The document RD Sharma Solutions -Ex-20.4, Mensuration, Class 6, Maths Class 6 Notes | EduRev is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.

All you need of Class 6 at this link: Class 6

**Question 1. Find the area of a rectangle, whose** **(i) Length = 6cm, breadth = 3 cm (ii) Length = 8 cm, breadth = 3cm**

Length = 6 cm

Breadth = 3 cm

Area of rectangle = 6 Ã— 3 = 18 cm

(ii) Area of a rectangle = Length Ã— Breadth

Length = 8 cm Breadth = 3 cm

Area of rectangle = 8 Ã— 3 = 24 cm

(iii) Area of a rectangle = Length Ã— Breadth

Length = 4.5 cm Breadth =2 cm

Area of rectangle = 4.5 Ã— 2 = 9 cm

**Question 2. Find the area of a square whose side is:** **(i) 5 cm (ii) 4.1 cm (iii) 5.5 cm (iv) 2.6 cm****Answer:** Area of a square = Side Ã— Side (i) Side of the square = 5 cm

Area of the square = 5 Ã— 5 = 25 cm^{2}

(ii) Side of the square = 4.1 cm

Area of the square = 4.1 Ã— 4.1 = 16.81 cm^{2}

(iii) Side of the square = 5.5 cm

Area of the square = 5.5 Ã— 5.5 = 30.25 cm^{2}

(iv) Side of the square = 2.6 cm

Area of the square = 2.6 Ã— 2.6 = 6.76 cm^{2}

**Question 3. The area of a rectangle is 49 cm ^{2} and its breadth is 2.8 cm. Find the length of the rectangle.**

Area of the rectangle = Length Ã— Breadth

Therefore, Length = Area / Breadth = 49 / 2.8 = 17.5 cm

**Question 4. The side of a square is 70 cm. Find its area and perimeter.****Solution: **Side of the square = 70 cm

Area of the square = Side Ã— Side = 70 Ã— 70 = 4900 cm^{2}

Perimeter of the square = 4 Ã— Side = 4 Ã— 70 = 280 cm

**Question 5. The area of a rectangle is 225 cm ^{2} and its one side is 25 cm, find its other side.**

**Question 6. What will happen to the area of a rectangle if its** **(i) Length and breadth are trebled (ii) Length is doubled and breadth is same**

Original area = l Ã— b = lb

Now, the length and breadth are trebled which means they become three times of their original value.

Therefore New length = 3l

New breadth = 3b New area = 3l Ã— 3b = 9lb

Thus, the area of the rectangle will become 9 times that of its original area.

(ii) If the length is doubled and the breadth is same. Let the initial length and breadth be l and b, respectively.

Original area = l Ã— b = lb

Now, length is doubled and breadth remains same. Therefore New length = 2l

New breadth = b

New area = 2l x b = 2 lb

Thus, the area of the rectangle will become 2 times that of its original area.

(iii) If the Length is doubled and breadth is halved.

Let the initial length and breadth be land b, respectively.

Original area =l Ã— b = lb Now, length is doubled and breadth is halved.

Therefore New length = 2l

New breadth = b / 2

New area = 2 l Ã— b / 2 = lb

New area is also lb. This means that the areas remain the same.

**Question 7. What will happen to the area of a square if its side is:** **(i) Tripled (ii) increased by half of it** **Solution: **(i) Let the original side of the square be s. Original area = s x s = s^{2}

If the side of a square is tripled, new side will be equal to 3s.

New area = 3s Ã— 3s = 9 s^{2} This means that the area becomes 9 times that of the original area.

(ii) Let the original side of the square be s. Original area = s x s = s^{2} If the side of a square is increased by half of it, new side =

New area =

This means that the area becomes times that of the original area.

**Question 8. Find the perimeter of a rectangle whose area is 500 cm ^{2} and breadth is 20 cm.**

Therefore Length = Area / Breadth = = 25 cm

Perimeter of a rectangle = 2 (Length + Breadth) = 2 (25 + 20) cm = 2 Ã— 45 cm = 90 cm

**Question 9. A rectangle has the area equal to that of a square of side 80 cm. If the breadth of the rectangle is 20 cm, Find its length.** **Solution: **Side of the square = 80 cm Area of square = Side Ã— Side = 80 x 80 = 6400 cm^{2}

Given that: Area of the rectangle = Area of the square = 6400 cm^{2}

Breadth of the rectangle = 20 cm

Applying the formula: Length of the rectangle = Area / Breadth We get: Length of the rectangle = = 320 cm

**Question 10. Area of a rectangle of breadth 17 cm is 340 cm ^{2} . Find the perimeter of the rectangle.**

Applying the formula: Length of a rectangle = Area / Breadth We get: Length of the rectangle = =20 cm

Perimeter of rectangle = 2 (Length + Breadth) = 2 (20 + 17) = 2 x 37 = 74 cm

**Question 11. A marble tile measures 15 cm Ã— 20cm. How many tiles will be required to cover a wall of size 4m Ã— 6m?** **Solution: **Dimensions of the tile = 15 cm Ã— 20 cm

