RS Aggarwal Maths Solutions Class 6

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

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 21E

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

Objective questions
Mark against the correct answer in each of the following :

Question 1.
Solution:
Ratio in the sides of a rectangle = 7 : 5
and perimeter = 96 cm

Question 2.
Solution:
Area of a rectangle = 650 cm²
and breadth (b) = 13 cm

Question 3.
Solution:
Length of a rectangular field (l) = 34 m
and breadth (b) = 18 m
Circumference = 2 (l + b)
= 2 (34 + 18)m
= 2 x 52
= 104 m
Rate of fencing = Rs. 22.50 per m
Total cost = Rs. 22.50 x 104
= Rs. 2340 (b)

Question 4.
Solution:
Total cost of fencing = Rs. 2400
Rate = Rs. 30 per m
Perimeter of the rectangular field

Question 5.
Solution:
Area of rectangular carpet =120 cm²
Perimeter = 46 m
Now 2 (l + b)
= 46 m

Question 6.
Solution:
Let width of a rectangle = x
Then length = 3x
and diagonal = 6√10 cm

Question 7.
Solution:
Ratio in length and perimeter of a rectangle = 1 : 3
Let length = x,
then perimeter = 3x

Question 8.
Solution:
Length of diagonal of a square = 20 cm

Question 9.
Solution:
Total cost of fencing around a square field = Rs. 2000
and rate = Rs. 25 per metre

Question 10.
Solution:
Circumference = πd
= \(\\ \frac { 22 }{ 7 } \) x 7
= 22 cm (b)

Question 11.
Solution:
(a) Diameter = \(\frac { circumference }{ \pi } \)
= \(\\ \frac { 88\times 7 }{ 22 } \)
= 28 cm

Question 12.
Solution:
Circumference = πd
= \(\\ \frac { 22 }{ 7 } \) x 70
= 220 cm

Question 13.
Solution:
Length of the lane = 150 m
Breadth of the lane = 9 m
Area of the lane = (150 x 9) m²
= 1350 m²
Area of the brick = 22.5 cm x 7.5 cm
= 168.75 cm²

Question 14.
Solution:
Length of a rectangular room (l) = 5 m 40 cm = 5.4 m
and breadth (b) = 4 m 50 cm
= 4.5 m
Area = l x b = 5.4 x 4.5 m²
= 24.3 m² (b)

Question 15.
Solution:
Length of a sheet (l) = 72 cm
and breadth (b) = 48 cm
Area = l x b = 72 x 48 cm²
Area of paper for one envelope = 18 x 12 cm²
No. of envelopes = \(\\ \frac { 72\times 48 }{ 18\times 12 } \) =16 (d)

Hope given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21E 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 21 Concept of Perimeter and Area Ex 21D

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 21D

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) = 46 cm
Breadth (b) = 25 cm
Area of rectangle = l x b
= 46 x 25 sq. cm
= 1150 sq.cm.
(ii) Length (l) = 9 m
Breadth (b) = 6 m
Area = l x b = 9 x 6
= 54 sq. metre Ans.
(iii) Length (l) = 14.5 m
Breadth (b) = 6.8 m
Area = l x b
= 14.5 x 6.8 sq. m
= 98.6 sq. m Ans.
(iv) Length (l) = 2m 5cm
= 2x 100 cm + 5 cm
= 200 cm + 5 cm
= 205 cm
Breadth = 60 cm
Area = l x b
= 205 cm x 60 cm
= 12300 cm²
(v) Length (l) = 3.5 km
Breadth (b) = 2 km
Area = l x b
= 3.5 x 2
= 7 sq. km. Ans.

Question 2.
Solution:
Side of a square plot = 14 m
Area = (Side)²
= (14)²
= 196 m²

Question 3.
Solution:
Length of top of table (l) = 2 m, 25 cm = 2.25 m
Breadth (l) = 1 m 20 cm = 1.20 m
Area of the top of the table = l x b
= (2.25 x 1.20) sq. m
= 2.7 sq. m. Ans.

Question 4.
Solution:
Length of carpet (l) = 30 m 75 cm
= 30.75 m
Breadth (b) = 80 cm = 0.80 m
Area of the carpet = l x b
= (30.75 x 0.80) sq. m
= 24.6 sq. m
Cost of one square metre = Rs. 20
Total cost = 24.6 x 150
= Rs. 3690. Ans

Question 5.
Solution:
Length of the sheet of paper 3 m 24 cm
= 300 cm + 24 cm
= 324 cm
Breadth of the sheet of the paper 1 m 72 cm
= 100 cm + 72 cm
= 172 cm
Area of the sheet of paper = (324 x 172) cm²
Also, area of the piece of paper required for an envelope = (18 x 12) cm².
Number of envelopes that can be made
= \(\\ \frac { 324\times 172 }{ 18\times 12 } \)
= 258

Question 6.
Solution:
Length of room (l) = 12.5 m
Breadth (b) = 8 m
Area = l x b
= (12.5 x 8) sq. m
= 100 sq. m
Side of square carpet (a) = 8 m
Area of carpet = a² = (8 x 8) sq. m.
= 64 square metre
Area left without carpet = 100 sq. m – 64 sq. m
= 36 sq. m Ans.

Question 7.
Solution:
Length of the lane = 150 m
Breadth of the lane = 9 m
Area of the lane = (150 x 9) m²
= 1350 m²
Area of the brick = 22.5 cm x 7.5 cm
= 168.75 cm²

Question 8.
Solution:
Length of room (l) = 13 m
Breadth (b) = 9 m
Area of floor or carpet = l x b
= 13 x 9
= 117 sq. m

Question 9.
Solution:
Let the length of the rectangular park = 5x metres
and the breadth of the rectangular park = 3x metres

Question 10.
Solution:
Side of the square plot = 64 m
Perimeter of the square plot = 4 x Side
= 4 x 64
= 256 m
Perimeter of the rectangular plot = Perimeter of the square plot = 256 m
Length of the rectangular plot = 70 m
Perimeter = 2 x (Length + Breadth)
256 = 2 (70 + b)
256 = 140 + 2b
=> 2b = 256 – 140
=> 2b = 116
b = \(\\ \frac { 116 }{ 2 } \) = 58 cm
Area of the rectangular plot = (length x breadth)]
= (70 x 58) m²
= 4060 m²
Area of the square plot = Side x Side
= (64 x 64) m²
= 4096 m².
Square plot has the greater area than that of the rectangular plot by
(4096 – 4060)
= 36 m².

Question 11.
Solution:
Total cost of cultivating the rectangular field = Rs. 71400
Rate of cultivating = Rs. 35 per sq. m

Question 12.
Solution:
Area of rectangle = 540 sq. cm
Length (l) = 36 cm

Question 13.
Solution:
Measure of a marble tile = 12cm x 10cm
Area of wall = 4 m x 3 m
= 12 m²
Area of one marble tile
= 12 x 10
= 120 cm²

Question 14.
Solution:
Area of a rectangle = 600 cm²
Breadth = 25 cm

Question 15.
Solution:
diagonal of square = 5 √2

Question 16.
Solution:
(i) We name the given region as shown in the figure

Question 17.
Solution:
Measures are in cm
(i) In the figure, there are three rectangles and one square



= 5(6 x 6) cm²
= 180 cm²

Hope given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21D 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 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.

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

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.

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

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.