NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area (2026-27)

Think and Reflect

Question 1: 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?

Image 1

Solution:
The width of each lane is 1.22 m. Therefore, the difference between the radii of the first and second lanes is:

Difference in radius = 1.22 m

The additional distance covered in the second lane occurs along the curved portions of the track. The two semicircular curves together form one complete circle.

The extra distance covered in Lane 2 is:

Extra distance = 2π × difference in radius

= 2 × 22/7 × 1.22

= 7.67 m approximately

Therefore, the runner in the second lane needs a stagger of approximately 7.67 m.

Yes, an equal stagger is required between the second and third lanes because the width of every lane is the same. Hence, the difference between the radii of Lane 2 and Lane 3 is also 1.22 m.

Think and Reflect 

Question 1: 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?

Solution:

Image 2

Unlike a rectangle, knowing only the lengths of the sides of a parallelogram is not sufficient to determine its area. This can be understood by examining the rigidity of the two shapes.

Imagine four sticks joined with hinges at their corners. Two sticks have a length of a units, while the other two have a length of b units.

When these sticks form a rectangle, all the angles are fixed at 90°. Therefore, the height is equal to one of the side lengths.

In Fig. (i):

• a = h

Therefore:

• Area = b × h = bh square units

However, when the same sticks form a parallelogram, the hinged corners can move. The side lengths a and b remain unchanged, but the shape can become more tilted.

As the parallelogram becomes thinner or more slanted, where 90° > x > y, its perpendicular height decreases even though its side lengths remain the same.

Since:

Area of a parallelogram = Base × Height

The area changes whenever the height changes.

In Fig. (ii):

• a > h₁

Therefore:

• Area = b × h₁ < bh

Further, in Fig. (iii):

• a > h₁ > h₂

Therefore:

• Area = b × h₂ < b × h₁

Hence, the area of a parallelogram cannot be calculated by knowing only the lengths of its sides because parallelograms with the same side lengths can have different perpendicular heights and, therefore, different areas.

Think and Reflect

Question 1: 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?

Image 3

Solution:
Yes, it is possible. A polygon can be divided into a finite number of smaller polygonal pieces using straight cuts. These pieces can then be rearranged to form another polygon of the same area.

Since ΔABD and ΔACD have equal areas, ΔABD can be cut into a finite number of pieces and rearranged to cover ΔACD exactly.

In this case, draw DE parallel to AC and cut ΔABD along DE. This divides ΔABD into two pieces: ΔAED and ΔBED.

Image 4

After rearrangement:

• ΔAED covers the region ΔDE′A.

• ΔBED covers the remaining region ΔDE′C.

Therefore, the two pieces obtained from ΔABD can be rearranged to cover ΔACD completely.

Think and Reflect 

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 a perimeter of 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: Rearranging Polygons of Equal Area

1. A square and a non-square rectangle with equal area

This case is easy to visualise. Consider a square of dimensions 4 × 4 units and a rectangle of dimensions 2 × 8 units.

Area of the square = 4 × 4 = 16 square units

Area of the rectangle = 2 × 8 = 16 square units

The square can be cut into suitable pieces, and those pieces can be rearranged to form the rectangle exactly.

2. Two triangles with different shapes but equal area

Two triangles may have completely different shapes. For example, one may be an acute-angled triangle, while the other may be a long and narrow obtuse-angled triangle.

Even then, each triangle can be divided into suitable pieces and rearranged to form the same rectangle. Since both triangles can be transformed into the same rectangle, the pieces of one triangle can also be rearranged to form the other triangle.

3. A triangle and a square with equal area

This transformation is more complex, but it is still possible. It can be completed in stages:

Triangle → Rectangle → Square

Image 5

The triangle can first be divided and rearranged into a rectangle. The rectangle can then be cut into suitable pieces and rearranged to form a square of the same area.

Conjecture:
Any two simple polygons P and Q can be subdivided into a finite number of congruent polygonal pieces if and only if they have the same area.

Part 2: Rectangles with a Perimeter of 40 Units

Let the length and breadth of a rectangle be x units and y units.

Given:

2(x + y) = 40

Therefore:

x + y = 20

So:

y = 20 − x

The area of the rectangle is:

A = xy

Substituting y = 20 − x:

A = x(20 − x)

1. How many such rectangles are there?

Since x can take any real value between 0 and 20, there are infinitely many rectangles with a perimeter of 40 units.

For example, their dimensions may be:

• 10 × 10 units

• 19 × 1 units

• 19.99 × 0.01 units

Therefore, infinitely many such rectangles are possible.

2. Which rectangle has the largest area?

The square has the largest area among all rectangles with the same perimeter.

For a square:

x = 10 and y = 10

Area = 10 × 10 = 100 square units

Therefore, the rectangle with the largest area has dimensions 10 units × 10 units, and its maximum area is 100 square units.

3. Which rectangle has the smallest area?

There is no single rectangle with the smallest area.

As x approaches 0, the value of y approaches 20, and the rectangle becomes increasingly thin. Its area gets closer and closer to 0.

For example, if:

x = 0.0001

then:

y = 19.9999

Its area is very close to 0.

However, x cannot be exactly 0 because the figure would no longer be a rectangle. Therefore, the area has an infimum of 0, but there is no rectangle with a minimum area. A thinner rectangle with a still smaller area can always be formed.

This may appear surprising because we often expect both a maximum and a minimum value to exist. The square is the most efficient rectangle because it encloses the greatest area for a fixed perimeter. However, the least efficient rectangle does not exist as a proper rectangle; it keeps becoming thinner and approaches a line segment.

Think and Reflect

Question 1: 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?

Solution: To square a triangle, also known as the quadrature of a triangle, we use a standard straightedge-and-compass construction.

Since a triangle cannot be converted directly into a square in a single step, a rectangle is used as an intermediate shape.

The procedure follows:

Triangle → Rectangle → Square

Step 1: Convert the Triangle into a Rectangle

First, construct a rectangle whose area is equal to the area of the given triangle.

Let triangle ABC have:

Base BC = b
Height AD = h

Therefore:

Area of ΔABC = 1/2 × BC × AD

= 1/2 × b × h

Now construct rectangle BCGF with:

One side = BC = b
Other side = DE = 1/2 h

The height AD can be bisected to obtain DE = 1/2 h.

