Class 6 - Mathematics
Chapter - Perimeter and Area : Exercise 11.3
Top Block 1
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 :
(a) Circumference of the circle = 2πr = 2×22⁄7×14 = 88 cm
(b) Circumference of the circle = 2πr = 2×22⁄7×28 = 176 mm
(c) 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 :
(a) Area of circle = πr2 = 22⁄7 × 14 × 14 = 22 x 2 x 14 = 616 mm2
(b) Diameter = 49 m
∴ radius = 49⁄2 = 24.5 m
∴ Area of circle = πr2 = 22⁄7 × 24.5 × 24.5 = 22 x 3.5 x 24.5 = 1886.5 m2
Area of circle = πr2 = 22⁄7 × 5 × 5 = 550⁄7cm2
Question : 3.If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22⁄7)
Answer :
Circumference of the circular sheet = 154 m
⇒ 2πr = 154 m
⇒ r = 154⁄2π
⇒ r = (154 x 7)⁄(2 x 22) = 24.5 m
Now Area of circular sheet = πr2 = 22⁄7×24.5 × 24.5
= 22 x 3.5 x 24.5 = 1886.5 m2
Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m2respectively.
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 Rs. 4 per meter. (Take π = 22⁄7)
Answer :
Diameter of the circular garden = 21 m
∴ Radius of the circular garden = 21⁄2 m
Now Circumference of circular garden = 2πr = 2×22⁄7×21⁄2
= 22 x 3 = 66 m
The gardener makes 2 rounds of fence so the total length of the rope of fencing
= 2 x 2πr = 2 x 66 = 132 m
Since the cost of 1 meter rope = Rs. 4
Therefore, cost of 132 meter rope = 4 c 132 = 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)
Answer :
Radius of circular sheet (R) = 4 cm and radius of removed circle (r) = 3 cm
Area of remaining sheet = Area of circular sheet – Area of removed circle
= πR2 – πr2 = π(R2 – r2)
= π(42 – 32) = π(16 – 9)
= 3.14 x 7 = 21.98cm2
Thus, the area of remaining sheet is 21.98cm2.
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 costs Rs. 15. (Take π = 3.14)
Answer :
Diameter of the circular table cover = 1.5 m
∴ Radius of the circular table cover = 1.5⁄2 m
Circumference of circular table cover = 2πr = 2 × 3.14 × 1.5⁄2 = 4.71 m
Therefore the length of required lace is 4.71 m.
Now the cost of 1 m lace = Rs. 15
Then the cost of 4.71 m lace = 15 x 4.71 = Rs. 70.65
Hence, the cost of 4.71 m lace is Rs. 70.65.
Question : 7.Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Answer :
Diameter = 10 cm
∴ Radius = 10⁄2 = 5 cm
According to Question,
Perimeter of figure = Circumference of semi-circle + diameter
= πr + D
= 22⁄7×5+10 = 110⁄7+10
= (110+70)⁄7 = 180⁄7 = 25.71 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 Rs. 15/m2. (Take π = 3.14)
Answer :
Diameter of the circular table top = 1.6 m
∴ Radius of the circular table top = 1.62= 0.8 m
Area of circular table top = πr2
= 3.14 x 0.8 x 0.8 = 2.0096 m2
Now cost of 1 m2 polishing = Rs. 15
Then cost of 2.0096m2 polishing = 15 x 2.0096 = Rs. 30.14 (approx.)
Thus, the cost of polishing a circular table top is Rs. 30.14 (approx.)
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 :
Total length of the wire = 44 cm
∴ the circumference of the circle = 2πr = 44 cm
⇒ 2 × 22⁄7 × r = 44
⇒ r = (44 x 7)⁄(2 x 22) = 7 cm
Now Area of the circle = πr2 = 22⁄7 × 7 × 7 = 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.
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)
Mddle block 1
Answer :
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) + (l × b)]
= 22⁄7 × 14 × 14 – [(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.
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 :
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 – 22⁄7 x 2 x 2
= 36 – 12.56
= 23.44 cm2
Therefore, the area of aluminium sheet left is 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)
Answer :
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 78.5cm2 respectively.
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 :
Diameter of the circular flower bed = 66 m
∴ Radius of circular flower bed (r) = 662 = 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.
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 :
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.
Question : 15.Find the circumference of the inner and the outer circles, shown in the adjoining figure. (Take π = 3.14)
Answer :
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.
Question : 16.How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22⁄7)
Answer :
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 = n × 2πr
⇒ 35200 = n × 2 × 22⁄7 × 28
⇒ n = (35200 × 7)⁄(2 × 22 × 28)
⇒ n = 200 revolutions
Thus wheel must rotate 200 times to go 352 m.
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 :
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.