Dimensions of the wall = 4 m Ã— 6 m = 400 cm Ã— 600 cm (Since, 1 m = 100 cm)

Area of the tile = 15 cm Ã— 20 cm = 300 cm^{2}Area of the wall = 400 cm Ã— 600 cm = 2,40,000 cm^{2}

Number of tiles required to cover the wall = = 800 tiles

**Question 12. A marble tile measures 10 cm Ã— 12 cm. How many tiles will be required to cover a wall of size 3m Ã— 4m? Also, find the total cost of the tiles at the rate of Rs 2 per tile. ****Solution: **Dimension of the tile = 10 cm Ã— 12 cm

Dimension of the wall =3 m Ã— 4 m = 300 cm Ã— 400 cm (Since, 1 m = 100 cm)

Area of the tile = 10 cm Ã— 12 cm = 120 cm^{2} Area of the wall = 300 cm Ã— 400 cm = 1,20,000 cm^{2}

Number of tiles required to cover the wall = Area of wall / Area of one tile = = 1,000 tiles

Cost of tiles at the rate of Rs. 2 per tile = 2 x 1,000 = Rs. 2,000

**Question 13. One tile of a square plot is 250 m, find the cost of leveling it at the rate of Rs 2 per square meter.****Solution: **Side of the square plot = 250 m Area of the square plot = Side Ã— Side = 250 Ã— 250 = 62,500 m^{2}

Rate of leveling the plot = Rs. 2 per m^{2} Cost of leveling the square plot = Rs. 62,500 Ã— 2 = Rs. 1,25,000

**Question 14. The following figures have been split into rectangles. Find the areas. (The measures are given in centimeters)**

**Solution: **(i) This figure consists of two rectangles II and IV and two squares I and III.

Area of square I = Side Ã— Side = 3 Ã— 3 = 9 cm=2 Similarity, area of rectangle II = (2 Ã— 1) = 2 cm^{2}

Area of square III = (3 Ã— 3) = 9 cm^{2} Area of rectangle IV = (2 x 4) = 8 cm^{2}

Thus, the total area of this figure = (Area of square I + Area of rectangle II + Area of square III + Area of rectangle IV) = 9 + 2 + 9 + 8 = 28 cm^{2}

(ii) This figure consists of three rectangles I, II and Ill.

Area of rectangle I = Length Ã— Breadth = 3 Ã— 1 = 3 cm^{2}

Similarly, area of rectangle II = (3 Ã— 1) = 3 cm^{2} Area of rectangle III = (3 Ã— 1) = 3 cm^{2}

Thus, the total area of this figure = (Area of rectangle I + area of rectangle II + area of rectangle III) = 3 + 3 + 3 = 9 cm^{2}

**Question 15. Split the following shapes into rectangles and find the area of each. (The measures are given in centimeters)**

**Solution: **(i) This figure consists of two rectangles I and II.

The area of rectangle I = Length Ã— Breadth = 10 Ã— 2 = 20 cm^{2}

Similarly, area of rectangle II = 10 Ã— 32 = 15 cm^{2} Thus, total area of this figure = (Area of rectangle I + Area of rectangle II) = 20 + 15= 35 cm^{2}

(ii) This figure consists of two squares I and III and one rectangle II. Area of square I = Area of square III = Side x Side = 7 x 7 = 49 cm^{2}

Similarly, area of rectangle II = (21 x 7) = 147 cm^{2}

Thus, total area of this figure = (Area of square I + Area of rectangle II + Area of square III) = 49 + 49 + 147 = 245 cm^{2}

(iii) This figure consists of two rectangles I and II. Area of rectangle I = Length x Breadth = 5 x 1 = 5 cm^{2}

Similarly, area of rectangle II = 4 Ã— 1 = 4 cm^{2}

Thus, total area of this figure = (Area of rectangle I + Area of rectangle II) = 5 + 4 = 9 cm^{2}

**Question 16. How many tiles with dimension 5 cm and 12 cm will be needed to fit a region whose length and breadth are respectively:** **(i) 100 cm and 144 cm (ii) 70 cm and 36 cm** **Solution: **(i) Dimension of the tile = 5cm Ã— 12 cm Dimension of the region = 100 cm Ã— 144 cm

Area of the tile = 5 cm Ã— 12 cm = 60 cm^{2} Area of the region = 100 cm Ã— 144 cm = 14,400 cm^{2}

Number of tiles required to cover the region = Area of the region / Area of one tile = = 240 tiles

(ii) Dimension of the tile = 5 cm Ã— 12 cm Dimension of the region = 70 cm Ã— 36 cm

Area of the tile = 5 cm Ã— 12 cm = 60 cm^{2} Area of the region = 70 cm Ã— 36 cm = 2,520 cm^{2}

Number of tiles required to cover the region = Area of the region / Area of one tile = = 42 tiles