Length of rectangular garden = 90 m and breadth of rectangular garden = 75 m Outer length of rectangular garden with path = 90 + 5 + 5 = 100 m Outer breadth of rectangular garden with path = 75 + 5 + 5 = 85 m Outer area of rectangular garden with path = length x breadth = 100 x 85 = 8,500 m² Inner aRead more
Length of rectangular garden = 90 m and breadth of rectangular garden = 75 m
Outer length of rectangular garden with path = 90 + 5 + 5 = 100 m
Outer breadth of rectangular garden with path = 75 + 5 + 5 = 85 m
Outer area of rectangular garden with path = length x breadth = 100 x 85 = 8,500 m²
Inner area of garden without path = length x breadth = 90 x 75 = 6,750 m²
Now, Area of path = Area of garden with path – Area of garden without path
= 8,500 – 6,750
= 1,750 m²
Since, 1m² = 1/10000 hectares
Therefore, 6,750 m² = 6750/10000 = 0.675 hectares
In 1 hour, minute hand completes one round means makes a circle. Radius of the circle (r) = 15 cm Circumference of circular clock = 2πr = 2 x 3.14 x 15 = 94.2 cm Therefore, the tip of the minute hand moves 94.2 cm in 1 hour. Class 7 Maths Chapter 11 Exercise 11.3 for more answers vist to: https://wwRead more
In 1 hour, minute hand completes one round means makes a circle.
Radius of the circle (r) = 15 cm
Circumference of circular clock = 2πr
= 2 x 3.14 x 15
= 94.2 cm
Therefore, the tip of the minute hand moves 94.2 cm in 1 hour.
Let wheel must be rotate n times of its circumference. Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm ∴ Distance covered by wheel = n x circumference of wheel ⇒ 35200 = nX2πr ⇒ 35200 = nx2x22/7x28 ⇒ n = 35200x7/2x2228 ⇒ n = 200 revolutions Thus, wheel must rotate 200 times to go 352 mRead more
Let wheel must be rotate n times of its circumference.
Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm
∴ Distance covered by wheel = n x circumference of wheel
⇒ 35200 = nX2πr
⇒ 35200 = nx2x22/7×28
⇒ n = 35200×7/2×2228
⇒ n = 200 revolutions
Thus, wheel must rotate 200 times to go 352 m.
Radius of outer circle (r) = 19 m ∴ Circumference of outer circle = 2πr = 2 x 3.14 x 19 = 119.32 m Now radius of inner circle (r') = 19 – 10 = 9 m ∴ Circumference of inner circle = 2πr' = 2 x 3.14 x 9 = 56.52 m Therefore, the circumferences of inner and outer circles are 56.52 m and 119.32 m respectRead more
Radius of outer circle (r) = 19 m
∴ Circumference of outer circle = 2πr = 2 x 3.14 x 19
= 119.32 m
Now radius of inner circle (r’) = 19 – 10 = 9 m
∴ Circumference of inner circle = 2πr’ = 2 x 3.14 x 9
= 56.52 m
Therefore, the circumferences of inner and outer circles are 56.52 m and 119.32 m
respectively.
Circular area by the sprinkler = πr² = 3.14 x 12 x 12 = 3.14 x 144 = 452.16 m² Area of the circular flower garden = 314 m² Since Area of circular flower garden is smaller than area by sprinkler. Therefore, the sprinkler will water the entire garden. Class 7 Maths Chapter 11 Exercise 11.3 for more anRead more
Circular area by the sprinkler = πr²
= 3.14 x 12 x 12
= 3.14 x 144
= 452.16 m²
Area of the circular flower garden = 314 m²
Since Area of circular flower garden is smaller than area by sprinkler.
Therefore, the sprinkler will water the entire garden.