Image 6

Here, ΔAIH and ΔBIF together exactly cover the region ΔCGH. Therefore, the triangle and the constructed rectangle have equal areas.

If the sides of the rectangle are:

x = b and y = 1/2 h

then:

Area of the rectangle = x × y

= b × 1/2 h

= 1/2 bh

This is equal to the area of triangle ABC.

Step 2: Convert the Rectangle into a Square

After constructing a rectangle with the same area as the triangle, follow the standard procedure for converting a rectangle into a square of equal area.

Thus, by first transforming the triangle into an equal-area rectangle and then transforming that rectangle into an equal-area square, we obtain a square whose area is exactly equal to the area of the given triangle.

Exercise Set 6.1 Class 9 Maths Solutions

Unless stated otherwise, use the approximation 22 7 for π.

Question 1: The perimeter of a circle is 44 cm. What is its radius? 

Solution:

Given:

Circumference of the circle, C = 44 cm

We know that:

C = 2πr

Therefore:

44 = 2 × 22/7 × r

r = 44 × 7/(2 × 22)

r = 7 cm

Hence, the radius of the circle 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:

The circumference of a circle is:

C = 2πr

(i) When r = 7 cm

C = 2 × 22/7 × 7

C = 44 cm

Correct to 3 significant figures, the circumference is 44.0 cm.

(ii) When r = 10 cm

C = 2 × 22/7 × 10

C = 440/7 cm

C = 62.857... cm

Correct to 3 significant figures, the circumference is 62.9 cm.

(iii) When r = 12 cm

C = 2 × 22/7 × 12

C = 528/7 cm

C = 75.428... cm

Correct to 3 significant figures, the circumference is 75.4 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:

The length of an arc with radius r and central angle θ is:

Arc length = θ/360° × 2πr

(i) When r = 3.5 cm and θ = 60°

Arc length = 60°/360° × 2 × 22/7 × 3.5

= 1/6 × 22

= 11/3 cm

= 3.666... cm

Therefore, the length of the arc is approximately 3.67 cm.

(ii) When r = 6.3 m and θ = 120°

Arc length = 120°/360° × 2 × 22/7 × 6.3

= 1/3 × 39.6

= 13.2 m

Therefore, the length of the arc is 13.2 m.

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:

The perimeter of a sector is the sum of its arc length and two radii.

Perimeter of sector = θ/360° × 2πr + 2r

Here:

r = 14 cm
θ = 75°

Therefore:

P = 75°/360° × 2 × 22/7 × 14 + 2 × 14

= 75/360 × 88 + 28

= 55/3 + 28

= 55/3 + 84/3

= 139/3 cm

= 46 1/3 cm

Hence, the perimeter of the sector is 46 1/3 cm.

Question 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): 

Image 7

(i)

The boundary consists of two semicircular arcs of equal radius and two straight sides of the rectangular portion.

Image 8

The two semicircular arcs together form one complete circle of radius 30 m.

Perimeter = 2πr + 2l

= 2 × 22/7 × 30 + 2 × 80

= 1320/7 + 160

= 1320/7 + 1120/7

= 2440/7 m

= 348 4/7 m

(ii)

The boundary consists of two semicircular arcs with radii 4 cm and 6 cm, along with two straight portions of 2 cm each.

Image 9

Perimeter = πr₁ + πr₂ + 2w

= 22/7 × 4 + 22/7 × 6 + 2 × 2

= 22/7 × 10 + 4

= 220/7 + 28/7

= 248/7 cm

= 35 3/7 cm

(iii)

The boundary consists of four semicircles, each having a radius of 5 cm.

Image 10

Perimeter = 4 × πr

= 4 × 22/7 × 5

= 440/7 cm

= 62 6/7 cm

(iv)

The boundary consists of three semicircular arcs of equal radius 6 cm.

Image 11

Perimeter = 3 × πr

= 3 × 22/7 × 6

= 396/7 cm

= 56 4/7 cm

(v)

The boundary consists of four quadrants of radius 14 cm and four semicircles of radius 7 cm.

Image 12

Perimeter of four quadrants:

= 4 × 1/4 × 2π × 14

= 28π

Perimeter of four semicircles:

= 4 × 1/2 × 2π × 7

= 28π

Total perimeter = 28π + 28π

= 56π

= 56 × 22/7

= 176 cm

(vi)

The boundary consists of one large semicircle with diameter 28 cm and four smaller semicircles, each with diameter 7 cm.

Image 13

Radius of the large semicircle = 14 cm

Radius of each small semicircle = 7/2 cm

Perimeter = π × 14 + 4 × π × 7/2

= 14π + 14π

= 28π

= 28 × 22/7

= 88 cm

(vii)

The boundary consists of three semicircles whose diameters are the sides of a right-angled triangle with perpendicular sides 6 cm and 8 cm.

Image 14

Using the Baudhayana-Pythagoras theorem:

Hypotenuse = √(6² + 8²)

= √(36 + 64)

= √100

= 10 cm

Therefore, the radii of the three semicircles are 3 cm, 4 cm and 5 cm.

Perimeter = πr₁ + πr₂ + πr₃

= π(r₁ + r₂ + r₃)

= 22/7 × (3 + 4 + 5)

= 22/7 × 12

= 264/7 cm

= 37 5/7 cm

(viii)

The boundary consists of one large semicircle with diameter 12 cm and three smaller semicircles, each having a diameter of 4 cm.

Image 15

Radius of the large semicircle = 6 cm

Radius of each small semicircle = 2 cm

Perimeter = πr₁ + 3πr₂

= 22/7 × 6 + 3 × 22/7 × 2

= 132/7 + 132/7

= 264/7 cm

= 37 5/7 cm

(ix)

The boundary consists of one large semicircle of radius 10 cm and two smaller semicircles, each having a diameter of 10 cm.

Image 16

Radius of each small semicircle = 5 cm

Perimeter = πr₁ + 2πr₂

= 22/7 × 10 + 2 × 22/7 × 5

= 220/7 + 220/7

= 440/7 cm

= 62 6/7 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:

(i)

The distance travelled by the tyre in one complete revolution is equal to its circumference.

