Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises

Perimeter is the distance around a boundary, while area is the space covered by a two-dimensional region.
Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises connect these ideas with circles, triangles, trapeziums, kites, semicircles and equal-area figures.

Measuring Space Perimeter and Area Class 9 brings together perimeter, circumference, arc length, area of triangles, Heron formula Class 9, area of circle Class 9, sectors, segments and visual area models. Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises include numerical questions, algebraic area models, proof-based geometry and figure-based area comparisons. These Class 9 Maths Chapter 6 End of Chapter Exercise Solutions follow the textbook order and keep equations copy-friendly. The textbook says to use 22/7 for π unless stated otherwise.

Key Takeaways

• Triangle Area: Area can be found using 1/2 × base × height or Heron’s formula.
• Circle Measures: Circumference, sector area and quadrant area use π.
• Trapezium and Kite: Their areas can be proved by splitting into triangles.
• Similar Figures: If lengths are multiplied by k, area is multiplied by k².

Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises Structure 2026

Exercise Type	Main Topic	Question Count
Numerical	Triangle, circle and wheel problems	9
Proof	Trapezium, kite and equal-area figures	10
Figure-Based	Shaded fractions, semicircles and circles	8

Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises for Triangle Area

The first six questions revise algebraic area models, isosceles triangles, right triangles and Heron’s formula. These questions check whether students can choose the right area method from the given data.

Q1. Draw figures for (a + b)(a - b) = a² - b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

The identities can be shown by splitting rectangles or squares into smaller area parts.

For:

(a + b)(a - b) = a² - b²

Draw a square of side a.

Remove a square of side b from it.

The remaining area is:

a² - b²

This remaining region can be rearranged into a rectangle with sides:

a + b and a - b

So:

(a + b)(a - b) = a² - b²

For:

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

Draw a square of side:

a + b + c

Split each side into lengths:

a, b and c

The square contains:

a², b², c², ab, ab, bc, bc, ca, ca

So:

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

Answer:

The area models show the identities by dividing one large shape into smaller squares and rectangles.

Q2. An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

The area of the triangle is 50√2 cm².

Given:

Equal sides = 15 cm each

Perimeter = 40 cm

Base = 40 - 15 - 15

Base = 10 cm

Half base = 5 cm

Let height be h.

Using Baudhāyana-Pythagoras theorem:

15² = h² + 5²

225 = h² + 25

h² = 200

h = 10√2 cm

Area = 1/2 × base × height

Area = 1/2 × 10 × 10√2

Area = 50√2 cm²

Answer:

The area of the triangle is 50√2 cm².

Q3. An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

Each equal side is 13 cm.

Given:

Base = 10 cm

Area = 60 cm²

Use:

Area = 1/2 × base × height

60 = 1/2 × 10 × h

60 = 5h

h = 12 cm

Half base = 5 cm

Let each equal side be x.

Using Baudhāyana-Pythagoras theorem:

x² = 12² + 5²

x² = 144 + 25

x² = 169

x = 13 cm

Answer:

Each equal side is 13 cm.

Q4. The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.

The perimeter is 36 cm.

Given:

Area = 54 cm²

One leg = 12 cm

Let the other leg be x.

Area = 1/2 × 12 × x

54 = 6x

x = 9 cm

Hypotenuse:

h² = 12² + 9²

h² = 144 + 81

h² = 225

h = 15 cm

Perimeter = 12 + 9 + 15

Perimeter = 36 cm

Answer:

The perimeter is 36 cm.

Q5. The sides of a triangle are in the ratio 2:3:4, and its perimeter is 45 cm. Find its area.

The area is 45√15/4 cm².

Let the sides be:

2x, 3x, 4x

Perimeter:

2x + 3x + 4x = 45

9x = 45

x = 5

So the sides are:

10 cm, 15 cm, 20 cm

Semi-perimeter:

s = 45/2

Use Heron’s formula:

Area = √[s(s - a)(s - b)(s - c)]

Area = √[(45/2)(45/2 - 10)(45/2 - 15)(45/2 - 20)]

Area = √[(45/2)(25/2)(15/2)(5/2)]

Area = √(84375/16)

Area = 75√15/4 cm²

Answer:

The area is 75√15/4 cm².

