**Exercise 20.4**

*Question 1. Find the area of a rectangle, whose*

* (i) Length = 6cm, breadth = 3 cm (ii) Length = 8 cm, breadth = 3cm*

* (iii) Length = 4.5 cm, breadth = 2 cm*

**Solution: **

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

Length = 6 cm Breadth = 3 cm

Area of rectangle = 6 Ã— 3 = 18 cm^{2}

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

Length = 8 cm Breadth = 3 cm

Area of rectangle = 8 Ã— 3 = 24 cm^{2}

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

Length = 4.5 cm

Breadth =2 cm

Area of rectangle = 4.5 Ã— 2 = 9 cm^{2}

*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.*

**Solution: **Area = 49 cm^{2} Breadth = 2.8 cm

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.*

*Solution: **Area = 225 cm ^{2} *

One of the sides = 25 cm

Area of the rectangle = Product of the lengths of its two side

Other side = Area / Side = 225 / 25 = 9 cm

*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*

* (iii) Length is doubled *

**Solution: **

(i) If the length and breadth of a rectangle are trebled.

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

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 = \(\left (s + \frac{1}{2}s \right ) = \frac{3}{2}s\)

New area = \(\frac{3}{2}s \times \frac{3}{2}s = \frac{9}{4}s\)

This means that the area becomes \( \frac{ 9 } { 4 }\)

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

**Solution: **

Area = 500 cm^{2}

Breadth = 20 cm

Area of rectangle = Length Ã— Breadth

Therefore Length = Area / Breadth

= \( \frac{ 500 } { 20 }\)

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 = \( \frac{ 6400 } { 20 }\)

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

**Solution: **

Area of the rectangle = 340 cm^{2}

Breadth of the rectangle = 17 cm

Applying the formula:

Length of a rectangle = Area / Breadth

We get:

Length of the rectangle = \( \frac{ 340 } { 17 }\)

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 = \(\frac{ Area of wall}{ Area of one tile }\)

= \( \frac{ 240000 } { 300 }\)

*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

= \( \frac{ 120000 } { 120 }\)

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

= \( \frac{ 14400 } { 60 }\)

(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

= \( \frac{ 2520 } { 42 }\)