Circumference = πd

Given:

d = 56 cm

Therefore:

C = 22/7 × 56

= 176 cm

Since 100 cm = 1 m:

176 cm = 1.76 m

Hence, the car travels 1.76 m when the tyre completes one revolution.

(ii)

Distance travelled by the car = 10 km

Converting kilometres into metres:

10 km = 10,000 m

Number of revolutions = Total distance travelled ÷ Distance covered in one revolution

= 10,000 ÷ 1.76

= 5681.8 approximately

Therefore, the tyre makes approximately 5,682 revolutions while travelling 10 km.

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

Image 17

Solution:

(i)

The four petals are formed by four semicircular arcs, each with a diameter of 14 cm.

Radius of each semicircle = 7 cm

Total perimeter = 4πr

= 4 × 22/7 × 7

= 88 cm

(ii)

The six petals are formed by six equal arcs. Each arc belongs to a circle of radius 42 cm and subtends an angle of 120° at the centre.

Total perimeter:

= 6 × 120°/360° × 2πr

= 6 × 1/3 × 2 × 22/7 × 42

= 2 × 264

= 528 cm

Question 8: The ratio of the perimeters of two circles is 5 : 4. What is the ratio of their radii?

Solution:

Let the radii of the two circles be r₁ and r₂.

Their circumferences are:

C₁ = 2πr₁

C₂ = 2πr₂

Given:

C₁ : C₂ = 5 : 4

Therefore:

2πr₁ : 2πr₂ = 5 : 4

Cancelling the common factor 2π:

r₁ : r₂ = 5 : 4

Hence, the ratio of their radii is 5 : 4.

Exercise Set 6.2 Class 9 Maths Solutions

Question 1: Find the area of triangle ADE in Fig. 6.31.

Image 18

Solution:

In ΔADE:

AD = BC = 8 cm

The perpendicular distance from E to AD is equal to DC:

Height = 10 cm

Area of a triangle = ½ × Base × Height

Area of ΔADE = ½ × 8 × 10

= 40 cm²

Therefore, the area of ΔADE is 40 cm².

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:

Image 19

The trapezium has parallel sides of 40 cm and 20 cm. Its two equal non-parallel sides are 26 cm each.

Draw perpendiculars from the ends of the shorter parallel side to the longer parallel side. This forms a rectangle and two identical right-angled triangles.

The base of each right-angled triangle is:

(40 − 20)/2 = 10 cm

Let the height of the trapezium be h.

Using the Baudhayana–Pythagoras theorem:

10² + h² = 26²

100 + h² = 676

h² = 576

h = 24 cm

Image 20

Area of a trapezium = ½ × Sum of parallel sides × Height

= ½ × (40 + 20) × 24

= ½ × 60 × 24

= 720 cm²

Therefore, the area of the trapezium is 720 cm².

Alternative method:

An isosceles trapezium is a cyclic quadrilateral. Therefore, Brahmagupta’s formula may also be used.

Its four sides are 40 cm, 26 cm, 20 cm and 26 cm.

Semi-perimeter:

s = (40 + 26 + 20 + 26)/2

= 56 cm

Area = √[(s − 40)(s − 26)(s − 20)(s − 26)]

= √[(56 − 40)(56 − 26)(56 − 20)(56 − 26)]

= √(16 × 30 × 36 × 30)

= 720 cm²

Hence, the area is 720 cm².

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:

Two sides of the triangle are:

a = 8 cm and b = 11 cm

Perimeter = 32 cm

The third side is:

c = 32 − (8 + 11)

= 13 cm

The semi-perimeter is:

s = 32/2

= 16 cm

Using Heron’s formula:

Area = √[s(s − a)(s − b)(s − c)]

= √[16(16 − 8)(16 − 11)(16 − 13)]

= √(16 × 8 × 5 × 3)

= √1920

= 8√30 cm²

Approximately:

8√30 ≈ 43.82 cm²

Therefore, the area of the triangle is 8√30 cm², or approximately 43.82 cm².

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

Solution:

Let the sides of the triangular plot be:

3x, 5x and 7x

According to the question:

3x + 5x + 7x = 300

15x = 300

x = 20

Therefore, the sides are:

3x = 60 m

5x = 100 m

7x = 140 m

The semi-perimeter is:

s = 300/2

= 150 m

Using Heron’s formula:

Area = √[s(s − a)(s − b)(s − c)]

= √[150(150 − 60)(150 − 100)(150 − 140)]

= √(150 × 90 × 50 × 10)

= 1500√3 m²

Approximately:

1500√3 ≈ 2598.08 m²

Therefore, the area of the triangular plot is 1500√3 m², or approximately 2598.08 m².

Question 5: One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has an area of 128 cm², find the length of the shorter diagonal.

Solution:

Let the shorter diagonal be x cm.

Then, the longer diagonal is 2x cm.

Area of a rhombus = ½ × d₁ × d₂

Therefore:

½ × x × 2x = 128

x² = 128

x = √128

x = 8√2 cm

Approximately:

8√2 ≈ 11.31 cm

Therefore, the length of the shorter diagonal is 8√2 cm, or approximately 11.31 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 (ΔPCD) : area (ΔQCD)?

Solution:

Image 21

Triangles PCD and QCD have the same base CD.

Since ABCD is a parallelogram:

AB ∥ CD

Points P and Q lie on AB. Therefore, their perpendicular distances from CD are equal.

Thus, ΔPCD and ΔQCD have the same base and equal heights.

Hence:

Area (ΔPCD) = Area (ΔQCD)

Therefore:

Area (ΔPCD) : Area (ΔQCD) = 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:

Draw the diagonal QS and let it intersect PR at T.

Image 22

The diagonals of a parallelogram bisect each other. Therefore:

QT = ST

Triangles PSO and PQO have the same base PO.

Also, S and Q are at equal perpendicular distances from the diagonal PR because T is the midpoint of QS.

Thus, the two triangles have the same base and equal heights.

Therefore:

Area (ΔPSO) = Area (ΔPQO)

Hence proved.

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:

Let ABCD be the given 4-gon.

Let P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.

Join P, Q, R and S.

