Concept of Perimeter and Area Exercise 21.1 |

Concept of Perimeter and Area Exercise 22.2 |

Concept of Perimeter and Area Exercise 22.3 |

1. Find the perimeter of a rectangle in which:

(i) Length = 16.8 cm and breadth = 6.2 cm

Answer:

Length = 16.8 cm

Breadth = 6.2 cm

Perimeter of a rectangle = \( 2 \times \left ( length + breadth \right ) \)

= \( 2 \times \left ( 16.8 + 6.2 \right ) = 46 cm \)

(ii) Length = 2 m 25 cm and breadth = 1 m 50 cm

Answer:

Length = 2 m 25 cm

= (200 + 25) cm (1 m = 100 cm)

= 225 cm

Breadth = 1 m 50 cm

= (100 + 50) cm (1 m = 100 cm)

= 150

Perimeter of a rectangle = \( 2 \times \left ( length + breadth \right ) \)

= \( 2 \times \left ( 225 + 150 \right ) = 750 cm \)

(iii) Length = 8 m 5 dm and breadth = 6 m 8 dm

Answer:

Length = 8 m 5 dm

= (80 + 5) dm (1 m = 10 dm)

= 85 dm

Breadth = 6 m 8 dm

= (60 + 8) dm (1 m = 10 dm)

= 68 dm

Perimeter of a rectangle = \( 2 \times \left ( length + breadth \right ) \)

= \( 2 \times \left ( 85 + 68 \right ) = 306 dm \)

2. Find the cost of fencing a rectangular field 62 m long and 33 m wide at Rs 16 per meter.

Answer:

Length of the field = 62 m

Breadth of the field = 33 m

Perimeter of a field = \( 2 \times \left ( length + breadth \right ) \)

= \( 2 \times \left ( 62 + 33 \right ) m = 190 m \)

Cost of fencing per meter = Rs 16

Total cost of fencing = Rs (16 × 190) = Rs 3040

3. The Length and the breadth of a rectangle field are in the ration 5 : 3. If its perimeter is 128 m, find the dimensions of the field.

Answer:

Let the length of the rectangle be 5x m.

Breadth of the rectangle = 3x m

Perimeter of a field = \( 2 \times \left ( length + breadth \right ) \)

= \( 2 \times \left ( 5x + 3x \right ) m = 16 m \)

It is given that the perimeter of the field is 128 m.

Therefore, \(16x = 128 \)

\( \Rightarrow x = \frac {128} {16} = 8 \)

Therefore, \(Length = \left ( 5 \times 8 \right ) = 40 m \)

\(breadth = \left ( 3 \times 8 \right ) = 24 m\)

4. The cost of fencing the rectangle field at Rs 18 per metre is Rs 1980. If the width of the field is 23 m, find its length.

Answer:

Total cost of fencing = Rs 1980

Rate of fencing = Rs 18 per meter

\( Perimeter \; of \; the \; field = \frac {Total \; cost} {Rate} = \frac {Rs 1980} {Rs 18/m} = \left ( \frac {1980} {18} \right ) m = 110 m \)

Let the length of the field be x metre.

Perimeter of a field = \( 2 \times \left ( x + 23 \right ) m \)

Therefore, \(2 \left ( x + 23 \right ) = 110 \)

\( \Rightarrow \left ( x + 23 \right ) = 55 \)

\( x = \left ( 55 – 23 \right ) = 32 \)

Hence, the length of the field is 32 m.

5. The length and the breadth of a rectangle field are in the ratio 7:4. The cost of fencing the field at Rs 25 per metre is Rs 3300. Find the dimension of the field.

Answer:

Total cost of fencing = Rs 3300

Rate of fencing = Rs 25/m

\(Perimeter \; of \; the \; field = \frac {Total \; cost} {Rate \; of \; fencing} = \left ( \frac {Rs \; 3300} {Rs \; 25/m} \right ) = \left ( \frac {3300} {25} \right ) m = 132 \; m\)

Let the length and the breadth of the rectangular field be 7x and 4x, respectively.

