NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 with Answers
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Perimeter and Area Class 7 Ex 11.1
Perimeter and Area Class 7 Ex 11.2
Perimeter and Area Class 7 Ex 11.4
Perimeter and Area Class 7 MCQ
Exercise 11.3
Question 1:
Find the circumference of the circles with the following radius: (Take π = 22/7)
(a) 14 cm (b) 28 mm (c) 21 cm
Answer 1:
When radius= 14cm
Circumference of the circle = 2πr
= (2×22/7×14)
= 88 cm
(b) when radius= 28 cm
Circumference of the circle = 2πr
= (2×22/7×28)
= 176 mm
(c) when radius = 21cm
Circumference of the circle = 2πr
= (2×22/7×21)
= 132 cm
Question 2:
Find the area of the following circles, given that: (take π= 22/7)
(a) radius = 14 mm (b) diameter = 49 m (c) radius 5 cm
Answer 2:
when radius= 14mm
Area of circle = πr²
= 22/7×(14) ²
=(22/7×196)
= 616 mm²
(b) When diameter = 49 m
Radius= (49/2) m
Therefore, area of circle = πr²
=22/7×(49/2) ²
=(22/7×49/2×49/2)
=(11×49×7) /2
= 11×49×3.5)
= 1886.5 m²
(c) when radius= 5 cm
Area of circle = πr²
=(22/7) ×(5) ²
=(22/7) ×25
=(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. (takre π=22/7)
Answer 3:
Let the radius of the circle= r metre
Therefore, circumference of the circular sheet = 2πr
→2πr = 154
→r = (154×7) /(22×2)
→ r = (49/2)
→r = 24.5 metre
Now Area of circular sheet = πr²
=(22/7) × (49/2) ²
= (22/7) ×(49/2) ×(49/2)
=(11×7×49) /2
=1886.5 m²
Thus, the radius and area of the circular sheet are 24.5 m and 1886.5 m2 respectively.
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 costs of the rope, if it cost ₹4 per meter. ( Take = 22/7)
Answer 4:
Diameter of the circular garden = 21 m
Therefore radius of the circular garden =
(21/2) metre
Now Circumference of circular garden = 2πr
= 2×(22/7) ×(21/2)
= 22×3
=66 metre
The gardener makes 2 rounds of fence so the total length of the rope of fencing
= 2 x circumference
= 2 x 66 = 132 metre
Since, the cost of rope is ₹ 4 per metre
Therefore, total cost of 132 metre 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)
Answer 5:
Radius of circular sheet (R) = 4 cm
Therefore area = πr²
=(3.14) ×(4) ²
=3.14×4×4
= 50.24 cm²
And radius of removed circle = 3 cm
Area of removed circle = πr²
=(3.14) × (3) ²
=(3.14×3×3)
=28.26 cm²
Area of remaining sheet
= Area of circular sheet – Area of removed circle
= (50.24-28.26)
=21.98 cm²
Thus, the area of the remaining sheet is 21.98 cm².
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1 with AnswersQuestion 6:
Saima wants to put a lace on the edge of a circular table cover of 1.5 m diameter. Find the length of the lace required and also find its cost if one meter of the lace costs ₹15. (Take π= 3.14)
Answer 6:
Diameter of the circular table cover = 1.5 metre
Radius of the circular table cover =
(1.5/2) metre = 0.75 metre
Circumference of circular table cover = 2πr
=(2×3.14×0.75) metre
= 4.71 metre
Therefore the length of necessary lace is 4.71 m.
Now the cost of lace is ₹ 15 per metre
Then the cost of 4.71 metre lace = ₹(15 x 4.71)
= ₹ 70.65
Hence, the cost of the required lace is ₹ 70.65.
Question 7:
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Answer 7:
Diameter of the semicircle= 10 cm
Radius = (10/2) cm
=5 cm
Perimeter of the given semicircle= (Circumference of semicircle + diameter)
= (2πr/2) +2r
= πr + 2r
= (3.14×5) +(2×5)
=15.70+10
=25.70 cm
Thus, the perimeter of the given figure is 25.71 cm.
Question 8:
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹15/m². (Take = 3.14)
Answer 8:
Diameter of the circular table top = 1.6 metre
Therefore, radius of the circular table top =
(1.6/2) metre
= 0.8 metre
Area of circular table top = πr²
=3.14×(0.8) ²
= 3.14 x 0.8 x 0.8
= 2.0096 m²
Now rate of polishing = ₹15 per square metre
Then cost of 2.0096 m² polishing = (15 x 2.0096) = ₹ 30.14 (approximately)
Thus, the total cost of polishing of the circular table top is ₹ 30.14.
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 encloses more area, the circle or the
square? (take π=22/7)
Answer 9:
length of the wire = 44 cm
Since by using this wire we make a circle, so circumference of the circle is equal to the length of the wire.