Q6. The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area in two different ways.

The area is 84 cm² by both methods.

Method 1: Right triangle method

Check:

7² + 24² = 25²

49 + 576 = 625

625 = 625

So it is a right-angled triangle.

Area = 1/2 × 7 × 24

Area = 84 cm²

Method 2: Heron’s formula

s = (7 + 24 + 25)/2

s = 28

Area = √[28(28 - 7)(28 - 24)(28 - 25)]

Area = √[28 × 21 × 4 × 3]

Area = √7056

Area = 84 cm²

Answer:

The area is 84 cm².

Ganita Manjari Class 9 Chapter 6 End of Chapter Exercises for Circles and Wheels

Questions 7 to 9 use circumference, quadrant area and wheel revolutions. A wheel covers one circumference in one complete turn.

Q7. 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.

The cyclist travels 1320/7 m.

Given:

Diameter = 60 cm

Number of rotations = 100

Circumference = πd

Circumference = 22/7 × 60

Circumference = 1320/7 cm

Distance in 100 rotations:

Distance = 100 × 1320/7

Distance = 132000/7 cm

Convert to metres:

Distance = 132000/(7 × 100)

Distance = 1320/7 m

Answer:

The cyclist travels 1320/7 m.

Q8. Find the area of a quadrant of a circle whose circumference is 66 cm.

The area of the quadrant is 693/8 cm².

Given:

Circumference = 66 cm

Use:

2πr = 66

2 × 22/7 × r = 66

44r/7 = 66

r = 66 × 7/44

r = 21/2 cm

Area of quadrant = πr²/4

Area = 1/4 × 22/7 × (21/2)²

Area = 1/4 × 22/7 × 441/4

Area = 693/8 cm²

Answer:

The area of the quadrant is 693/8 cm².

Q9. 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.

The car travels 176 cm in one turn, and the wheel turns 6250/11 times in 1 km.

Given:

Radius = 28 cm

Circumference = 2πr

Circumference = 2 × 22/7 × 28

Circumference = 176 cm

So one complete turn covers:

176 cm

Now:

1 km = 100000 cm

Number of turns = 100000 ÷ 176

Number of turns = 100000/176

Number of turns = 6250/11

Number of turns ≈ 568.18

Answer:

The car travels 176 cm per turn, and the wheel turns 6250/11 times.

Class 9 Maths Chapter 6 End of Chapter Exercise Solutions for Trapezium, Kite and Similar Figures

Questions 10 to 15 are proof-based. They use area of rectangle, area of parallelogram, area of trapezium, area of kite Class 9 and scale factor reasoning.

Q10. Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Yes, the rectangles are congruent.

Let the first rectangle have sides a and b.

Let the second rectangle have sides c and d.

Same perimeter gives:

2(a + b) = 2(c + d)

a + b = c + d

Same area gives:

ab = cd

So the two pairs of side lengths have the same sum and same product.

Therefore:

{a, b} = {c, d}

Answer:

The rectangles have the same side lengths, so they are congruent.

Q11. Using the parallelogram area, show that the area of a trapezium is 1/2(a + b)h.

The area of the trapezium is 1/2(a + b)h.

Let the parallel sides be a and b.

Let the height be h.

Complete the trapezium into a parallelogram of height h.

The parallelogram has base:

a + b

So:

Area of parallelogram = (a + b)h

The trapezium is half of this parallelogram.

Therefore:

Area of trapezium = 1/2(a + b)h

Answer:

Area of trapezium = 1/2(a + b)h.

Q12. By dividing a trapezium into two triangles, show that its area is half the sum of the parallel sides multiplied by the height.

The area of the trapezium is 1/2(a + b)h.

Let the parallel sides be a and b.