Diameter of the circular flower bed = 66 m ∴ Radius of circular flower bed (r) = 66/2 = 33 m ∴ Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m According to the question, Area of path = Area of bigger circle – Area of smaller circle = πR² - πr² = π(R²-r²) = π[(37)²-(33)² = 3.14 [Read more
Diameter of the circular flower bed = 66 m
∴ Radius of circular flower bed (r) = 66/2 = 33 m
∴ Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m
According to the question,
Area of path = Area of bigger circle – Area of smaller circle
= πR² – πr² = π(R²-r²)
= π[(37)²-(33)²
= 3.14 [ (37 + 33) (37 – 33)] [∵ a²-b²=(a+b)(a+b)]
= 3.14 x 70 x 4
= 879.20 m²
Therefore, the area of the path is 879.20 m².
The circumference of the circle = 31.4 cm ⇒ 2πr = 31.4 ⇒ 2 x 3.14 x r = 31.4 ⇒ r = 31x4/2 3.14 = 5 cm Then area of the circle = πr² = 3.14 x 5 x 5 = 78.5 cm² Therefore, the radius and the area of the circle are 5 cm and 78.5 cm² respectively. Class 7 Maths Chapter 11 Exercise 11.3 for more answers vRead more
The circumference of the circle = 31.4 cm
⇒ 2πr = 31.4
⇒ 2 x 3.14 x r = 31.4
⇒ r = 31×4/2 3.14 = 5 cm
Then area of the circle = πr² = 3.14 x 5 x 5
= 78.5 cm²
Therefore, the radius and the area of the circle are 5 cm and 78.5 cm² respectively.
Radius of circle = 2 cm and side of aluminium square sheet = 6 cm According to question, Area of aluminium sheet left = Total area of aluminium sheet – Area of circle = side x side -πr² = 6 x 6 – 22/7x 2 x 2 = 36 – 12.56 = 23.44 cm² Therefore, the area of aluminium sheet left is 23.44 cm². Class 7 MRead more
Radius of circle = 2 cm and side of aluminium square sheet = 6 cm
According to question,
Area of aluminium sheet left = Total area of aluminium sheet – Area of circle
= side x side -πr²
= 6 x 6 – 22/7x 2 x 2
= 36 – 12.56
= 23.44 cm²
Therefore, the area of aluminium sheet left is 23.44 cm².
Radius of circular sheet (R) = 14 cm and Radius of smaller circle (r) = 3.5 cm Length of rectangle (l) = 3 cm and breadth of rectangle (b) = 1 cm According to question, Area of remaining sheet=Area of circular sheet– (Area of two smaller circle + Area of rectangle) = πR² - [2(πR²) + (lxb) = 22/7x14xRead more
Radius of circular sheet (R) = 14 cm and Radius of smaller circle (r) = 3.5 cm
Length of rectangle (l) = 3 cm and breadth of rectangle (b) = 1 cm
According to question,
Area of remaining sheet=Area of circular sheet– (Area of two smaller circle + Area
of rectangle)
= πR² – [2(πR²) + (lxb)
= 22/7x14x14 – [(2×22/7×3.5×3.5)-(3×1)]
= 22 x 14 x 2 – [44 x 0.5 x 3.5 + 3]
= 616 – 80
= 536 cm²
Therefore the area of remaining sheet is 536 cm².
Total length of the wire = 44 cm ∴ the circumference of the circle = 2πr = 44 cm ⇒ 2x22/7xr=44 ⇒ r = 44x7/2x22 = 7 cm Now Area of the circle =πr² = 22/7x7x7=154 cm² Now the wire is converted into square. Then perimeter of square = 44 cm ⇒ 4 x side = 44 ⇒ side = 44/4 = 11 cm Now area of square = sideRead more
Total length of the wire = 44 cm
∴ the circumference of the circle = 2πr = 44 cm
⇒ 2×22/7xr=44
⇒ r = 44×7/2×22 = 7 cm
Now Area of the circle =πr²
= 22/7x7x7=154 cm²
Now the wire is converted into square.
Then perimeter of square = 44 cm
⇒ 4 x side = 44
⇒ side = 44/4 = 11 cm
Now area of square = side x side = 11 x 11 = 121 cm²
Therefore, on comparing, the area of circle is greater than that of square, so the circle
enclosed more area.
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectares.
