Students can Download Chapter 11 Perimeter and Area Ex 11.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

Question 1.

Find the circumference of the circles with the following radius :

(Take n = \(\frac{22}{7}\))

a) 14 cm

radius = r = 14 cm

∴ circumference of a circle = 2πr

b) 28 mm

radius = r = 28 cm

∴ circumference of a circle = 2πr

c) 21 cm

radius = r = 21 cm

circumference of a circle = 2πr

Question 2.

Find the area of the following circles, given that:

a) radius = 14 mm (Take π = \(\frac{22}{7}\))

radius = 14 mm

∴ Area of the circle = πr^{2}

= 616 sq.cms

b) diameter = 49 m

Area of the circle = πr^{2}

= 1,886.5 sq.metrs

c) radius = 5 cm

radius = r = 5 cm

∴ Area of the circle = πr^{2}

Question 3.

If the circumference of a circular sheet is 154m, find its radius. Also find the area of the sheet. (Take π = \(\frac{22}{7}\))

Solution:

C = circumference of a circle = 154m = 2πr

Area of the sheet = πr^{2}

= 1,886.5 sq.metrs

Question 4.

A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per meter.

(Take π = \(\frac{22}{7}\))

d = diameter = 21 m

∴ radius of the circular garden

circumference of the circular garden = 2πr

Length of the rope needed to make one round of fence } = 66 mts

Length of the rope needed to make 2 rounds =

66 × 2 = 132 mts

cost of rope per metre = Rs. 4

Total cost of rope = 132 × 4 = Rs. 528

Question 5.

From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.

(Take π = 3.14)

Solution:

Radius of the outer circle = 4 cms.

∴ Area of the outer circle = πr^{2}

= 3.14 × 4 × 4

= 28.26 sq.cms

= 50.24 sq.cms

π = 3.14

Radius of the inner circle = 3 cms

∴ Area of the inner circle = πr^{2}

= 3.14 × 3 × 3 = 28.26 sq. cms

Area of the remaining part = 50.24 – 28.26 = 21.98 sq.cms

Question 6.

Saima wants to put lace on the edge of a circular table cover of diameter 1.5m. Find the length of the lace required and also find its cost if one meter of the lace

costs ₹ 15. (Take π = 3.14)

Solution:

Diameter of the table cover = 1.5m

Circumference of the table cover }

2πr – πd = 3.14 × 1.5 (∵ d = 2r) = 4.710 mts

∴ Length of the lace required = 4.71 mt

Cost of lace per meter = Rs. 15

∴ Total cost of the lace =4.71 × 15 = Rs 70.65

Question 7.

Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Solution:

Diameter of the semi-circle = 10 cm

Circumference of semi-circle =

Perimeter of the semi-circle(including its diameter) } = 15.7 + 10 = 25.7 cms

Question 8.

Find the cost of polishing a circular table-top of diameter 1.6m, if the rate of polishing is ? 15/m^{2}. (Take π = 3.14)

Solution:

Diameter of the Table Top = 1.6 m

Area of the Table Top = πr^{2}

Rate of polishing = Rs. 15/sq.m

Total cost of polishing = 2.0096 × 15

= Rs.30.14

Question 9.

Shazil took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides ? Which figure encloses more area, the circle or the square? (Take π = \(\frac{22}{7}\))

Solution:

Length of the wire = 44 cms

Total length of the circle = perimeter of the circle

∴ radius = 7 cms.

∴ Length of wire = perimeter of the square = 154 sq.cms

44 = 44

Area of the square = a^{2} = 11 × 11 = 121 sq.cms.

∴ The area of the circle is more means the circle encloses more area.

Question 10.

From a circular card sheet of radius 14cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = \(\frac{22}{7}\))

Solution:

Radius of the circular Card = 14 cm

Area of the Card

Area of small 2 circles =

= 77 sq. cms.

Area of rectangle = l × b = 3 × 1 = 3 sq. cms

Area of the remaining sheet

= 616 – (77 + 3)

= 616 = 80 = 536 sq.cm

Question 11.

A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the leftover aluminium sheet? (Take π = 3.14)

Solution:

Area of square aluminium sheet = 6 × 6 = 36 sq. cms

Side = 6 cm radius = 2 cm

Area of circle =

Area of the left over Aluminium sheet = 36 – 12.56 = 23.44sq.cm

Question 12.

The circumference of a circle is 31.4. Find the radius and the area of the circle? (Take π = 3.14)

Solution:

The circumference of a circle = 2πr = 31.4 cm

Area of the cirlce = πr^{2}

= 3.14 × 5 × 5

= 78.5 sq.cms

Question 13.

A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66m. What is the area of this path? (π = 3.14)

Solution:

Diameter of the flower bed = 66 m

Radius = 66/2 = 33 m

Area of the flower bed = πr^{2}

= 3.14 × 33 × 33

= 3419.46 sq.cm

Radius of the flower bed with path = 33 + 4 = 37 ms.

Area of the flower bed with path =

= 3.14 × 37 × 37

= 4298.66 sq.cm

Area of the path = Area of the flower bed with path – Area of the flower bed

= 4298.66 – 3419.46

= 879.2 sq.mts

Question 14.

A circular flower garden has an area of 314 m^{2}. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will sprinkler water the

entire garden? (Take π = 3.14)

Solution:

Area of the circular flower garden = 314 sq.m

Radius of the sprinkler = 12m

Area of the sprinkler = πr^{2} = 3.14 × 12 × 12

= 452.16 sq.mt

∴ 452.16 > 314

∴ Yes, the sprinkler water the entire garden.

Question 15.

Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Solution:

Circumference of the outer circle = 2πr

radius = 19 m

= 2 × 3.14 × 19

= 119.32 mts

Circumference of the inner circle = 2πr

radius = 19 – 10 = 9 cm.

= 2 × 3.14 × 9

= 56.52 cm

Question 16.

How many times a wheel of radius 25cm must to go 352 m ? (Take π = \(\frac{22}{7}\))

Solution:

radius of the wheel = r = 28 cm

circumference of the wheel = 2πr

= 176 cms

To cover the distance of 352 mtrs (∵ 1 m = 100 cm)

No. of rotation

Question 17.

The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)

Solution:

Length of the minute hand = 15 cm

1 hour = 1 round = circumference = 2πr

= 2 × 3.14 × 15

= 94.2 cms

∴ The tip of the minute hand in one hour moves 94.2 cms.