Let height be h.

Draw a diagonal.

The trapezium becomes two triangles.

Area of first triangle = 1/2 × a × h

Area of second triangle = 1/2 × b × h

Total area:

Area = 1/2ah + 1/2bh

Area = 1/2(a + b)h

Answer:

Area of trapezium = 1/2(a + b)h.

Q13. Show how two identical copies of a trapezium make a parallelogram.

Two identical trapeziums make a parallelogram with base a + b and height h.

Place one trapezium beside an identical inverted copy.

The combined figure is a parallelogram.

Its base is:

a + b

Its height is:

h

Area of parallelogram = (a + b)h

So area of one trapezium:

Area = 1/2(a + b)h

Answer:

Area of trapezium = 1/2(a + b)h.

Q14. Show that the area of a kite is half the product of its diagonals using algebra and geometry.

The area of a kite is 1/2 × d1 × d2.

Let the diagonals be d1 and d2.

In a kite, one diagonal bisects the other at right angles.

Let the bisected diagonal have parts d2/2 and d2/2.

Using algebra:

Area of kite = sum of two triangle areas

Area = 1/2 × d1 × d2/2 + 1/2 × d1 × d2/2

Area = d1d2/4 + d1d2/4

Area = d1d2/2

Using geometry:

Two congruent triangles combine to form the kite.

The total base-height product becomes:

d1 × d2

So:

Area = 1/2 × d1 × d2

Answer:

Area of kite = 1/2 × d1 × d2.

Q15. Three problems about fitting congruent shapes together.

Scaling all side lengths by k multiplies area by k².

Q15(i). Rectangles with sides a, b and 2a, 2b

Area of ABCD = ab

Area of PQRS = 2a × 2b

Area of PQRS = 4ab

So:

Area of PQRS = 4 × Area of ABCD

Four copies of ABCD fit into PQRS in a 2 by 2 arrangement.

Answer:

PQRS has 4 times the area of ABCD.

Q15(ii). Triangles with sides a, b, c and 2a, 2b, 2c

All side lengths are doubled.

Scale factor = 2

Area factor = 2²

Area factor = 4

So:

Area of ∆PQR = 4 × Area of ∆ABC

Four copies can be arranged to form the larger similar triangle.

Answer:

∆PQR has 4 times the area of ∆ABC.

Q15(iii). Triangles with sides a, b, c and 3a, 3b, 3c

All side lengths are tripled.

Scale factor = 3

Area factor = 3²

Area factor = 9

So:

Area of ∆PQR = 9 × Area of ∆ABC

Nine copies can be arranged to form the larger similar triangle.

Answer:

∆PQR has 9 times the area of ∆ABC.

Measuring Space Perimeter and Area Class 9: Shaded Fractions and Circle Coverage

Questions 16 to 18 use area fractions from the diagrams. Fig. 6.43, Fig. 6.44, Fig. 6.45 and Fig. 6.46 ask for shaded fractions and circle coverage in rectangles.

Q16. What fraction of the triangle and square is shaded in Fig. 6.43 and Fig. 6.44?

In Fig. 6.43, the shaded fraction of the triangle is 3/4.

The unshaded top triangle is formed by joining midpoints of two sides.

So its area is:

1/4 of the full triangle

Therefore:

Shaded area = 1 - 1/4

Shaded area = 3/4

In Fig. 6.44, the shaded fraction of the square is 1/5.

The central shaded square is formed by lines from corners to side midpoints.

A coordinate-area calculation gives:

Shaded area = 1/5 × area of square

Answer:

Triangle shaded fraction = 3/4

Square shaded fraction = 1/5

Q17. What fraction of each rectangle is covered by the circles in Fig. 6.45 and Fig. 6.46?

The fraction covered by the circles is π/4 in both figures.

Let each circle have radius r.

For n equal circles in one row:

Total circle area = nπr²

Rectangle height = 2r

Rectangle width = 2nr

Rectangle area = 2r × 2nr

Rectangle area = 4nr²

Fraction covered:

Fraction = nπr² / 4nr²

Fraction = π/4

Answer:

The covered fraction is π/4.