Perimeter of the field = 2(7x + 4x) = 22x

It is given that the perimeter of the field is 132 m.

Therefore, \(22x = 132 \)

\( \Rightarrow x = \frac {132} {22} = 6 \)

Therefore, \(Length \; of \; the \; field = \left ( 7 \times 6 \right ) m = 42 \; m \)

\( Breadth \; of \; the \; field = \left ( 4 \times 6 \right ) m = 24 \; m \)

6. Find the perimeter of a square, each of whose sides measures:

(i) 3.8 cm

Answer:

Side of the square = 3.8 cm

Perimeter of the square = (4 × side)

= ( 4 × 3.8 ) = 15.2 cm

(ii) 4.6 m

Answer:

Side of the square = 4.6 m

Perimeter of the square = (4 × side)

= ( 4 × 4.6 ) = 18.4 m

(iii) 2 m 5 dm

Answer:

Side of the square = 2 m 5 dm

= (20 + 5) dm (1 m = 10 dm)

= 25 dm

Perimeter of the square = (4 × side)

= ( 4 × 25 ) = 100 dm

7. The cost of putting a fence around a square field at Rs 35 per metre is Rs 4480. Find the length of each side of the field.

Answer:

Total cost of fencing = Rs 4480

Rate of fencing = Rs 35/m

\( Perimeter \; of \; the \; field = \frac {Total \; cost} {Rate } = \left ( \frac {Rs \; 4480} {Rs \; 35/m} \right ) = \left ( \frac {4480} {35} \right ) m = 128 \; m\)

Let the length of the each side of the field be x metres.

Perimeter = (4x) metres

Therefore, \(4x = 128 \)

\( \Rightarrow x = \frac {128} {4} = 32 \)

Hence, the length of each side of the field is 32 m.

8. Each side of the square field measure 21 m. Adjacent to this field, there is a rectangular field having its sides in the ratio 4 : 3. If the perimeters of both the field are equal, find the dimensions of the rectangular field.

Answer:

Side of the square field = 21 m

Perimeter of the square field = (4 × 21) m

= 84 m

Let the length and breadth of the rectangular field be 4x and 3x, respectively.

Perimeter of the rectangular field = Perimeter of the square field

Therefore, \(14x = 84 \)

\( \Rightarrow x = \frac {84} {14} = 6 \)

Therefore, Length of the rectangular field = \(\left ( 4 \times 6 \right ) m = 24 \; m \)

Breadth of the rectangular field = \( \left ( 3 \times 6 \right )m = 18 \; m \)

9. Find the perimeter of

(i) A rectangle of sides 7.8 cm, 6.5 cm and 5.9 cm

Answer:

Sides of the triangle are 7.8 cm, 6.5 cm and 5.9 cm

Perimeter of a triangle = (First side + Second side + Third side)

(7.8 + 6.5 + 5.9) cm

= 20.2 cm

(ii) An equilateral triangle of side 9.4 cm

Answer:

In an equilateral triangle, all sides are equal.

Length of each sides of the triangle = 9.4 cm

Therefore, Perimeter of the triangle = \(\left ( 3 \times sides \right ) \;cm \)

= (3 × 9.4) cm

= 28.2 cm

(iii) An isosceles triangle with equal sides 8.5 cm each and third side 7 cm.

Answer:

Length of two equal sides = 8.5 cm

Length of the third side = 7 cm

Therefore, Perimeter of the triangle = \( \left ( 2 \times equal \; sides \right ) + third \; side\; cm \)

= {(2 × 8.5) + 7} cm

= 24 cm

10. Find the perimeter of

(i) A regular pentagon of side 8 cm

Answer:

Length of each sides of the given pentagon = 8 cm

Therefore, perimeter of the pentagon = (5 × 8) cm

= 40 cm

(ii) A regular octagon of side 4.5 cm

Answer:

Length of each sides of the given octagon = 4.5 cm

Therefore, perimeter of the octagon = (4.5 × 8) cm

= 36 cm

(iii) A regular decagon of side 3.6 cm

Answer:

Length of each sides of the given decagon = 3.6 cm

Therefore, perimeter of the decagon = (10 × 3.6) cm

= 36 cm

10. Find the perimeter of each of the following figures:

(i)

Perimeter of the figure = sum of all the sides

= (27 + 35 + 35 + 45) cm

= 142 cm

(ii)

Perimeter of the figure = sum of all the sides

= (18 + 18 + 18 + 18) cm

= 72 cm

(iii)

Perimeter of the figure = sum of all the sides

= (8 + 16 + 4 + 12 + 12 + 16 + 4) cm

= 72 cm

## Exercise 21B

1. Find the circumference of the circle whose radius is

(i) 28 cm

Solution:

Radius, r = 28 cm

Therefore, Circumference of the circle, C = 2.pi.r

= \( \left ( 2 \times \frac {22} {7} \times 28 \right ) \)

= 176 cm

Hence, the circumference of the given circle is 176 cm.

(ii) 10.5 cm

Solution:

Radius, r = 10.5 cm

Therefore, Circumference of the circle, C = 2.pi.r

= \( \left ( 2 \times \frac {22} {7} \times 10.5 \right ) \)

= 66 cm

Hence, the circumference of the given circle is 66 cm.

(iii) 3.5 m

Solution:

Radius, r = 3.5 m

Therefore, Circumference of the circle, C = 2.pi.r

= \( \left ( 2 \times \frac {22} {7} \times 3.5 \right ) \)

= 22 m

Hence, the circumference of the given circle is 22 m.

2. Find the radius of the circle whose circumference is 176 cm.

Solution:

Let the radius of the given circle be r cm.

Circumference of the circle = 176 cm

Circumference = 2πr

Therefore, 2πr = 176

\( \Rightarrow r = \frac {176} {2 \pi } \)

\( \Rightarrow r = \left ( \frac {176} {2} \times \frac {7} {22} \right ) \)

r = 28

The radius of the given circle is 28 cm.

4. Find the diameter of the wheel whose circumference is 264 cm

Solution:

Let the radius of the given circle be r cm.

Diameter = 2 × Radius = 2r cm

Circumference of the wheel = 264 cm

Circumference of the wheel = 2πr

Therefore, \(2 \pi r = 264 \)

\( \Rightarrow 2 r = \frac {264} {\pi } \)

\( \Rightarrow 2 r = \left ( 264 \times \frac {7} {22} \right ) \)

\( \Rightarrow 2 r = 84 \)

Diameter of the given wheel is 84 cm.

5. Find the distance covered by the wheel of a car is 500 revolution if the diameter of the wheel is 77 cm.

Solution:

Radius of the wheel = (Diameter of the wheel)\2

\( \Rightarrow r = \frac {77}{2} \)

Circumference of the wheel = 2πr

= \( \left ( 2 \times \frac {22} {7} \times \frac {77} {2} \right ) \)

= 242 cm

In 1 revolution the wheel covers a distance equal to its circumference.

Therefore, distance covered by the wheel in 1 revolution = 242 cm

Therefore, distance covered by the wheel in 500 revolution = (500 × 242) cm

= 121000 cm

= 1210 m (100 cm = 1 m)

= 1.21 km (1000 m = 1 km)

6. The diameter of the wheel of a car is 70 cm. how many revolutions will it make to travel 1.65 km?

Solution:

Radius of the wheel ( r ) = (Diameter of the wheel)\2

\( r = \frac {70} {2} cm = 35 cm \)

Circumference of the wheel = 2πr = \( \left ( 2 \times \frac {22} {7} \times 35 \right ) \)

= 220 cm

In 1 revolution the wheel covers a distance equal to its circumference.

Therefore, 220 cm distance = 1 revolution

Therefore, 1 cm distance = 1/220 revolution

Therefore, \(1 km \left ( 100000 \;cm \right ) = \frac {1 \times 100000} {220}\; revolutions\)

Therefore, \(1.65 \;km = \frac {1.65 \times 100000} {220}\; revolutions \)

= 750 revolutions

Thus, the wheel will make 750 revolutions to travel 1.65 km