## 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.

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