Image 23

In ΔABC, P and Q are the midpoints of AB and BC. By the midpoint theorem:

PQ ∥ AC

Similarly, in ΔADC:

SR ∥ AC

Therefore:

PQ ∥ SR

In ΔABD:

PS ∥ BD

In ΔBCD:

QR ∥ BD

Therefore:

PS ∥ QR

Hence, PQRS is a parallelogram.

Image 24

Now, ΔPBQ is similar to ΔABC, with the corresponding sides in the ratio 1 : 2.

Therefore:

Area (ΔPBQ) = ¼ Area (ΔABC) …… (i)

Similarly:

Area (ΔSDR) = ¼ Area (ΔADC) …… (ii)

Adding equations (i) and (ii):

Area (ΔPBQ) + Area (ΔSDR)

= ¼[Area (ΔABC) + Area (ΔADC)]

= ¼ Area (4-gon ABCD) …… (iii)

Similarly:

Area (ΔPAS) = ¼ Area (ΔBAD)

and:

Area (ΔQCR) = ¼ Area (ΔBCD)

Therefore:

Area (ΔPAS) + Area (ΔQCR)

= ¼[Area (ΔBAD) + Area (ΔBCD)]

= ¼ Area (4-gon ABCD) …… (iv)

Adding equations (iii) and (iv):

Area of the four corner triangles

= ¼ Area (ABCD) + ¼ Area (ABCD)

= ½ Area (ABCD)

Therefore:

Area (PQRS)

= Area (ABCD) − Area of the four corner triangles

= Area (ABCD) − ½ Area (ABCD)

= ½ Area (ABCD)

Hence:

Area of parallelogram PQRS = ½ Area of 4-gon ABCD

Question 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).

Image 25

Solution:

Given:

D is the midpoint of BC.

Therefore:

BD = DC

Since AD is the median of ΔABC, it divides the triangle into two triangles of equal area.

Thus:

Area (ΔABD) = Area (ΔACD) …… (i)

In ΔPBC, PD is also a median because D is the midpoint of BC.

Therefore:

Area (ΔPBD) = Area (ΔPCD) …… (ii)

Subtracting equation (ii) from equation (i):

Area (ΔABD) − Area (ΔPBD)

= Area (ΔACD) − Area (ΔPCD)

Therefore:

Area (ΔABP) = Area (ΔACP)

Hence proved.

Question 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)?

Image 26

Solution:

Let the side of square ABCD be a units.

Let the perpendicular distances of P from AB and CD be h₁ and h₂, respectively.

Image 27

Since AB and CD are opposite sides of the square:

h₁ + h₂ = a

The total area of the red region is:

Area (ΔPAB) + Area (ΔPCD)

= ½ × a × h₁ + ½ × a × h₂

= ½a(h₁ + h₂)

= ½a × a

= ½a²

Similarly, let the perpendicular distances of P from BC and AD be k₁ and k₂.

Since BC and AD are opposite sides of the square:

k₁ + k₂ = a

The total area of the green region is:

Area (ΔPBC) + Area (ΔPDA)

= ½ × a × k₁ + ½ × a × k₂

= ½a(k₁ + k₂)

= ½a × a

= ½a²

Thus, the red and green regions have equal areas.

Therefore:

Area of red region: Area of green region = 1 : 1

Question 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) = ½ Area (ΔABC).

Image 28

Solution:

Given:

D is the midpoint of AB.

PD ∥ CQ

Image 29

We have to prove:

Area (ΔBPQ) = ½ Area (ΔABC)

Join DC.

Since D is the midpoint of AB, CD is a median of ΔABC.

A median divides a triangle into two triangles of equal area.

Therefore:

Area (ΔBDC) = ½ Area (ΔABC) …… (i)

Triangles ΔDPC and ΔDPQ have the same base DP.

Also, C and Q lie on the line CQ, which is parallel to DP. Therefore, the two triangles lie between the same parallels.

Hence:

Area (ΔDPC) = Area (ΔDPQ) …… (ii)

Now:

Area (ΔBDC)

= Area (ΔBPD) + Area (ΔDPC)

Using equation (ii):

Area (ΔBDC)

= Area (ΔBPD) + Area (ΔDPQ)

= Area (ΔBPQ) …… (iii)

From equations (i) and (iii):

Area (ΔBPQ) = ½ Area (ΔABC)

Hence proved.

Exercise Set 6.3 Class 9 Maths Solutions

Unless stated otherwise, use the approximation 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:

Given:

Radius, r = 7 cm

Angle of the sector, θ = 60°

Image 30

Area of a sector = θ/360° × πr²

= 60°/360° × 22/7 × 7²

= 1/6 × 22/7 × 49

= 77/3

= 25⅔ cm²

Therefore, the area of the sector is 25⅔ cm².

Question 2: Find the area of a quadrant of a circle whose circumference is 44 cm.

Solution:

Image 31

Given the circumference of the circle, C = 44 cm.

We know that:

2πr = 44

2 × 22/7 × r = 44

r = 7 cm

A quadrant is one-fourth of a circle.

Area of the quadrant = 1/4 × πr²

= 1/4 × 22/7 × 7²

= 77/2

= 38.5 cm²

Therefore, the area of the quadrant is 38.5 cm².

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:

The length of the minute hand represents the radius.

Therefore:

r = 7 cm

The minute hand completes 360° in 60 minutes.

Angle swept in 10 minutes:

θ = 10/60 × 360°

= 60°

Image 32

Area swept by the minute hand:

A = θ/360° × πr²

= 60°/360° × 22/7 × 7²

= 77/3

= 25⅔ cm²

Thus, the area swept by the minute hand in 10 minutes is 25⅔ cm².

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:

Given:

Radius, r = 10 cm

Image 33

(i) Area of the minor sector

Angle of the minor sector, θ = 90°

Area = θ/360° × πr²

= 90°/360° × 3.14 × 10²

= 1/4 × 3.14 × 100

= 78.5 cm²

Therefore, the area of the minor sector is 78.5 cm².

(ii) Area of the major sector

Angle of the major sector, θ = 270°

Area = θ/360° × πr²

= 270°/360° × 3.14 × 10²

= 3/4 × 314

= 235.5 cm²

Therefore, the area of the major sector is 235.5 cm².

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:

Given:

Image 34

Radius, r = 15 cm

Angle at the centre, θ = 60°

Since △OAB is an isosceles triangle with its vertex angle equal to 60°, all its angles are 60°. Therefore, △OAB is an equilateral triangle.

