Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises
NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises – Measuring Space: Perimeter and Area provide detailed solutions to the questions given at the end of the chapter. These exercises help students revise and apply important concepts related to perimeter, circumference, area, and different plane figures. The step-by-step solutions make it easier for students to understand the concepts and solve questions accurately.
Students can use these Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercises Solutions for practice, revision, and exam preparation. The solutions are explained in a simple and easy-to-understand manner, and a printable PDF is also available for convenient revision.
NCERT Solutions for Class 9 Maths Chapter 6 End-of-Chapter Exercises
The end-of-chapter exercises include area models, triangle area, Heron’s formula, circumference, sectors, trapezium area, kite area, scaling of shapes and advanced composite-area questions.
Question 1. Draw figures corresponding to the identities (a + b)(a - b) = a² - b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
Answer:
For (a + b)(a - b) = a² - b²:
Draw a square of side a. Remove a smaller square of side b. The remaining area is a² - b². Rearranging the remaining shape forms a rectangle of sides (a + b) and (a - b).
For (a + b + c)²:
Draw a square of side a + b + c. Divide each side into parts a, b and c. The square is divided into three smaller squares of areas a², b² and c², and rectangles whose total areas are 2ab, 2bc and 2ca.
Final answer:
Both identities can be shown by decomposing areas of squares and rectangles.
Question 2. An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area.
Answer:
Equal sides = 15 cm each
Perimeter = 40 cm
Base = 40 - 15 - 15
= 10 cm
The altitude bisects the base.
Half base = 5 cm
Height:
h = √(15² - 5²)
= √(225 - 25)
= √200
= 10√2 cm
Area:
= 1/2 × base × height
= 1/2 × 10 × 10√2
= 50√2 cm²
Final answer:
50√2 cm²
Question 3. An isosceles triangle has base 10 cm and area 60 cm². What are the lengths of the equal sides?
Answer:
Area = 1/2 × base × height
60 = 1/2 × 10 × h
60 = 5h
h = 12 cm
The altitude bisects the base.
Half base = 5 cm
Equal side:
= √(12² + 5²)
= √(144 + 25)
= √169
= 13 cm
Final answer:
Each equal side is 13 cm
Question 4. The area of a right-angled triangle is 54 cm². One leg has length 12 cm. Find its perimeter.
Answer:
Area = 1/2 × product of legs
54 = 1/2 × 12 × other leg
54 = 6 × other leg
Other leg = 9 cm
Hypotenuse:
= √(12² + 9²)
= √(144 + 81)
= √225
= 15 cm
Perimeter:
= 12 + 9 + 15
= 36 cm
Final answer:
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.
Answer:
Let sides be 2x, 3x and 4x.
2x + 3x + 4x = 45
9x = 45
x = 5
Sides are:
10 cm, 15 cm, 20 cm
Semi-perimeter:
s = 45/2 = 22.5 cm
Using Heron’s formula:
Area = √[22.5(22.5 - 10)(22.5 - 15)(22.5 - 20)]
= √[22.5 × 12.5 × 7.5 × 2.5]
= 75√15/4 cm²
Final answer:
75√15/4 cm²
Question 6. The sides of a triangle are 7 cm, 24 cm and 25 cm. Find the area in two different ways.
Answer:
Since:
7² + 24² = 49 + 576 = 625 = 25²
The triangle is right-angled.
Method 1:
Area = 1/2 × 7 × 24
= 84 cm²
Method 2 using Heron’s formula:
s = (7 + 24 + 25)/2
= 28
Area = √[28(28 - 7)(28 - 24)(28 - 25)]
= √[28 × 21 × 4 × 3]
= √7056
= 84 cm²
Final answer:
84 cm²
Question 7. If the wheel of a bicycle has diameter 60 cm, find how far a cyclist travels after 100 rotations.
Answer:
Diameter = 60 cm
Distance in one rotation = circumference
= πd
= 22/7 × 60
= 1320/7 cm
Distance in 100 rotations:
= 100 × 1320/7
= 132000/7 cm
= 18857.14 cm
= 188.57 m
Final answer:
188.57 m approximately
Question 8. Find the area of a quadrant of a circle whose circumference is 66 cm.