Let radius of the circle is r cm
Therefore, the circumference of the circle → 2πr = 44
→r = (44×7) /(2×22)
→r = 7 cm
Now Area of the circle = πr²
= (22/7) ×(7) ²
=(22×7)
=154 cm²
Now since the same wire bent to form a square, hence the perimeter of the square is equal to the length of the wire.
Therefore, perimeter of square = 44 cm
→4 x side = 44
→side = (44/4)
→side = 11cm
Now area of square = side x side = 11 x 11 = 121 cm²
Therefore the area of the circle is 154 cm² and that of the square is 121 cm². So, the circle has a greater area than the square.
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. (Take π = 22/7)
Answer 10:
Radius of circular cardboard sheet (R) = 14 cm
Therefore area of the cardboard sheet = πr²
=(22/7) ×(14) ²
=(22×14×2)
=616 cm²
and Radius of smaller circle(r) = 3.5 cm
Area of smaller circle = πr²
= (22/7) × (3.5) ²
= (22×0.5×3.5)
= 38.5 cm²
Area of two smaller circle =(38.5×2) =77 cm²
Now, Length of rectangle = 3cm
Breadth of the rectangle = 1cm
Therefore area of rectangle = (length×breadth)
= (3×1)
=3 cm²
According to question,
Area of remaining cardboard sheet=Area of circular cardboard sheet– (Area of two smaller circle + Area
of rectangle)
= 616 - (77 + 3)
= (616 - 80)
= 536 cm²
Therefore the area of the remaining sheet is 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)
Answer 11:
Side of square sheet = 6cm.
Area of square sheet =( side ×side)
= (6×6) cm²
= 36 cm²
Radius of circle = 2 cm
Area of the circle = πr²
= 3.14 × (2) ²
= (3.14 × 4)
= 12.56 cm²
According to question,
Area of remaining aluminium sheet = Total area of aluminium sheet – Area of circle
= (36 – 12.56)
= 23.44 cm²
Therefore, the area of aluminium sheet remains is 23.44 cm².
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1 with AnswersQuestion 12:
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle.
(Take π= 3.14)
Answer 12:
The circumference of the circle = 31.4 cm
Let the radius of the circle = r cm
Therefore, 2πr = 31.4
→ r = 31.4 /(3.14 × 2)
→r = 5 cm
Then area of the circle = πr²
= 3.14 × (5) ²
= 3.14 × 25
= 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)
Answer 13:
Diameter of the circular flower bed (2r) = 66 m
Radius of circular flower bed(r) = (66/2) =33m
Area of circular bed = πr²
= 3.14 × (33) ²
= 3.14 × 33 ×33
= 3419.46 m²
Radius of circular flower bed with path (R) = 33 + 4 = 37 m
Area of circular flower bed with path = π R²
= 3.14 × (37)²
= 3.14 × 37 × 37
= 4298.66 m²
According to the question,
Area of path = Area of circular flower bed with path – Area of circular flower bed
= (4298.66 - 3419.46)
879.2 m²
Therefore, the area of the path is 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)
Answer 14:
Radius of circular area covered by the sprinkler(r) = 12 m
Therefore area covered by sprinkler = πr²
= 3.14 × (12) ²
= 3.14 × 144
= 452.16 m²
Area of the circular flower garden = 314 m²
Since Area of the circular flower garden is smaller than the area by sprinkler.
Therefore, the sprinkler will water the entire garden.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1 with AnswersQuestion 15:
Find the circumference of the inner and the outer circles, shown in the adjoining figure. (Take π = 3.14)
Answer 15:
Radius of outer circle(R) = 19 m
Therefore circumference of outer circle = 2πR
= (2×3.14×19)
=119.32 metre
Now radius of inner circle(r) =(19 – 10) = 9 metre
Circumference of inner circle = 2πr
= 2 x 3.14 x 9
= 56.52 metre
Therefore, the circumference of the inner circle is 56.52 metre and the circumference of outer circles is 119.32 metre.
Question 16:
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
Answer 16:
Radius of the wheel(r) = 28 cm
Circumference of the wheel = 2πr
= 2 × (22/7) × 28 cm
= (2×22×4) cm
= 176 cm
When a wheel rotates one time then it covers distance which is equal to its circumference.
Therefore distance covered by one time rotation = 176 cm
Now total distance = 352 metre = 35200 cm
Therefore number of rotation to complete total distance = (35200/176)
= 200 times.
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 )
Answer 17:
In 1 hour, one hour hand completes one complete rotation that makes a circle.
Now radius of that circle(r) = 15 cm
Circumference of circular clock = 2πr
= (2×3.14÷15)
= 94.2 cm
Therefore, the tip of the minute hand moves 94.2 cm in one hour.
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