(i) Area of the minor segment

Area of the minor segment
= Area of the minor sector − Area of △OAB

Area of the sector:

= 60°/360° × 3.14 × 15²

= 1/6 × 3.14 × 225

= 117.75 cm²

Area of the equilateral triangle:

= √3/4 × 15²

= 1.73/4 × 225

= 97.3125 cm²

Therefore:

Area of the minor segment

= 117.75 − 97.3125

= 20.4375 cm²

≈ 20.44 cm²

(ii) Area of the major segment

Area of the circle:

= πr²

= 3.14 × 15²

= 706.5 cm²

Area of the major segment:

= Area of the circle − Area of the minor segment

= 706.5 − 20.44

= 686.06 cm²

Therefore, the areas of the minor and major segments are 20.44 cm² and 686.06 cm², respectively.

Question 6: A car has two wipers that 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:

Given:

Radius, r = 28 cm

Angle swept by each wiper, θ = 120°

Image 35

Area cleaned by one wiper:

A = θ/360° × πr²

= 120°/360° × 22/7 × 28²

= 1/3 × 22/7 × 784

= 2464/3

= 821⅓ cm²

Since there are two wipers:

Total area cleaned

= 2 × 821⅓

= 1642⅔ cm²

Thus, the total area cleaned during each sweep is 1642⅔ 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 πr²(1/6 − √3/4π).

Solution:

The angle subtended by the chord at the centre is 60°.

Image 36

Since OA = OB = r, △OAB is an isosceles triangle.

Therefore:

∠OAB = ∠OBA

= (180° − 60°)/2

= 60°

Hence, △OAB is an equilateral triangle.

Area of the minor segment
= Area of the minor sector − Area of △OAB

= 60°/360° × πr² − √3/4 × r²

= πr²/6 − √3r²/4

Taking πr² as the common factor:

Area of the minor segment

= πr²(1/6 − √3/4π)

Hence proved.

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 3√3/4π ≈ 0.413.

Solution:

Let △ABC be an equilateral triangle inscribed in a circle of radius r and centre O.

Image 37

The centre O divides the median AD in the ratio 2 : 1.

Therefore:

AD = 3/2 OA

= 3r/2

For an equilateral triangle with side a:

AD = √3/2 a

Therefore:

3r/2 = √3a/2

a = √3r

Area of △ABC:

= √3/4 × a²

= √3/4 × (√3r)²

= 3√3/4 r²

Area of the circle:

= πr²

Required ratio:

Area of triangle/Area of circle

= (3√3r²/4)/(πr²)

= 3√3/4π

Using √3 = 1.73 and π = 3.14:

3√3/4π ≈ 0.413

Hence, the required ratio is approximately 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 2/π ≈ 0.637.

Solution:

Let ABCD be a square inscribed in a circle of radius r.

The diagonal of the square is equal to the diameter of the circle.

Image 38

Therefore:

Diagonal = 2r

If the side of the square is a, then:

Diagonal = √2a

Thus:

√2a = 2r

a = √2r

Area of the square:

= a²

= (√2r)²

= 2r²

Area of the circle:

= πr²

Required ratio:

Area of square/Area of circle

= 2r²/πr²

= 2/π

≈ 0.637

Hence, the required ratio is approximately 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 3√3/2π ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?

Solution:

Let ABCDEF be a regular hexagon inscribed in a circle of radius r and centre O.

Image 39

A regular hexagon inscribed in a circle can be divided into six equilateral triangles, each having side r.

Area of one equilateral triangle:

= √3/4 r²

Area of the hexagon:

= 6 × √3/4 r²

= 3√3/2 r²

Area of the circle:

= πr²

Required ratio:

Area of hexagon/Area of circle

= (3√3r²/2)/(πr²)

= 3√3/2π

≈ 0.827

In Question 8, the inscribed equilateral triangle consists of three such smaller equilateral triangles, whereas the regular hexagon consists of six.

Since 6 is twice 3, the ratio obtained for the hexagon is exactly twice the ratio obtained for the equilateral triangle.

Class 9 Maths Chapter 6 End of Chapter Exercises Solutions

In the problems below, unless stated otherwise, use the approximation 22/7 for π.

Question 1: Identities in algebra can sometimes be shown as area relationships. For example:

Image 40

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 square of side (a + b) can be divided into four parts:

a square of area a², a square of area b², and two rectangles, each having area ab.

Therefore:

(a + b)² = a² + ab + ab + b²

= a² + 2ab + b²

For the identity:

(a + b)(a − b) = a² − b²

Take a square of side a and remove a smaller square of side b. The remaining region has area a² − b².

The remaining pieces can be rearranged to form a rectangle whose length is (a + b), and breadth is (a − b).

Image 41

Therefore:

(a + b)(a − b) = a² − b²

For the identity:

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

Image 42

Draw a square of side (a + b + c) and divide it into nine parts.

Its total area is:

a² + b² + c² + ab + ab + ac + ac + bc + bc

= a² + b² + c² + 2ab + 2ac + 2bc

Hence, the identity is represented through the area model.

Question 2: An isosceles triangle has a perimeter of 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

Solution:

The equal sides are 15 cm each.

Perimeter = 40 cm

Third side:

= 40 − 15 − 15

= 10 cm

Semi-perimeter:

s = 40/2

= 20 cm

Using Heron’s formula:

Area = √[s(s − a)(s − b)(s − c)]

= √[20(20 − 15)(20 − 15)(20 − 10)]

= √(20 × 5 × 5 × 10)

= √5000

= 50√2 cm²

≈ 70.7 cm²

Therefore, the area of the triangle is 50√2 cm² or approximately 70.7 cm².

Question 3: An isosceles triangle has a base of 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

Solution:

Let each equal side be a.

The base is 10 cm, so half of the base is:

b = 5 cm

Image 43

For an isosceles triangle:

Area = b√(a² − b²)

Given area = 60 cm².

Therefore:

60 = 5√(a² − 5²)

12 = √(a² − 25)

Squaring both sides:

144 = a² − 25

a² = 169

a = 13 cm

Therefore, the lengths of the two equal sides are 13 cm each.

