HBSE 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.3

Haryana State Board HBSE 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 11 Perimeter and Area  Exercise 11.3

Question 1.
Find the circumference of the circles with the following radii :
(a) 10 cm
(b) 28 mm
(c) 20 cm[Take π=22/7]
Solution:
Circumference of the circle = 2πr
or C = 2πr
(a) r = 10 cm
C = 2πr
= 2 × \(\frac{22}{7}\) × 10 = \(\frac{440}{7}\) = 62.86

(b) r = 28 mm
C = 2 × \(\frac{22}{7}\) x 28 = 176cm

(c) r = 20 cm
C = 2 × \(\frac{22}{7}\) x 20
= \(\frac{880}{7}\) = 125.7 cm

Question 2.
Find the area of the following circles:
(a) radius = 14mm
(b) diameter =49 m
(c) radius =5 cm [Take π= 22/7]
Solution:
Area of circle = πr², where π= \(\frac{22}{7}\)
(a) r = 14mm
A = \(\frac{22}{7}\) × (14)²
=\(\frac{22}{7}\) × 14 × 14 = 316 mm².

(b) d = 49 cm, r = \(\frac{49m}{2}\)
∴ A = \(\frac{22}{7} \times \frac{49}{2} \times \frac{49}{2}\)
= 1886.5m².

(c) r = 5cm
∴ A= \(\frac{22}{7}\) × (5)²
= \(\frac{22}{7}\) × 5 × 5 = 78.57 cm².

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet.
[Tube π = 22/7]
Solution:
Circumference of circular sheet = 2πr
or, 154m = 2 × \(\frac{22}{7}\) × r
or, \(\frac{44}{7}\) r = 154 m.
or, 44r = 154 × 7
∴ r = \(\frac{154 \times 7}{44}=\frac{49}{2}\)
r = 24.50cm
Area of circular sheet = 2πr²
= \(\frac{22}{7}\) × 24.50 × 24.50 = 1886.5 cm².

Question 4.
A gardener wants to fence a circular gardezi of thc diameter 21 ni. Find the length of the rope he needs to purchase, if lic makes 2 rounds of feiice. Also find the cost of the rope, ¡lit cost Rs. 4 per meter.
[Take π = 22/7]
Solution:
d = 21 m
∴ r = \(\frac{21}{2}\)
Circumference of circular
garden = 2πr
or,length of the rope
= Circumference of garden = 2πr = 2 × \(\frac{22}{7} \times \frac{21}{2}\) = 66m
∴ length of the 2 rounds of the fence
= 2 × 66 rn = 132 m
The cost of 132 m rope = 132 × 4 = Rs.528.

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed (Fig. 11.30). Find the area of the remaining sheet.
[Take π = 22/7]
im 1
Solution:
Radius of the circular sheet (R) = 4 cm
Radius of the removed circular sheet (inner circle)
(r) = 3 cm
Since
Area of a circle = π × (Radius)²
Area of the remaining sheet = Area of outer circle – Area of inner circle = πR² – πr² = π[R² – r²]
= π(R + r) (R – r)
= 3.14 (4 + 3) (4 – 3) cm²
= 3.14 (7) (1) cm² = 21.98 cm²

Question 6.
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace cost Rs. 15. [Take π = 3.14]

Solution:
d = 1.5 m,
r = \(\frac{1.5}{2}\) m
=> Circumference of the table cover = 2πr
= 2 × 3.14 × \(\frac{1.5}{2}\)m
= 4.71m
∴ Length of the lace required = 4.71 m
∴ Cost of the lace = Rs. 4.71 × 15 = Rs. 70.65.

Question 7.
Find the perimeter of the adjoining Fig. , which is a semicircle including its diameter.

Solution:
Perimeter or circumference of semicircle = \(\frac{2 \pi r}{2}\)
= πr
= \(\frac{22}{7}\) × 10 cm
= \(\frac{220}{7}\) cm = 31.43 cm.

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs. 15 per sq. m.
[Take π = 3.14]
Solution:
Diameter of the table-top = 1.6 m
∴ Radius = \(\frac{1.6}{2}\)
= 0.8 m
∴ Area of the table-top = πr²
= 3.14 × (0.8)² m²
= 3.14 × 0.64 m²
= 2.0096 m²
∴ Cost of polishing the table-top
= Rs. 2.0096 × 15
= Rs. 30.144.

