} \pi r^2-c\right)+\left(\frac{1}{2} \pi r^2-c\right)\) + c + d = πr²
⇒ d – c = 0
⇒ d = c

Area of A = Area of B

Question 22.
In Fig. (below), 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:

Area of 4 petals = 8 × area of minor segment APO
= \(8 \times\left(\frac{1}{4} \pi \times 1^2-\frac{1}{2} \times 1 \times 1\right)\)
= \(8 \times\left(\frac{\pi}{4}-\frac{1}{2}\right)\)
= \(\frac{8(\pi-2)}{4}\)
= 2(π – 2) sq units

Question 23.
In Fig., 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 \(\frac{1}{4} \pi l^2\).

Solution:
Here, OA = r, OC = R
Area of shaded part = π(R² – r²) …… (1)
Now in ∆OAC,
r² + AC² = OC²
⇒ r² + \(\frac{l^2}{4}\) = R²
⇒ R² – r² = \(\frac{l^2}{4}\)

Substituting value of R² – r² in (1)
Area of shaded part = π(R² – r²)
= π × \(\frac{l^2}{4}\)
= \(\frac{1}{4} \pi l^2\)

Question 24.
In the figure, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).

Solution:

Here, A + B = Area of semicircle PQR + area of semicircle RST + area ∆PRT – area of semicircle PRT
= \(\frac{\pi a^2}{2}+\frac{\pi b^2}{2}+C-\frac{1}{2} \pi c^2\)
= \(\frac{1}{2} \pi\left(a^2+b^2\right)+C-\frac{1}{2} \pi c^2\)
= \(\frac{1}{2} \pi c^2+C-\frac{1}{2} \pi c^2\)
= C
= area (C)

Question 25.
Fig. (below) 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:

AB = BC = CA
∆ABC is an equilateral triangle
∴ ∠A = 60°
Required area = 2 × area of ∆ABC + 4 × area of segment APC
= \(2 \times \frac{r^2 \sqrt{3}}{4}+\frac{r^2}{2}\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)\)
= \(r^2 \frac{\sqrt{3}}{2}+\frac{r^2}{2}\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)\)
= \(\frac{r^2}{2}\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\) sq units

Question 26.
In Fig. (below), 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 \(\frac{2(A+C)(B+C)}{C}\).

Solution:
Let the adjacent sides of the rectangle be (a + b) and (c + d) as shown in the figure.

So, area of rectangle (a + b) (c + d)
For triangle A, base = c and height = a
So, Area A = \(\frac {1}{2}\)ac
For triangle C, base = b and height = c
So, Area C = \(\frac {1}{2}\)bc
For triangle B, base = b and height = d
So, Area B = \(\frac {1}{2}\)bd
Now, \(\frac{2(A+C)(B+C)}{C}\) = \(\frac{2\left(\frac{1}{2} a c+\frac{1}{2} b c\right)\left(\frac{1}{2} b d+\frac{1}{2} b c\right)}{\frac{1}{2} b c}\)
= \(\frac{2 \times \frac{1}{2} c(a+b) \times \frac{1}{2} b(d+c)}{\frac{1}{2} b c}\)
= (a + b) (c + d)
= Area of rectangle

Question 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.
Solution:
In ∆AOB, r² + r² = AB²
⇒ AB² = 2r²
⇒ AB = r√2
Now, AD = \(\frac{1}{2} \mathrm{AB}=\frac{r}{\sqrt{2}}\)
Area AEBF = area of semicircle AEB – area of minor segment AFB
= \(\frac{1}{2} \pi\left(\frac{r}{\sqrt{2}}\right)^2-\frac{r^2}{2}\left[\frac{\pi^2}{2}-1\right]\)
= \(\frac{\pi r^2}{4}-\frac{\pi r^2}{4}+\frac{r^2}{2}\)
= \(\frac{r^2}{2}\)
Area ∆AOB = \(\frac {1}{2}\)(r)(r) = \(\frac{r^2}{2}\)
∴ The areas of the shaded regions are equal.