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

Solution:

Given:

Area = 54 cm²

One leg = 12 cm

Image 44

Let the other leg be b.

Area of a right-angled triangle:

54 = 1/2 × 12 × b

54 = 6b

b = 9 cm

Using the Baudhayana-Pythagoras theorem:

Hypotenuse = √(12² + 9²)

= √(144 + 81)

= √225

= 15 cm

Perimeter:

= 12 + 9 + 15

= 36 cm

Therefore, the perimeter of the triangle is 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 the sides be 2x, 3x and 4x.

According to the question:

2x + 3x + 4x = 45

9x = 45

x = 5

Therefore, the sides are:

10 cm, 15 cm and 20 cm.

Semi-perimeter:

s = 45/2

= 22.5 cm

Using Heron’s formula:

Area = √[s(s − a)(s − b)(s − c)]

= √[22.5(22.5 − 10)(22.5 − 15)(22.5 − 20)]

= √(22.5 × 12.5 × 7.5 × 2.5)

= 18.75√15

≈ 72.62 cm²

Therefore, the area of the triangle is approximately 72.62 cm².

Question 6: The sides of a triangle have lengths 7 cm, 24 cm, and 25 cm. Find the area of the triangle in two different ways.

Solution:

First, observe that:

7² + 24² = 49 + 576 = 625

Image 45

Also:

25² = 625

Therefore:

7² + 24² = 25²

Hence, the triangle is right-angled.

Method 1: Using Base and Height

Area = 1/2 × base × height

= 1/2 × 24 × 7

= 84 cm²

Method 2: Using Heron’s Formula

Semi-perimeter:

s = (7 + 24 + 25)/2

= 28 cm

Area:

= √[28(28 − 7)(28 − 24)(28 − 25)]

= √(28 × 21 × 4 × 3)

= √7056

= 84 cm²

Therefore, the area of the triangle is 84 cm² by both methods.

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:

During one complete rotation, the wheel covers a distance equal to its circumference.

Image 46

Diameter, d = 60 cm

Circumference:

C = πd

= 22/7 × 60

= 1320/7 cm

For 100 rotations:

Distance = 100 × 1320/7

= 132000/7 cm

= 1320/7 m

≈ 188.57 m

Therefore, the cyclist travels approximately 188.57 m.

Question 8: Find the area of a quadrant of a circle whose circumference is 66 cm.

Solution:

Image 47

Given circumference:

2πr = 66

2 × 22/7 × r = 66

r = 66 × 7/44

= 21/2 cm

= 10.5 cm

Area of a quadrant:

= 1/4 × πr²

= 1/4 × 22/7 × (21/2)²

= 693/8

= 86⅝ cm²

Therefore, the area of the quadrant is 86⅝ cm².

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:

Image 48

Radius of the wheel, r = 28 cm

Distance covered in one complete turn:

C = 2πr

= 2 × 22/7 × 28

= 176 cm

Therefore, the car travels 176 cm in one complete turn.

Now:

1 km = 1,00,000 cm

Number of turns:

= Total distance/Distance covered in one turn

= 1,00,000/176

= 6250/11

= 568 2/11

Therefore, the wheel turns 568 2/11 times, or approximately 568.18 times, during a journey of 1 km.

Question 10: Two rectangles have the same area and the same perimeter. Does this mean that they are congruent with each other?

Solution:

Yes, the two rectangles must be congruent.

Let the length and breadth of one rectangle be l₁ and b₁, and those of the other rectangle be l₂ and b₂.

Since their areas are equal:

l₁b₁ = l₂b₂

Since their perimeters are equal:

2(l₁ + b₁) = 2(l₂ + b₂)

Therefore:

l₁ + b₁ = l₂ + b₂

The sum and product of the length and breadth are the same for both rectangles. Hence, both rectangles have the same pair of dimensions, possibly interchanged.

Therefore, the two 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., 1/2(a + b)h.

Image 49

Solution:

The trapezium shown in the figure can be divided into a parallelogram and a triangle.

For the parallelogram:

Base = a

Height = h

Area = ah

The total lower parallel side of the trapezium is b. The part used by the parallelogram is a.

Therefore, the base of the triangle is:

b − a

Area of the triangle:

= 1/2 × (b − a) × h

Total area of the trapezium:

= ah + 1/2(b − a)h

= 1/2[2ah + bh − ah]

= 1/2(a + b)h

Hence, the area of the trapezium is:

1/2 × Sum of parallel sides × Height.

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:

Consider a trapezium ABCD whose parallel sides are AB = a and DC = b, and whose height is h.

Image 50

Draw diagonal BD. It divides the trapezium into △ABD and △BCD.

Area of △ABD:

= 1/2 × a × h

= 1/2 ah

Area of △BCD:

= 1/2 × b × h

= 1/2 bh

Therefore, the area of the trapezium:

= 1/2 ah + 1/2 bh

= 1/2(a + b)h

Hence proved.

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:

Take two identical trapeziums whose parallel sides are a and b and whose height is h.

Rotate one trapezium through 180° and place it beside the other. Together, the two trapeziums form a parallelogram.

Image 51

The base of the parallelogram is:

a + b

Its height is: h

Therefore:

Area of the parallelogram = (a + b)h

The parallelogram consists of two identical trapezoids.

Hence:

Area of one trapezium = 1/2 × Area of parallelogram

= 1/2(a + b)h

Thus, the area of a trapezium is half the sum of its parallel sides multiplied by its height.

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:

Let the diagonals of the kite be d₁ and d₂.

The diagonals of a kite intersect at right angles.

(i) Using Algebra

Image 52

Suppose the longer diagonal d₁ is divided into lengths h₁ and h₂.

Therefore:

h₁ + h₂ = d₁

The shorter diagonal d₂ forms the common base of two triangles.

Area of the first triangle:

= 1/2 d₂h₁

Area of the second triangle:

= 1/2 d₂h₂

Total area of the kite:

= 1/2 d₂h₁ + 1/2 d₂h₂

= 1/2 d₂(h₁ + h₂)

= 1/2 d₂d₁

Therefore:

Area of kite = 1/2 d₁d₂

(ii) Using Geometry

Image 53

Draw a rectangle around the kite such that its length and breadth are equal to the two diagonals of the kite.