Q18. Make and prove a conjecture for 10, 20 and 50 circles fitted into a rectangle.

The fraction of the rectangle covered is always π/4.

For 10 circles:

Fraction = π/4

For 20 circles:

Fraction = π/4

For 50 circles:

Fraction = π/4

General proof:

For n circles of radius r:

Total area of circles = nπr²

Area of rectangle = 4nr²

Fraction = nπr² / 4nr²

Fraction = π/4

Answer:

The area fraction occupied by circles is always π/4.

Class 9 Maths Area Solutions for Figure-Based Proof Questions

Questions 19 to 23 use stacked rectangles, trisection, semicircles, flowers and concentric circles. The figures require area comparison rather than long calculation.

Q19. Nine identical rectangles form a large rectangle of area 72 cm². Find the perimeter of each small rectangle.

The perimeter of each small rectangle is 18√10/5 cm.

Let each small rectangle have sides x and y.

There are 9 identical rectangles.

Total area = 72 cm²

So:

9xy = 72

xy = 8

From the arrangement:

4x = 5y

So:

x = 5y/4

Substitute:

(5y/4)y = 8

5y²/4 = 8

y² = 32/5

y = 4√10/5

Then:

x = 5y/4

x = √10

Perimeter of each small rectangle:

P = 2(x + y)

P = 2(√10 + 4√10/5)

P = 2(9√10/5)

P = 18√10/5 cm

Answer:

The perimeter of each small rectangle is 18√10/5 cm.

Q20. Show that the shaded blue triangle and shaded red triangle have equal areas.

The blue and red triangles have equal areas.

The base is divided into three equal parts.

Both shaded triangles stand between consecutive trisection points.

So their bases are equal.

They also have the same vertex above the same straight base line.

So their heights are equal.

Therefore:

Area of blue triangle = Area of red triangle

Answer:

The two shaded triangles have equal areas.

Q21. Show that regions A and B in Fig. 6.49 have equal area.

Regions A and B have equal area.

Let the side of the square be 2r.

The quarter circle has radius 2r.

Each semicircle has radius r.

When the areas of the two semicircles and the quarter circle are compared, the extra part in one region matches the missing part in the other.

So:

Area A = Area B

Answer:

The two shaded regions A and B have equal area.

Q22. Find the perimeter and area of the 4-petalled flower in Fig. 6.50.

The perimeter is 4π units, and the area is 2π - 4 square units.

The square side is 2 units.

Each semicircle has radius 1 unit.

Each petal is made of two quarter-circle arcs.

Perimeter of one petal:

Perimeter = π/2 + π/2

Perimeter = π

For four petals:

Total perimeter = 4π units

Area of one petal:

Area = two 90° sectors - two right triangles

Area = π/2 - 1

For four petals:

Total area = 4(π/2 - 1)

Total area = 2π - 4

Answer:

Perimeter = 4π units

Area = 2π - 4 square units

Q23. Show that the area between two concentric circles is 1/4πl².

The green region has area 1/4πl².

Let the larger radius be R.

Let the smaller radius be r.

Chord BC touches the smaller circle at A.

So OA is perpendicular to BC.

Since BC = l:

AB = l/2

In right triangle OAB:

OB² = OA² + AB²

R² = r² + (l/2)²

R² - r² = l²/4

Area of green region:

Area = πR² - πr²

Area = π(R² - r²)

Area = π × l²/4

Area = 1/4πl²

Answer:

The green region has area 1/4πl².

Class 9 Maths Ganita Manjari Chapter 6 Solutions for Semicircle Area Problems

Questions 24 to 27 use semicircle areas, circle overlap and area equality. These starred questions combine circle formulas with Baudhāyana-Pythagoras theorem and equal-area reasoning.

Q24. Semicircles are drawn on all sides of a right-angled triangle. Show that Area(A) + Area(B) = Area(C).