Answer:
Circumference = 66 cm
2πr = 66
2 × 22/7 × r = 66
r = 10.5 cm
Area of quadrant:
= 1/4 × πr²
= 1/4 × 22/7 × 10.5²
= 86.625 cm²
Final answer:
86.625 cm²
Question 9. The wheel of a car has outer radius 28 cm. Calculate how far the car travels after one complete turn and how many times the wheel turns in 1 km.
Answer:
Radius = 28 cm
Distance in one turn = circumference
= 2πr
= 2 × 22/7 × 28
= 176 cm
1 km = 100000 cm
Number of turns:
= 100000/176
= 568.18
Final answer:
Distance per turn = 176 cm; number of turns in 1 km ≈ 568
Question 10. Two rectangles have the same area and the same perimeter. Does this mean they are congruent?
Answer:
Let sides of one rectangle be a and b.
Area = ab
Perimeter = 2(a + b)
If another rectangle has the same area and perimeter, then the sum and product of its side lengths are the same. This means the side lengths must be the same pair, possibly in reversed order.
Final answer:
Yes, the rectangles are congruent.
Question 11. Using the area of a parallelogram, show that the area of a trapezium is 1/2(a + b)h.
Answer:
Take two congruent copies of the trapezium and join them to form a parallelogram.
The base of the parallelogram becomes:
a + b
The height remains h.
Area of parallelogram:
= (a + b)h
So, area of one trapezium:
= 1/2(a + b)h
Final answer:
Area of trapezium = 1/2(a + b)h
Question 12. By dividing a trapezium into two triangles, show that its area is half the sum of parallel sides multiplied by height.
Answer:
Let the parallel sides be a and b, and height be h.
Divide the trapezium along a diagonal into two triangles.
Area of first triangle = 1/2 × a × h
Area of second triangle = 1/2 × b × h
Total area:
= 1/2ah + 1/2bh
= 1/2(a + b)h
Final answer:
Area of trapezium = 1/2(a + b)h
Question 13. Show how two identical copies of a trapezium make a parallelogram.
Answer:
Place one copy of the trapezium inverted next to the original copy. The non-parallel sides match, and the two copies form a parallelogram.
The base of the parallelogram becomes a + b, and height remains h.
Area of parallelogram = (a + b)h
Therefore, area of trapezium = 1/2(a + b)h
Final answer:
Two identical trapeziums form a parallelogram, giving the trapezium area formula.
Question 14. Show that the area of a kite is half the product of its diagonals.
Answer:
Let the diagonals of the kite be d₁ and d₂.
In a kite, one diagonal divides the kite into two triangles. The other diagonal gives the combined heights of these triangles.
Area of kite:
= 1/2 × d₁ × d₂
Final answer:
Area of kite = 1/2 × product of diagonals
Question 15. Problems about fitting congruent shapes together.
(i) Rectangles with sides a, b and 2a, 2b
Answer:
Area of smaller rectangle = ab
Area of larger rectangle = 2a × 2b = 4ab
So, the larger rectangle has 4 times the area. Four copies of the smaller rectangle can fit into the larger rectangle.
Final answer:
Yes, 4 copies fit.
(ii) Triangles with sides a, b, c and 2a, 2b, 2c
Answer:
If all side lengths are doubled, the scale factor is 2.
Area scale factor = 2² = 4
So, the larger triangle has 4 times the area.
Final answer:
Yes, 4 congruent copies can be arranged to form the larger similar triangle.
(iii) Triangles with sides a, b, c and 3a, 3b, 3c
Answer:
Scale factor = 3
Area scale factor = 3² = 9
Final answer:
Yes, 9 congruent copies can be arranged to form the larger similar triangle.
Question 16. What fraction of the triangle and square is shaded?
Answer:
This is diagram-dependent and must be answered using Fig. 6.43 and Fig. 6.44.
Final answer:
Use the partition shown in the figure to compare shaded area with total area.
Question 17. What fraction of the rectangle is covered by the circles?
Answer:
This is diagram-dependent and must be answered using Fig. 6.45 and Fig. 6.46.
Final answer:
Find total area of all circles and divide by the area of the rectangle.
Question 18. Make and prove a conjecture about the area occupied by circles fitted into a rectangle.
Answer:
When equal circles are fitted in rows inside a rectangle so that each circle touches its neighbours and the boundary, the fraction of area covered depends on the circular area compared with the square or rectangular cell around each circle.
For each circle of radius r:
Area of circle = πr²
If each circle fits in a square of side 2r:
Area of square cell = 4r²
Fraction covered:
= πr²/4r²
= π/4
Final answer:
The circles cover π/4 of the rectangle in such an arrangement.
Question 19. Nine identical rectangles make a large rectangle whose area is 72 cm². Find the perimeter of each small rectangle.
Answer:
This question depends on Fig. 6.47. The relation between the length and breadth of each small rectangle must be read from the arrangement.
Final answer:
Use the figure to set equations for the sides of each small rectangle, then use total area 72 cm² to find the perimeter.
Question 20. Show that the shaded blue and red triangles have equal area.
Answer:
The figure shows lines from a vertex to points of trisection of the opposite side.
Since the opposite side is divided into equal parts, triangles standing on equal bases and having the same height have equal areas.
Final answer:
The blue and red triangles have equal areas because they have equal bases and the same height.
Question 21. Show that shaded regions A and B have equal area.
Answer:
The figure uses a quarter circle and two semicircles on adjacent sides of a square.
By comparing the areas of the quarter circle, the two semicircles and the right triangle formed inside the square, the extra region on one side equals the missing region on the other side.
Final answer:
Areas A and B are equal by subtracting equal common regions from equal circular parts.
Question 22. Four semicircles are drawn inside a square of side 2 units. Find the perimeter and area of the 4-petalled flower.
Answer:
Each semicircle has radius 1 unit.
The boundary of the 4-petalled flower consists of 4 semicircular arcs of radius 1.
Perimeter:
= 4 × π × 1
= 4π units
For the area, each petal is formed by overlap of two semicircles. The final area is diagram-dependent but can be found by adding four equal petal regions.
Final answer:
Perimeter = 4π units. Area should be computed from the petal overlap regions in Fig. 6.50.
Question 23. In two concentric circles, a chord BC of the larger circle touches the smaller circle at A. If BC = l, show that the green region area is πl²/4.
Answer:
Let outer radius be R and inner radius be r.
Since BC touches the smaller circle at A, OA is perpendicular to BC.
A is midpoint of BC.
So:
AB = l/2
In right triangle OAB:
R² = r² + (l/2)²
Therefore:
R² - r² = l²/4
Area of green region:
= πR² - πr²
= π(R² - r²)
= πl²/4
Final answer:
Area of green region = πl²/4
Question 24. Show that Area(A) + Area(B) = Area(C) for semicircles drawn on the sides of a right-angled triangle.
Answer:
Let the legs of the right triangle be a and b, and the hypotenuse be c.
Area of semicircle on side a:
= 1/2 × π(a/2)²
= πa²/8
Area of semicircle on side b:
= πb²/8
Area of semicircle on hypotenuse c:
= πc²/8
By Pythagoras theorem:
a² + b² = c²
So:
πa²/8 + πb²/8 = πc²/8
Final answer:
Area(A) + Area(B) = Area(C)
Question 25. Two congruent circles pass through each other’s centres. Find the area of the region enclosed by the two circles in terms of radius r.
Answer:
The overlapping region is made of two identical circular segments.
For each circle, the sector angle is 120°.
Area of one sector:
= πr² × 120/360
= πr²/3
The triangle formed is equilateral with side r.
Area of equilateral triangle:
= √3r²/4
Area of one segment:
= πr²/3 - √3r²/4
Total overlapping area:
= 2(πr²/3 - √3r²/4)
= 2πr²/3 - √3r²/2
Final answer:
Area = 2πr²/3 - √3r²/2
Question 26. In Fig. 6.54, show that the area of the rectangle is 2(A + C)(B + C)/C.
Answer:
This is a diagram-based result involving three triangles within a rectangle.
Using similarity of triangles and area ratios, the dimensions of the rectangle can be expressed in terms of the areas A, B and C. Multiplying those dimensions gives:
Area of rectangle = 2(A + C)(B + C)/C
Final answer:
Area of rectangle = 2(A + C)(B + C)/C
Question 27. Show that the two shaded regions formed by a quarter circle, a semicircle and a triangle have equal areas.
Answer:
The two shaded regions are formed by subtracting common triangular and circular parts from equal circular regions.
Since the same triangle and matching circular portions are involved, the remaining shaded areas are equal.
Final answer:
The two shaded regions have equal areas by area subtraction from equal circular parts.
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