Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3
Question 1.
Find the circumference of the circles with the following radius: ( Taken π = \(\frac { 22 }{ 7 }\) )
(a) 14 cm
(b) 28 mm
(c) 21 cm
Solution:
(a) Here, radius (r) = 14 cm
∴ Circumference (C) = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 14 cm = 2 x 22 x 2 cm
= 88 cm
(b) Here, radius (r) = 28 mm
∴ Circumference (C) = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 28 mm
= 2 x 22 x 4 mm = 176 mm
(c) Here, radius (r) = 21 cm
∴ Circumference (C) = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 21 cm
= 2 x 22 x 3 cm = 132 cm
Question 2.
Find the area of the following circles, given that: ( Taken π = \(\frac { 22 }{ 7 }\) )
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm
Solution:
(a) Here, radius (r) = 14 mm
∴ Area of the circle = π²
= \(\frac { 22 }{ 7 }\) x (14)² mm²
= \(\frac { 22 }{ 7 }\) x 14 x 14 mm²
= 22 x 2 x 14 mm² = 616 mm²
(b) Here, diameter = 49 m
∴ Radius (r) = \(\frac { 49 }{ 2 }\)m
Now, area of the circle = πr²
= \(\frac { 22 }{ 7 }\) x \(\frac { 49 }{ 2 }\)² m²
= \(\frac { 22 }{ 7 }\) x \(\frac { 49 }{ 2 }\) x \(\frac { 49 }{ 2 }\)m²
= 22 x \(\frac { 7 }{ 2 }\) x \(\frac { 49 }{ 2 }\) m²
= \(\frac { 11 ×7×49 }{ 2 }\)m²
= \(\frac { 3773 }{ 2 }\)m²
= 1886.5 m²
(c) Here, radius (r) = 5 cm
Area of the circle = πr²
= \(\frac { 22 }{ 7 }\) x (5 cm)²
= \(\frac { 22 }{ 7 }\) x 5 x 5 cm²
= \(\frac { 550 }{ 7 }\)cm²
Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. ( Taken π = \(\frac { 22 }{ 7 }\) )
Solution:
Here, the circumference of the circle = 154 m.
Consider radius = r m
Question 4.
A gardener wants to fence a circular garden of diameter 21 m. 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 metre. ( Taken π = \(\frac { 22 }{ 7 }\) )
Solution:
Diameter of the circular garden = 21 m
∴ Radius (r) = \(\frac { 21 }{ 2 }\)
∵ Circumference (C) = 2πr
∴ Circumference of the circular garden
= 2 x \(\frac { 22 }{ 7 }\) x \(\frac { 21 }{ 2 }\) m
= 22 x 3 m
= 66 m
Since, the rope makes 2 rounds of fence.
∴ Length of the rope = 2 x [Circumference of the garden]
= 2 x 66 m = 132 m
Thus, the length of the required rope = 132 m Now,cost of the rope = ₹ 4 x 132 = ₹ 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 circular sheet (R) = 4 cm.
Radius of the removed circular sheet (i.e. inner circle) (r)= 3 cm
Since, Area of a circle = π x (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(1)(7) 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 metre of the lace costs ₹ 15. (Take π = 3.14)
Solution:
Diameter of the circular table cover = 1.5 m.
∴ Radius (r) = \(\frac { 1.5 }{ 2 }\)m
∴ Circumference (C) = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x \(\frac { 1.5 }{ 2 }\)m
= \(\frac { 22×1.5 }{ 7 }\)m = 4.71m
Thus, the length of the lace required = 4.71 m.
Now, cost of lace = ₹ 15 x 4.71 = ₹ 70.71
Question 7.
Find the perimeter of the figure, which is a semicircle including its diameter.
Solution:
Diameter of the semicircle = 10 cm
∴ Radius (r) = \(\frac { 10 }{ 2 }\) = 5 cm
Now, circumference of the semicircle = \(\frac { 2πr }{ 2 }\) = πr
= \(\frac { 22 }{ 7 }\) x 5 cm
= \(\frac { 110 }{ 7 }\) cm = 15.71 cm
∴ Perimeter of the semicircle = Circumference + Diameter
= 15.71 cm + 10 cm = 25.71 cm
Question 8.
Find the cost of polishing a circular tabletop of diameter 1.6 m, if the rate of polishing is ₹ 15/m². (Take π = 3.14)
Solution:
Diameter of the circular table-top = 1.6 m
∴ Radius (r) = \(\frac { 1.6 }{ 2 }\)m
Now, area of the circular table-top = πr²
Since rate of polishing = ₹ 15/m²
∴ Total cost of polishing the circular table-top = ₹ 15 x [1.57 x 0.8 x 1.6] = ₹ 30.144.
Thus, the required cost of polishing the circular table-top = ₹ 30.144.
Question 9.
Shazli 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 enclose more area, the circle or the square? ( Taken π = \(\frac { 22 }{ 7 }\) )
Solution:
Here, length of the wire = 44 cm
∵ The wire is bent in the form of a circle.
∴ Circumference of the circle so formed = 44 cm
Let the radius of the circle be r
∴ 2πr = 44
or 2 x \(\frac { 22 }{ 7 }\) x r = 44
or r = \(\frac { 44×7 }{ 2×22 }\) cm = 7 cm.