Area of the rectangle:

= d₁d₂

The kite occupies exactly half of this rectangle because the outer triangular pieces can be rearranged to cover the kite.

Therefore:

Area of kite = 1/2 d₁d₂

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) Rectangles

Image 54

Area of rectangle ABCD:

= a × b

= ab

Let this area be A.

Area of rectangle PQRS:

= 2a × 2b

= 4ab

= 4A

Therefore, PQRS has four times the area of ABCD.

Yes, four copies of rectangle ABCD can be arranged to fit exactly inside rectangle PQRS.

(ii) Triangles with sides doubled

Let the semi-perimeter of △ABC be s.

Using Heron’s formula:

A = √[s(s − a)(s − b)(s − c)]

For △PQR, all the sides and the semi-perimeter are doubled.

Area of △PQR:

= √[2s(2s − 2a)(2s − 2b)(2s − 2c)]

= √[2 × 2 × 2 × 2 × s(s − a)(s − b)(s − c)]

= 4A

Therefore, △PQR has four times the area of △ABC.

Yes, four copies of △ABC can fit inside △PQR. Three copies may be placed at the vertices, while the fourth fits upside down in the centre.

Image 55

(iii) Triangles with sides tripled

For △PQR, all sides and the semi-perimeter are three times those of △ABC.

Area of △PQR:

= √[3s(3s − 3a)(3s − 3b)(3s − 3c)]

= √[3 × 3 × 3 × 3 × s(s − a)(s − b)(s − c)]

= 9A

Therefore, △PQR has nine times the area of △ABC.

Yes, nine copies of △ABC can be arranged inside △PQR.

Image 56

Question 16.

Image 57

Solution:

Fig. 6.43

Let the area of the complete triangle be A.

Draw a line to divide the shaded region into two parts, b and c.

A median divides a triangle into two triangles of equal area. From the divisions shown in the figure:

Image 58

Area a = Area b

Also:

Area c = Area d

The required shaded area is:

Area b + Area c

According to the equal-area divisions in the figure:

Area b = A/6

Area c = A/3

Therefore:

Shaded area = A/6 + A/3

= A/6 + 2A/6

= A/2

Required fraction:

= (A/2)/A

= 1/2

Thus, the shaded region occupies 1/2 of the whole triangle.

Image 59

Draw lines parallel to the sides of the square through the vertices of the shaded region.

The square is divided into 25 equal parts. The shaded region covers 5 of these parts.

Therefore:

Fraction of the square that is shaded

= 5/25

= ⅕

Question 17.

Image 60

Solution:

Fig. 6.45

Let the radius of each circle be r.

The rectangle has:

Length = 6r

Breadth = 2r

Area of the rectangle:

= 6r × 2r

= 12r²

There are three circles.

Total area of the circles:

= 3πr²

Required fraction:

= 3πr²/12r²

= π/4

Fig. 6.46

The rectangle has:

Length = 8r

Breadth = 2r

Area of the rectangle:

= 8r × 2r

= 16r²

There are four circles.

Total area of the circles:

= 4πr²

Required fraction:

= 4πr²/16r²

= π/4

Therefore, in both figures, the circles occupy π/4 of the rectangle.

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:

Conjecture:

When identical circles are placed side by side inside a rectangle as shown, the fraction of the rectangle covered by the circles is always π/4.

Case 1: 10 Circles

Radius of each circle = r

Total area of 10 circles:

= 10πr²

Length of the rectangle:

= 10 × 2r

= 20r

Breadth of the rectangle:

= 2r

Area of the rectangle:

= 20r × 2r

= 40r²

Required fraction:

= 10πr²/40r²

= π/4 

Case 2: 20 Circles

Total area of the circles:

= 20πr²

Length of the rectangle:

= 20 × 2r

= 40r

Area of the rectangle:

= 40r × 2r

= 80r²

Required fraction:

= 20πr²/80r²

= π/4

Case 3: 50 Circles

Total area of the circles:

= 50πr²

Length of the rectangle:

= 50 × 2r

= 100r

Area of the rectangle:

= 100r × 2r

= 200r²

Required fraction:

= 50πr²/200r²

= π/4

General Proof

Suppose n circles of radius r are fitted side by side inside the rectangle.

Total area of n circles:

= nπr²

Length of the rectangle:

= 2nr

Breadth of the rectangle:

= 2r

Area of the rectangle:

= 2nr × 2r

= 4nr²

Therefore:

Fraction covered by the circles

= nπr²/4nr²

= π/4

Hence, regardless of the number of circles, they always occupy π/4 of the rectangle.

Question 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.

Image 61

Solution:

Let the length and breadth of each small rectangle be a and b, respectively.

There are nine identical rectangles, and their total area is 72 cm².

Therefore:

9ab = 72

ab = 8 …(i)

From the arrangement shown in the figure:

4a = 5b

Therefore:

b = 4a/5

Substituting this value in equation (i):

a × 4a/5 = 8

4a²/5 = 8

a² = 10

a = √10 cm

Therefore:

b = 4√10/5 cm

Perimeter of each small rectangle:

= 2(a + b)

= 2(√10 + 4√10/5)

= 2(9√10/5)

= 18√10/5 cm

Therefore, the perimeter of each small rectangle is 18√10/5 cm.

Question 20: Show that the areas of the shaded blue (B) triangle and the shaded red (R) triangle are equal.

Image 62

Find a way of cutting up the blue (B) triangle into some number of pieces and rearranging the pieces to cover the red (R) triangle.

Image 63

Solution:

The opposite side is trisected. Therefore, the bases of the blue and red triangles are equal.

Let the base of each triangle be b.

Both triangles have the same top vertex, and their bases lie on the same straight line. Hence, they also have the same perpendicular height h.

Area of the blue triangle:

= 1/2 bh

Area of the red triangle:

= 1/2 bh

Therefore:

Area of blue triangle = Area of red triangle

For the rearrangement, divide the blue triangle along the lines indicated in the figure and shift the resulting pieces without changing their areas until they completely cover the red triangle.

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.

Image 64

Solution:

Let the side of the square be s.

The radius of the quarter circle is s.

Area of the quarter circle:

= 1/4 πs²

Each semicircle has a diameter s.

Therefore, the radius of each semicircle is s/2.

