Radius of each pehholder r = 3 cm and height h = 10.5 cm The penholder is open at the top, therefore, the area of cardboard for 1 penholder = 2πrh + πr² = 2 × π × 3 × 10.5 + π × 3² = 63π + 9π = 72π cm² So, the area of cardboard for 35 penholders = 35 × 72π cm² = 35 × 72 × 22/7 = 5 × 72 × 22 = 7920 cRead more
Radius of each pehholder r = 3 cm and height h = 10.5 cm
The penholder is open at the top, therefore, the area of cardboard for 1 penholder = 2πrh + πr²
= 2 × π × 3 × 10.5 + π × 3² = 63π + 9π = 72π cm²
So, the area of cardboard for 35 penholders = 35 × 72π cm²
= 35 × 72 × 22/7 = 5 × 72 × 22 = 7920 cm²
So, Vidhyalaya has to purchase 7920 cm² cardboard for the competition.
Radius of cone r = 10.5/2 = 5.25 cm and slant height l = 10 cm Curved surface area of cone = πrl = 22/7 × 5.25 × 10 = 22 × 0.75 × 10 = 165 cm² Hence, the curved surface area of cone is 165 cm².
Radius of cone r = 10.5/2 = 5.25 cm and slant height l = 10 cm
Curved surface area of cone = πrl
= 22/7 × 5.25 × 10
= 22 × 0.75 × 10
= 165 cm²
Hence, the curved surface area of cone is 165 cm².
Radius of cylindrical petrol storage tank r = 4.2/2 = 2.1 m and height h = 4.5 m. Curved surface area of cylindrical petrol storage tank = 2πrh = 2 × 22/7 × 2.1 × 4.5 = 2 × 22 × 0.3 × 4.5 = 59.4 m² Hence, the curved surface area of cylindrical petrol storage tank is 59.4 m².
Radius of cylindrical petrol storage tank r = 4.2/2 = 2.1 m and height h = 4.5 m.
Curved surface area of cylindrical petrol storage tank = 2πrh
= 2 × 22/7 × 2.1 × 4.5 = 2 × 22 × 0.3 × 4.5 = 59.4 m²
Hence, the curved surface area of cylindrical petrol storage tank is 59.4 m².
(i) Radius of circular well r = 3.5/2 m and depth h = 10 m Inner curved surface area of circular well = 2πrh = 2 × 22/7 × 3.5/2 × 10 = 22 × 0.5 × 10 = 110 m² Hence, the inner curved surface area of circular well is 110 m². (ii) The cost of plastering this curved surface at the rate of Rs 40 per m² =Read more
(i) Radius of circular well r = 3.5/2 m and depth h = 10 m
Inner curved surface area of circular well = 2πrh
= 2 × 22/7 × 3.5/2 × 10 = 22 × 0.5 × 10 = 110 m²
Hence, the inner curved surface area of circular well is 110 m².
(ii) The cost of plastering this curved surface at the rate of Rs 40 per m² = Rs 110 × 40 = Rs 4400
Hence, the cost of plastering this curved surface at the rate of 40 per m² is RS 4400.
Radis of cylindrical pipe r = 5/2 cm = 2.5 cm = 0.025 m and lenght h = 28 m Total surface area of cylindrical pipe = 2πr (r + h) = 2 × 22/7 × 0.025 × (0.025 + 28) = 2 × 22/7 × 0.025 × 28.025 = 4.4 m² (approx) Hence, the total radiating surface in the system is 4.4 m².
Radis of cylindrical pipe r = 5/2 cm = 2.5 cm = 0.025 m and lenght h = 28 m
Total surface area of cylindrical pipe = 2πr (r + h)
= 2 × 22/7 × 0.025 × (0.025 + 28) = 2 × 22/7 × 0.025 × 28.025 = 4.4 m² (approx)
Hence, the total radiating surface in the system is 4.4 m².
Radius of roller r = 84/2 = 42 cm = 0.42 m and length h = 120 cm = 1.2 m outer curved surface area of roller = 2πrh = 2 × 22/7 × 0.42 × 1.2 = 2 × 22 × 0.06 × 1.2 = 3.168 m² Area of ground levelled in on revolution = 3.168 m² Therefore, area of ground levelled in 500 revolutions = 500 × 3.168 = 1584Read more
Radius of roller r = 84/2 = 42 cm = 0.42 m and length h = 120 cm = 1.2 m
outer curved surface area of roller = 2πrh
= 2 × 22/7 × 0.42 × 1.2 = 2 × 22 × 0.06 × 1.2 = 3.168 m²
Area of ground levelled in on revolution = 3.168 m²
Therefore, area of ground levelled in 500 revolutions = 500 × 3.168 = 1584 m²
Hence, the area of playground is 1584 m².
Radius of cylindrical pillar r = 50/2 = 25 cm = 0.25 m and height h = 3.5 m Curved surface area of cylindrical pillar = 2πrh = 2 × 22/7 × 0.25 × 3.5 = 2 × 22 × 0.25 × 0.5 = 5.5 m² Cost of painting 1 m² area = Rs. 12.50 Therefore, cost of painting 5.5 m² area = Rs. 12.50 × 5.5 = Rs 68.75 Hence, the tRead more
Radius of cylindrical pillar r = 50/2 = 25 cm = 0.25 m and height h = 3.5 m
Curved surface area of cylindrical pillar = 2πrh
= 2 × 22/7 × 0.25 × 3.5 = 2 × 22 × 0.25 × 0.5 = 5.5 m²
Cost of painting 1 m² area = Rs. 12.50
Therefore, cost of painting 5.5 m² area = Rs. 12.50 × 5.5 = Rs 68.75
Hence, the total cost of painting the cylindrical pillar is Rs 68.75.