Thus, the required radius of the circle is 7 cm.
Now, area of the circle so formed = πr²
= \(\frac { 22 }{ 7 }\) x (7 cm)²
= \(\frac { 22 }{ 7 }\) x 7 x 7 cm² = 154 cm²
Since, the wire is rebent to form a square.
∴ [Perimeter of the square so formed] = [Circumference of the circle]
or
Perimeter of the square = 44 cm
Let the side of the square be x cm.
∴ 4 x x = 44 cm
(∵ Perimeter of a square = 4 x Side)
or x = \(\frac { 44 }{ 4 }\) cm = 11 cm
Now, area of the square = Side x Side
= 11 cm x 11 cm = 121 cm²
Since 154 cm² >121 cm²
∴ The circle encloses greater area.
Question 10.
From a circular card sheet of radius 14 cm, 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. ( Taken π = \(\frac { 22 }{ 7 }\) )
Solution:
Radius of the circular sheet = 14 cm
∴ Area of the sheet = πr²
= \(\frac { 22 }{ 7 }\) x (14 cm)²
= \(\frac { 22 }{ 7 }\) x 14 x 14 cm²
= 22 x 2 x 14 cm² = 616 cm²
Now, radius of a small circle = 3.5 cm
∴ Area of a small circle = \(\frac { 22 }{ 7 }\) x 3.5 x 3.5 cm²
= 22 x 0.5 x 3.5 cm²
So, area of two small circles
= 2 x 22 x 0.5 x 3.5 cm²
= 22 x 1 x 3.5 cm² = 77 cm²
Again, length of small rectangle = 3 cm. and
breadth of the small rectangle = 1 cm
Area of the small rectangle = 3 x 1 cm² = 3 cm²
∴ Total area removed from the sheet = 77 cm² + 3 cm² = 80 cm²
Area of the remaining sheet
= 616 cm² – 80 cm² = 536 cm²
Thus, the required area of the remaining sheet = 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:
Side of the square = 6 cm
Area of the square = Side x Side = 6 cm x 6 cm = 36 cm²
Radius of the circle cut out from the sheet, r = 2 cm
∴ Area of the circle = πr² = 3.14 x (2 cm)²
= 3.14 x 2 x 2 cm² = 12.56 cm²
Now, area of the remaining sheet = 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:
Here, circumference of the circle = 31.4 cm
Let radius of the circle be r
∴ 2πr =31.4
or 2 x 3.14 x r = 31.4
or r = \(\frac { 31.4 }{ 2×3.14 }\) cm = 5 cm
Thus, radius of the circle (r)= 5 cm
Now, area of the circle = πr²
= 3.14 x (5 cm)²
= 3.14 x 5 x 5 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?
Solution:
∵ Diameter of the flower bed = 66 m
∴ Radius (r) = \(\frac { 66 }{ 2 }\) m = 33 m
Since width of the surrounding path = 4 m
∴ Radius of the outer circle (R)
= 33 m + 4 m = 37 m
Now, area of the outer circle = πR²
and area of the inner circle = πr²
∴ Area of the path = πR² – πr²
= π[R² – r²]
= π[(R + r)(R – r)]
= 3.14[(37 + 33)(37 – 33)] m²
= 3.14[70 x 4] m²
= 3.14 x 280 m²
= 879.2 m²
Thus, the area of the path = 879.2 m²
Question 14.
A circular flower garden has an area of 314 m². A sprinkler at the centre 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 the circular garden = 314 m²
Let the radius of the garden be r m
∴ πr² = 314
or 314 x r² = 314
or \(\frac { 314 }{ 100 }\) x r² = 314
or r² = 314 x \(\frac { 100 }{ 314 }\) = 100
r² = 10² ⇒ r = 10 m
Now, radius of the area covered by the sprinkler = 12 m
Since 12 m > 10 m
∴ The sprinkler covers an area beyond the garden or we can say, yes, the entire garden is covered by the sprinkler.
Question 15.
Find the circumference of the inner and the outer circles, shown in the figure? (Take π = 3.14)
Solution:
Radius of the outer circle (R) = 19 m
∴ Circumference of the outer circle = 2πR
= 2 x 3.14 x 19 m = 119.32 m
Again, Radius of the inner circle (r)
= 19 m – 10 m = 9 m
∴ Circumference of the inner circle = 2πr
= 2 x 3.14 x 9 m = 56.52 m
Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? ( Taken π = \(\frac { 22 }{ 7 }\) )
Solution:
Radius of the wheel (r) = 28 cm
∴ Circumference of the wheel = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 28 cm
= 2 x 22 x 4 cm = 176 cm
So, the wheel can cover 176 cm of distance in a rotation.
Total distance to be covered = 352 m
= 352 x 100 cm
∴ Number of rotations = \(\frac { 352×100 }{ 176 }\)
= 2 x 100 = 200
Thus, the distance of 352 m will be covered in 200 turns.
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
∴ Radius of the circle made by the tip of the minute hand (r) = 15 cm.
Perimeter of the circle so formed = 2πr
= 2 x (3.14) x (15 cm) = 94.2 cm
Since, the minute hand describes one complete revolution in one hour.
Thus, the distance covered by the tip of the minute hand = 94.2 cm.