Area of each semicircle:

= 1/2 × π × (s/2)²

= πs²/8

Total area of the two semicircles:

= πs²/8 + πs²/8

= πs²/4

Thus, the total area of the two semicircles is equal to the area of the quadrant.

From the overlapping regions shown in the figure:

Area of B
= Area of quadrant − (Total area of the two semicircles − Area of A)

= πs²/4 − (πs²/4 − Area of A)

= Area of A

Therefore:

Area A = Area B

Question 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.

Image 65

Solution:

Each semicircle has a diameter of 2 units.

Therefore:

Radius, r = 1 unit

The boundary of the four petals is formed by four complete semicircular arcs.

Total perimeter:

= 4πr

= 4 × 22/7 × 1

= 88/7

= 12⅘? Wait: 88/7 = 12 4/7 units

Image 66

Therefore, the perimeter of the flower is 88/7 units or 12 4/7 units.

Area of the square:

= 2²

= 4 square units

The combined area of regions I and III is:

Area of square − Area of two opposite semicircles

= 4 − 2 × 1/2 × π × 1²

= 4 − π

Using π = 22/7:

= 4 − 22/7

= 6/7 square unit

Similarly, the combined area of regions II and IV is also 6/7 square units.

Therefore, the area of the four petals:

= 4 − 6/7 − 6/7

= 4 − 12/7

= 16/7 square units

Thus:

Perimeter of the flower = 88/7 units

Area of the flower = 16/7 square units

Question 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 1/4πl².

Image 67

Solution:

Join O to A and B.

Let:

OA = x

OB = y

Since BC touches the smaller circle at A, OA is perpendicular to BC.

Also, the perpendicular drawn from the centre of a circle to a chord bisects the chord.

Therefore:

AB = AC = l/2

Image 68

In right-angled △OAB:

OA² + AB² = OB²

x² + (l/2)² = y²

Therefore:

y² − x² = l²/4

Area of the green region:

= Area of larger circle − Area of smaller circle

= πy² − πx²

= π(y² − x²)

= π × l²/4

= 1/4πl²

Hence proved.

Question 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).

Image 69

Solution:

Let the sides of the right-angled triangle be 2a, 2b and 2c, where 2c is the hypotenuse.

By the Baudhayana-Pythagoras theorem:

(2a)² + (2b)² = (2c)²

Therefore:

a² + b² = c²

The areas of the semicircles drawn on the three sides are proportional to a², b² and c².

From the figure:

Area (A) + Area (B)

= Area of the two smaller semicircles + Area of the triangle − Area of the largest semicircle

= 1/2πa² + 1/2πb² + Area (C) − 1/2πc²

= 1/2π(a² + b²) + Area (C) − 1/2πc²

Since a² + b² = c²:

Area (A) + Area (B)

= 1/2πc² + Area (C) − 1/2πc²

= Area (C)

Hence proved.

Question 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.

Image 70

Solution:

Let the centres of the two circles be A and B, and let C be one of their points of intersection.

Image 71

Since AC, BC and AB are all radii of the two congruent circles:

AC = BC = AB = r

Therefore, △ABC is an equilateral triangle.

The area of the upper half of the common region is:

Area of two 60° sectors − Area of equilateral △ABC

= 2 × 60°/360° × πr² − √3/4 r²

= πr²/3 − √3r²/4

The complete common region consists of two halves.

Therefore, its total area is:

= 2(πr²/3 − √3r²/4)

= (2π/3 − √3/2)r²

Thus, the area enclosed by the two circles is:

(2π/3 − √3/2)r²

Question 26: In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, and C, as marked. Show that the area of the rectangle is 2(A + C)(B + C)/C.

Image 72

Solution:

Let the adjacent sides of the rectangle be:

a + b and c + d

Image 73

Therefore:

Area of rectangle = (a + b)(c + d)

For triangle A:

Base = c

Height = a

A = 1/2 ac

For triangle C:

Base = b

Height = c

C = 1/2 bc

For triangle B:

Base = b

Height = d

B = 1/2 bd

Now:

2(A + C)(B + C)/C

= 2(1/2ac + 1/2bc)(1/2bd + 1/2bc)/(1/2bc)

= 2[1/2c(a + b)][1/2b(d + c)]/(1/2bc)

= (a + b)(c + d)

This is equal to the area of the rectangle.

Hence:

Area of rectangle = 2(A + C)(B + C)/C

Question 27: In the figure, we see two shaded regions formed by a quarter circle, a semicircle, and a triangle.

Image 74

Show that the areas of the two shaded regions are equal.

Solution:

Let:

OA = OB = r

Since ∠AOB = 90°, by the Baudhayana-Pythagoras theorem:

AB = √(OA² + OB²)

= √(r² + r²)

= √2r

Area of △OAB:

= 1/2 × OA × OB

= 1/2 × r × r

= 1/2r²

The diameter of the semicircle is AB = √2r.

Therefore, its radius is:

√2r/2 = r/√2

Area of the semicircle:

= 1/2 × π × (r/√2)²

= 1/2 × π × r²/2

= 1/4πr²

Area of the quarter circle:

= 1/4πr²

Therefore, the area of the shaded portion AFBE:

= Area of semicircle + Area of △OAB − Area of quarter circle

= 1/4πr² + 1/2r² − 1/4πr²

= 1/2r²

Thus:

Area of shaded portion AFBE = Area of △OAB

Hence, the two shaded regions have equal areas.

Learn Class 9 Maths Ganita Manjari Chapter 6 with Vedantu

Students can learn NCERT Class 9 Maths Ganita Manjari Chapter 6, Measuring Space: Perimeter and Area, with Vedantu through clear explanations and step-by-step solutions. The chapter covers important concepts such as the perimeter and area of geometrical figures, circles, sectors, triangles, parallelograms, trapeziums, rhombuses and other related shapes.

Vedantu’s NCERT Solutions for Class 9 Maths Chapter 6 provide easy-to-understand answers for Think and Reflect questions, Exercise Sets 6.1, 6.2 and 6.3, and the End of Chapter Exercises. These solutions help students understand formulas, improve calculation skills, revise the complete chapter and prepare confidently for school examinations.

Access Exercise Wise NCERT Solutions for Chapter 6 Maths Class 9

CBSE Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

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