Question 9.
Shazli took a wire of length 44 cm and bent into the shape of a circle. Find the radius of that circle. Also find the area. If the same wire is bent into the shape of a square, what will be the length of each of its side ? Which figure encloses more area circle or square ? [Take π = 22 / 7]

Solution:
Circumference of circle = 2πr
or, 44 cm = 2 × \(\frac{22}{7}\) x r,
or, 44 r = 44 × 7
∴ r = \(\frac{44 \times 7}{44}\) = 7 cm
∴ radius of circle = 7 cm
∴ area of circle = πr²
= \(\frac{22}{7}\) × 7 × 7 = 154 cm²
Now, circumference of circle = perimeter of square
or 44 cm = 4 × side,
4 × side = 44 cm
∴ side = \(\frac{44}{4}\) = 11 cm.
Area of square = side × side = 11 cm × 11 cm = 121 cm².
Hence, area of circle is larger than area of square.
More area of circle = 154 – 121 = 33 cm².

Question 10.
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length cm and breadth 1 m are removed, (as shown in Fig.) Find the area of the remaining sheet.

[Take π = 22/7]
Solution:
From the figure,
Area of circular card sheet = πr²
= \(\frac{22}{7}\) × 14 × 14 [∵ r = 14 cm]
= 616 cm².
Area of two circles

Area of rectangle
= l × b = 3 × 1 = 3 cm².
[∵ l = 3 cm, b = 1 cm]
hence area of the remaining sheet
= Area of circular card sheet – (Area of two circles + area of rectangle)
= 616 cm² – (77 cm² + 3 cm²)
= 616 cm² – 80 cm² = 536 cm².

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 left over aluminium sheet ?

[Take π = 3.14}
Solution:
Area of square piece = side × side
= 6 × 6 = 36 cm².
Area of circle = πr² = 3.14 × 2 × 2
= 12.56 cm².
Area of the left over aluminium sheet = Area of square piece – Area of circle
= 36 cm² – 12.56 cm² = 23.44 cm².

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle ? [Take π = 3.14]
Solution:
Circumference of a circle = 31.4 cm
2πr = 31.4 cm
2 × \(\frac{22}{7}\) × r = 31.4cm
∴ r = 31.4 × \(\frac{7}{22} \times \frac{1}{2} \mathrm{cm}\)
r = 5 cm
Area of the circle = πr²
= 3.14 × 5 × 5 cm² = 3.14 × 25 cm² = 78.5 cm².

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path ?
[Take π = 3.14]

Solution:
Wide of path = 4 m
Outer circular diameter = 66 m + 4 + 4 = 74 m
∴ r = \(\frac{74}{2}\)= 37 m.
Inner circular diameter = 66 m
∴ r = \(\frac{66}{2}\)= 33 m.
The area of this path = Area of outer circular flower bed – Area of inner circular flower, bed.
= π[(37 m)² – (33 m)²]
= \(\frac{22}{7}\) [1369 – 1089] m²
= \(\frac{22}{7}\) × 280 m² = 880 m²
or, 3.14 × 280 m² = 879.2 m².

Question 14.
A circular flower garden has an area of about 314 m². A sprinkler at the center of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden ? [Take π =3.14]
Solution:
Area of circular flower garden = πr²
314 m² = πr²
\(r^{2}=\frac{314 \mathrm{~m}^{2}}{\pi}=\frac{314}{3.14}=\frac{31400}{314}=100 \mathrm{~m}^{2}\)
∴ r = 10 m
But radius of sprinkler water = 12 m
∴ Area of sprinkler water = πr²
= 3.14 × 12 × 12 = 452.16 m².
Yes, because the area of sprinkler water is greater than area of circular flower garden.

Question 15.
Find the circumference of the inner and the outer circle as shown in
[Take π = 3.14]

Solution:
Radius of the outer circle = 19 m
∴ Circumference of the outer circle = 2π r = 2 × 3.14 × 19 m = 119.32 m
Radius of the inner circle = 19 m – 10 m = 9 m
∴ Circumference of the inner circle = 2π r = 2 × 3.14 × 9m = 56.52 m.

Question 16.
How many times the wheel of radius 28 cm must rotate to go 352 m.
[Take π = 22/7]
Sol. Radius of the wheel (r) = 28 cm
Circumference of the wheel = 2πr cm = 2x \(\frac{22}{7}\) × 28cm = 176 cm
∴ Number of times the wheel must rotate to go 352 m
= \(\frac{352 \times 100}{176}\) = 200

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:
Circumference of circular clock = 2πr
= 2 ×3.14 × 15 [π = 15 cm]
= 94.20 cm
1 hour = 60 minute
In one hour the tip of the minute hand complete the one revolution
= The circumference of the circular clock.
hence the distance of the moving minute hand = 94.20 cm.