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

## 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

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.

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