Area(A) + Area(B) = Area(C).

Let the right triangle have legs a and b.

Let the hypotenuse be c.

Area of semicircle on side a:

Area(A) = 1/2 × π(a/2)²

Area(A) = πa²/8

Area of semicircle on side b:

Area(B) = πb²/8

Area of semicircle on side c:

Area(C) = πc²/8

By Baudhāyana-Pythagoras theorem:

a² + b² = c²

Multiply by π/8:

πa²/8 + πb²/8 = πc²/8

Therefore:

Area(A) + Area(B) = Area(C)

Answer:

Area(A) + Area(B) = Area(C).

Q25. Two congruent circles of radius r pass through each other’s centres. Find the area enclosed by the two circles.

The common enclosed area is r²(2π/3 - √3/2).

Let the two centres be A and B.

Since each circle passes through the other centre:

AB = r

The two intersection points form equilateral triangles with the centres.

Each sector angle is:

120°

Area of one 120° sector:

Sector area = πr² × 120/360

Sector area = πr²/3

Area of one equilateral triangle:

Triangle area = √3r²/4

One segment area:

Segment area = πr²/3 - √3r²/4

The common enclosed region has two such segments.

Area = 2(πr²/3 - √3r²/4)

Area = r²(2π/3 - √3/2)

Answer:

The enclosed area is r²(2π/3 - √3/2).

Q26. In Fig. 6.54, show that the rectangle area is 2(A + C)(B + C)/C.

The rectangle area is 2(A + C)(B + C)/C.

Let the rectangle have length L and height H.

The three triangles have areas A, B and C.

Using the shared altitude and base relations from the rectangle:

A + C represents one half-product involving one side.

B + C represents the matching half-product involving the other side.

The central triangle C is counted in both comparisons.

Combining the two half-products gives:

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

Answer:

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

Q27. Show that the two shaded regions in Fig. 6.55 have equal areas.

The two shaded regions have equal areas.

Both shaded parts are formed from the same quarter circle, semicircle and triangle.

Let the common triangle area be T.

Let the quarter circle area be Q.

Let the semicircle area be S.

One shaded region can be written as:

Q - T - common unshaded part

The other shaded region can be written as:

S - T - matching common part

The circle construction makes the remaining circular parts equal.

Therefore:

Area of first shaded region = Area of second shaded region

Answer:

The two shaded regions have equal areas.

Measuring Space Perimeter and Area Class 9 Concepts Used in End Exercises

The End of Chapter Exercises combine formulas from the full chapter. The main ideas are perimeter, circumference, triangle area, Heron’s formula, trapezium area, kite area, circle area and semicircle area.

Perimeter and Area Class 9

Perimeter measures boundary length.

Copy-friendly formula:

Perimeter of rectangle = 2(l + b)

Area measures covered region.

Copy-friendly formula:

Area of rectangle = l × b

Heron Formula Class 9

Heron’s formula finds triangle area from three sides.

s = (a + b + c)/2

Area = √[s(s - a)(s - b)(s - c)]

Area of Trapezium Class 9

Area of trapezium uses the two parallel sides and height.

Area = 1/2(a + b)h

Area of Kite Class 9

Area of a kite is half the product of its diagonals.

Area = 1/2 × d1 × d2

Area of Circle Class 9

Area of a circle depends on the square of its radius.

Area = πr²

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises

Concept	Copy-Friendly Formula	Used In
Heron’s formula	Area = √[s(s - a)(s - b)(s - c)]	Q5, Q6
Trapezium area	Area = 1/2(a + b)h	Q11, Q12, Q13
Circle area	Area = πr²	Q8, Q21, Q22, Q25

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6

Section	NCERT Solutions
Class 9 Maths Ganita Manjari 2026	NCERT Class 9 Maths Ganita Manjari 2026
Chapter 6	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6
Exercise 6.1	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.1
Exercise 6.2	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.2
Exercise 6.3	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3
End of Chapter Exercises	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises