CBSE Class 6

RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21C

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21A
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21B
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21C
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21D
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21E

Question 1.
Solution:
1. In the given figure, there are 12 complete squares, so its area is 12 cm².
2. In the given figure, there are 18 complete squares, so its area is 18 cm².
3. In the given figure, there are 14.5 complete squares, so its area is 14.5 cm² .
4. In the given figure, there are 6 complete squares and four half parts of a square, the area of the figure is 8 cm².
5. The given figure contains 9 complete squares and 6 half parts of a square. So the area of the figure is
\(\left( 9+\frac { 6 }{ 2 } \right) \) = 9 + 3 = 12 cm².
6. The given figure contains 16 complete squares. So, its area is 16 cm².
7. The given figure contains 4 complete squares, 8 more than half parts and 4 less than half parts of a square. Neglecting the less than half parts and considering the more than half parts as complete squares, the approximate area of the figure is 12 cm².
8. The given figure contains 7 complete squares and 5 more than half parts and some less than half parts of a square. Neglecting the less than half parts and considering more than half parts as complete square, the area of the figure is 12 cm² approximately.
9. The given figure contains 14 complete squares and four half parts of a square. So, the area of the figure is
\(\left( 14+\frac { 4 }{ 2 } \right) \) cm² = (14 + 2) cm² = 16 cm².

Hope given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21C are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21B

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21A
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21B
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21C
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21D
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21E

Question 1.
Solution:
(i) Radius of the circle (r) = 28 cm
Circumference = 2 πr
= 2 x \(\\ \frac { 22 }{ 7 } \) x 28 cm
= 176 cm Ans.

Question 2.
Solution:
(i) Diameter of the circle (d) = 14 cm
Circumference = πd
= \(\\ \frac { 22 }{ 7 } \) x 14
= 44 cm

Question 3.
Solution:
Circumference of the circle = 176 cm
Let r be the radius, then

Question 4.
Solution:
Circumference of a wheel = 264 cm
Let d be its diameter, then
πd = 264
=> \(\\ \frac { 22 }{ 7 } d\)
= 264

Question 5.
Solution:
Diameter of the wheel (d) = 77 cm
Circumference = πd

Question 6.
Solution:
Diameter of the wheel = 70 cm
circumference = πd

Hope given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21B are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21A

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21A
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21B
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21C
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21D
• RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21E

Question 1.
Solution:
(i) Length (l) = 16.8 cm
Breadth (b) = 6.2 cm
Perimeter = 2 (l + b)
= 2 (16.8 + 6.2) cm
= 2 x 23
= 46 cm


= 30m 6 dm

Question 2.
Solution:
Length of rectangular field (l) = 62 m
and breadth (b) = 33 m

Question 3.
Solution:
Perimeter of field = 128 m
Length + Breadth = \(\\ \frac { 128 }{ 2 } \) = 64 m
Ratio in length and breadth = 5:3
Let length (l) = 5x
Then breadth = 3x
5x + 3x = 64
=> 5x = 64
=> x = \(\\ \frac { 64 }{ 8 } \) = 8
Length of the field = 5x = 5 x 8 = 40m
and breadth = 3x = 3 x 8 = 24m

Question 4.
Solution:
Cost of fencing a rectangular field
= Rs. 18 per m
Total cost = Rs. 1980

Question 5.
Solution:
Total cost of fencing a rectangular field
= Rs 3300
Rate of fencing = Rs 25 per m

Question 6.
Solution:
(i) Side of square = 3.8 cm
Perimeter = 4 x side
= 4 x 3.8 cm
= 15.2 cm
(ii) Side of a square = 4.6 m
Perimeter = 4 x side
= 4 x 4.6 m
= 18.4 m
(iii) Side of a square = 2 m 5 dm
= 2.5 m
Perimeter = 4 x side
= 4 x 2.5 m
= 10 m

Question 7.
Solution:
Total cost of fencing a square field = Rs. 4480
Rate of fencing = Rs. 35 per m

Question 8.
Solution:
Side of a square field (a) = 21 m
Perimeter = 4a = 4 x 21 = 84m
Perimeter of rectangular field = 84 m
Ratio in length and breadth = 4 : 3
Let length (l) = 4x
and breadth (b) = 3x
Perimeter = 2 (l + b)

Question 9.
Solution:
(i) Sides of a triangle are 7.8 cm, 6.5 cm and 5.9 cm

Question 10.
Solution:
(i) Each side of a regular pentagon
= 8 cm
Perimeter = 5 x Side
= 5 x 8
= 40 cm
(ii) Each side of an octagon = 4.5 cm
Perimeter = 8 x Side
= 8 x 4.5
= 36 cm
(iii) Each side of a regular decagon = 3.6 cm
Perimeter = 10 x Side
= 10 x 3.6
= 36 cm

Question 11.
Solution:
We know that perimeter of a closed figure or a polygon = Sum of its sides
(i) In the figure, sides of a quadrilateral are 45 cm, 35 cm, 27 cm, 35 cm
Its perimeter = Sum of its sides
= (45 + 35 + 27 + 35) cm
= 142 cm
(ii) Sides of a quadrilateral (rhombus) are 18 cm, 18 cm, 18 cm, 18 cm
i.e., each side = 18 cm Perimeter
= 4 x Side
= 4 x 18
= 72 cm
(iii) Sides of the polygon given are 16 cm, 4 cm, 12 cm, 12 cm, 4 cm, 16 cm and 8 cm
Its perimeter = Sum of its sides
= (16 + 4 + 12 + 12 + 4 + 16 + 8) cm
= 72 cm

Hope given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 20 Two-Dimensional Reflection Symmetry (Linear Symmetry) Ex 20

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 20 Two-Dimensional Reflection Symmetry (Linear Symmetry) Ex 20

Mark against the correct answer in each of Q. 1 to Q. 8.

Question 1.
Solution:
(d)∵ A square has four lines of symmetry, two diagonals and two lines joining the mid-points of opposite sides.

Question 2.
Solution:
(c) ∵ A rectangle has two lines of symmetry, each one of which being the line joining of mid-points of opposite sides.

Question 3.
Solution:
(b) ∵ A rhombus has two lines of symmetry namely two diagonals.

Question 4.
Solution:
(d) Each diameter of a circle is its line of symmetry which are unlimited numbers.

Question 5.
Solution:
(a) ∵ A scalene triangle has no line of symmetry.

Question 6.
Solution:
(a) ∵ It is a figure of kite ; so AC is its line of symmetry.

Question 7.
Solution:
(c) ∵ Letter O has two lines of symmetry, one vertical and second horizontal

Question 8.
Solution:
(a) ∵ Letter Z has no line of symmetry.

Question 9.
Solution:

Question 10.
Solution:
(i) True (T) Parallelogram has no line of symmetry.
(ii) True (T) Bisector of an angle of*equal sides is the line of symmetry.
(iii) True (T)  Perpendiculars from each vertices’s of an equilateral-triangle to its opposite side is its line of symmetry.
(iv) False (F) Rhombus has two lines of symmetry which are its -diagonals.
(v) True (T) Square has four lines of symmetry, two diagonals and two perpendicular bisectors of opposite sides.
(vi) True (T) A rectangle has two lines of symmetry which are the perpendicular bisectors of its opposite sides.
(vii) True (T) H, I, O and X has two lines of symmetry.

Hope given RS Aggarwal Solutions Class 6 Chapter 20 Two-Dimensional Reflection Symmetry (Linear Symmetry) Ex 20 are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 19 Three-Dimensional Shapes Ex 19

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 19 Three-Dimensional Shapes Ex 19

Tick the correct answer in each of Q. 1 to Q. 6.

Question 1.
Solution:
(c) ∵ A cuboid has three dimensions, length, breadth and height or depth.

Question 2.
Solution:
(b) ∵ Its six faces arc of square.

Question 3.
Solution:
(d) ∵ Its shape is of a cylinder as it is round in shape on either sides/faces.

Question 4.
Solution:
(c) ∵ Football is round as sphere.

Question 5.
Solution:
(b) A brick has length, breadth and height.

Question 6.
Solution:
(d) ∵ Its shape is like a cone. Ans.

Question 7.
Solution:
(i) solid
(ii) 6, 12 and 18
(iii) opposite
(iv) sphere
(v) cube
(vi) 4, 8
(vii) 3, 6
(viii) 6, 3, 2, 9 Ans.

Question 8.
Solution:
(a) A cone:
(i) Conical cup
(ii) An ice cream cup
(iii) Conical tent house
(iv) Conical vessel.
(b) A cuboid :
(i) A book,
(ii) A brick,
(iii) a box,
(iv) a briefcase.
(c) A cylinder
(i) Circular pipe
(ii) A jar or tumbler
(iii) A round powder tin
(iv) Circular pillar.

Hope given RS Aggarwal Solutions Class 6 Chapter 19 Three-Dimensional Shapes Ex 19 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.