Length of rectangular garden = 90 m and breadth of rectangular garden = 75 m Outer length of rectangular garden with path = 90 + 5 + 5 = 100 m Outer breadth of rectangular garden with path = 75 + 5 + 5 = 85 m Outer area of rectangular garden with path = length x breadth = 100 x 85 = 8,500 m² Inner aRead more
Length of rectangular garden = 90 m and breadth of rectangular garden = 75 m
Outer length of rectangular garden with path = 90 + 5 + 5 = 100 m
Outer breadth of rectangular garden with path = 75 + 5 + 5 = 85 m
Outer area of rectangular garden with path = length x breadth = 100 x 85 = 8,500 m²
Inner area of garden without path = length x breadth = 90 x 75 = 6,750 m²
Now, Area of path = Area of garden with path – Area of garden without path
= 8,500 – 6,750
= 1,750 m²
Since, 1m² = 1/10000 hectares
Therefore, 6,750 m² = 6750/10000 = 0.675 hectares
Class 7 Maths Chapter 11 Exercise 11.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
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)
In 1 hour, minute hand completes one round means makes a circle. Radius of the circle (r) = 15 cm Circumference of circular clock = 2πr = 2 x 3.14 x 15 = 94.2 cm Therefore, the tip of the minute hand moves 94.2 cm in 1 hour. Class 7 Maths Chapter 11 Exercise 11.3 for more answers vist to: https://wwRead more
In 1 hour, minute hand completes one round means makes a circle.
Radius of the circle (r) = 15 cm
Circumference of circular clock = 2πr
= 2 x 3.14 x 15
= 94.2 cm
Therefore, the tip of the minute hand moves 94.2 cm in 1 hour.
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
How many times a wheel of radius 28 cm must rotate to go 352 m? ( Take π = 22/7)
Let wheel must be rotate n times of its circumference. Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm ∴ Distance covered by wheel = n x circumference of wheel ⇒ 35200 = nX2πr ⇒ 35200 = nx2x22/7x28 ⇒ n = 35200x7/2x2228 ⇒ n = 200 revolutions Thus, wheel must rotate 200 times to go 352 mRead more
Let wheel must be rotate n times of its circumference.
Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm
∴ Distance covered by wheel = n x circumference of wheel
⇒ 35200 = nX2πr
⇒ 35200 = nx2x22/7×28
⇒ n = 35200×7/2×2228
⇒ n = 200 revolutions
Thus, wheel must rotate 200 times to go 352 m.
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
Find the circumference of the inner and the outer circles, shown in the adjoining figure. ( Take π = 3.14)
Radius of outer circle (r) = 19 m ∴ Circumference of outer circle = 2πr = 2 x 3.14 x 19 = 119.32 m Now radius of inner circle (r') = 19 – 10 = 9 m ∴ Circumference of inner circle = 2πr' = 2 x 3.14 x 9 = 56.52 m Therefore, the circumferences of inner and outer circles are 56.52 m and 119.32 m respectRead more
Radius of outer circle (r) = 19 m
∴ Circumference of outer circle = 2πr = 2 x 3.14 x 19
= 119.32 m
Now radius of inner circle (r’) = 19 – 10 = 9 m
∴ Circumference of inner circle = 2πr’ = 2 x 3.14 x 9
= 56.52 m
Therefore, the circumferences of inner and outer circles are 56.52 m and 119.32 m
respectively.
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
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)
Circular area by the sprinkler = πr² = 3.14 x 12 x 12 = 3.14 x 144 = 452.16 m² Area of the circular flower garden = 314 m² Since Area of circular flower garden is smaller than area by sprinkler. Therefore, the sprinkler will water the entire garden. Class 7 Maths Chapter 11 Exercise 11.3 for more anRead more
Circular area by the sprinkler = πr²
= 3.14 x 12 x 12
= 3.14 x 144
= 452.16 m²
Area of the circular flower garden = 314 m²
Since Area of circular flower garden is smaller than area by sprinkler.
Therefore, the sprinkler will water the entire garden.
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
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)
Diameter of the circular flower bed = 66 m ∴ Radius of circular flower bed (r) = 66/2 = 33 m ∴ Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m According to the question, Area of path = Area of bigger circle – Area of smaller circle = πR² - πr² = π(R²-r²) = π[(37)²-(33)² = 3.14 [Read more
Diameter of the circular flower bed = 66 m
∴ Radius of circular flower bed (r) = 66/2 = 33 m
∴ Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m
According to the question,
Area of path = Area of bigger circle – Area of smaller circle
= πR² – πr² = π(R²-r²)
= π[(37)²-(33)²
= 3.14 [ (37 + 33) (37 – 33)] [∵ a²-b²=(a+b)(a+b)]
= 3.14 x 70 x 4
= 879.20 m²
Therefore, the area of the path is 879.20 m².
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. ( Take π = 3.14)
The circumference of the circle = 31.4 cm ⇒ 2πr = 31.4 ⇒ 2 x 3.14 x r = 31.4 ⇒ r = 31x4/2 3.14 = 5 cm Then area of the circle = πr² = 3.14 x 5 x 5 = 78.5 cm² Therefore, the radius and the area of the circle are 5 cm and 78.5 cm² respectively. Class 7 Maths Chapter 11 Exercise 11.3 for more answers vRead more
The circumference of the circle = 31.4 cm
⇒ 2πr = 31.4
⇒ 2 x 3.14 x r = 31.4
⇒ r = 31×4/2 3.14 = 5 cm
Then area of the circle = πr² = 3.14 x 5 x 5
= 78.5 cm²
Therefore, the radius and the area of the circle are 5 cm and 78.5 cm² respectively.
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-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)
Radius of circle = 2 cm and side of aluminium square sheet = 6 cm According to question, Area of aluminium sheet left = Total area of aluminium sheet – Area of circle = side x side -πr² = 6 x 6 – 22/7x 2 x 2 = 36 – 12.56 = 23.44 cm² Therefore, the area of aluminium sheet left is 23.44 cm². Class 7 MRead more
Radius of circle = 2 cm and side of aluminium square sheet = 6 cm
According to question,
Area of aluminium sheet left = Total area of aluminium sheet – Area of circle
= side x side -πr²
= 6 x 6 – 22/7x 2 x 2
= 36 – 12.56
= 23.44 cm²
Therefore, the area of aluminium sheet left is 23.44 cm².
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
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)
Radius of circular sheet (R) = 14 cm and Radius of smaller circle (r) = 3.5 cm Length of rectangle (l) = 3 cm and breadth of rectangle (b) = 1 cm According to question, Area of remaining sheet=Area of circular sheet– (Area of two smaller circle + Area of rectangle) = πR² - [2(πR²) + (lxb) = 22/7x14xRead more
Radius of circular sheet (R) = 14 cm and Radius of smaller circle (r) = 3.5 cm
Length of rectangle (l) = 3 cm and breadth of rectangle (b) = 1 cm
According to question,
Area of remaining sheet=Area of circular sheet– (Area of two smaller circle + Area
of rectangle)
= πR² – [2(πR²) + (lxb)
= 22/7x14x14 – [(2×22/7×3.5×3.5)-(3×1)]
= 22 x 14 x 2 – [44 x 0.5 x 3.5 + 3]
= 616 – 80
= 536 cm²
Therefore the area of remaining sheet is 536 cm².
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
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)
Total length of the wire = 44 cm ∴ the circumference of the circle = 2πr = 44 cm ⇒ 2x22/7xr=44 ⇒ r = 44x7/2x22 = 7 cm Now Area of the circle =πr² = 22/7x7x7=154 cm² Now the wire is converted into square. Then perimeter of square = 44 cm ⇒ 4 x side = 44 ⇒ side = 44/4 = 11 cm Now area of square = sideRead more
Total length of the wire = 44 cm
∴ the circumference of the circle = 2πr = 44 cm
⇒ 2×22/7xr=44
⇒ r = 44×7/2×22 = 7 cm
Now Area of the circle =πr²
= 22/7x7x7=154 cm²
Now the wire is converted into square.
Then perimeter of square = 44 cm
⇒ 4 x side = 44
⇒ side = 44/4 = 11 cm
Now area of square = side x side = 11 x 11 = 121 cm²
Therefore, on comparing, the area of circle is greater than that of square, so the circle
enclosed more area.
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/