Curved surface area of cylinder 4.4 m² and radius r = 0.7 m Let, the height of cylinder = h m Curved surface area of cylinder = 2πrh ⇒ 4.4 = 2 × 22/7 × 0.7 × h ⇒ 4.4 = 4.4h ⇒ h = 1 m Hence, the height of the cylinder is 1 m.
Curved surface area of cylinder 4.4 m² and radius r = 0.7 m
Let, the height of cylinder = h m
Curved surface area of cylinder = 2πrh
⇒ 4.4 = 2 × 22/7 × 0.7 × h ⇒ 4.4 = 4.4h ⇒ h = 1 m
Hence, the height of the cylinder is 1 m.
Curved surface area of cylinder 88 cm² and height h = 14 cm Let, the radius of base of cylinder = r cm Curved surface area of cylinder = 2πrh ⇒ 88 = 2 × 22/7 × r × 14 ⇒ 88 = 88r ⇒ r = 1 cm Hence, the diameter of base of cylinder = 2r = 2 × 1 = 2 cm.
Curved surface area of cylinder 88 cm² and height h = 14 cm
Let, the radius of base of cylinder = r cm
Curved surface area of cylinder = 2πrh
⇒ 88 = 2 × 22/7 × r × 14 ⇒ 88 = 88r ⇒ r = 1 cm
Hence, the diameter of base of cylinder = 2r = 2 × 1 = 2 cm.
Radius of base of cylinder r = 140/ 2 = 70 cm 0.7 m and height h = 1 m Total surface area of cylinder = 2πr(r + h) = 2 × 22/7 × 0.7(0.7 + 1) = 4.4 × 1.7 = 7.48 m² Hence, 7.48 square meter metal sheet is required to make the closed cylindrical tank.
Radius of base of cylinder r = 140/ 2 = 70 cm 0.7 m and height h = 1 m
Total surface area of cylinder = 2πr(r + h)
= 2 × 22/7 × 0.7(0.7 + 1) = 4.4 × 1.7 = 7.48 m²
Hence, 7.48 square meter metal sheet is required to make the closed cylindrical tank.
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Radius of each pehholder r = 3 cm and height h = 10.5 cm The penholder is open at the top, therefore, the area of cardboard for 1 penholder = 2πrh + πr² = 2 × π × 3 × 10.5 + π × 3² = 63π + 9π = 72π cm² So, the area of cardboard for 35 penholders = 35 × 72π cm² = 35 × 72 × 22/7 = 5 × 72 × 22 = 7920 cRead more
Radius of each pehholder r = 3 cm and height h = 10.5 cm
See lessThe penholder is open at the top, therefore, the area of cardboard for 1 penholder = 2πrh + πr²
= 2 × π × 3 × 10.5 + π × 3² = 63π + 9π = 72π cm²
So, the area of cardboard for 35 penholders = 35 × 72π cm²
= 35 × 72 × 22/7 = 5 × 72 × 22 = 7920 cm²
So, Vidhyalaya has to purchase 7920 cm² cardboard for the competition.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Radius of cone r = 10.5/2 = 5.25 cm and slant height l = 10 cm Curved surface area of cone = πrl = 22/7 × 5.25 × 10 = 22 × 0.75 × 10 = 165 cm² Hence, the curved surface area of cone is 165 cm².
Radius of cone r = 10.5/2 = 5.25 cm and slant height l = 10 cm
See lessCurved surface area of cone = πrl
= 22/7 × 5.25 × 10
= 22 × 0.75 × 10
= 165 cm²
Hence, the curved surface area of cone is 165 cm².
Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
Radius of cylindrical petrol storage tank r = 4.2/2 = 2.1 m and height h = 4.5 m. Curved surface area of cylindrical petrol storage tank = 2πrh = 2 × 22/7 × 2.1 × 4.5 = 2 × 22 × 0.3 × 4.5 = 59.4 m² Hence, the curved surface area of cylindrical petrol storage tank is 59.4 m².
Radius of cylindrical petrol storage tank r = 4.2/2 = 2.1 m and height h = 4.5 m.
See lessCurved surface area of cylindrical petrol storage tank = 2πrh
= 2 × 22/7 × 2.1 × 4.5 = 2 × 22 × 0.3 × 4.5 = 59.4 m²
Hence, the curved surface area of cylindrical petrol storage tank is 59.4 m².
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs 40 per m2.
(i) Radius of circular well r = 3.5/2 m and depth h = 10 m Inner curved surface area of circular well = 2πrh = 2 × 22/7 × 3.5/2 × 10 = 22 × 0.5 × 10 = 110 m² Hence, the inner curved surface area of circular well is 110 m². (ii) The cost of plastering this curved surface at the rate of Rs 40 per m² =Read more
(i) Radius of circular well r = 3.5/2 m and depth h = 10 m
See lessInner curved surface area of circular well = 2πrh
= 2 × 22/7 × 3.5/2 × 10 = 22 × 0.5 × 10 = 110 m²
Hence, the inner curved surface area of circular well is 110 m².
(ii) The cost of plastering this curved surface at the rate of Rs 40 per m² = Rs 110 × 40 = Rs 4400
Hence, the cost of plastering this curved surface at the rate of 40 per m² is RS 4400.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Radis of cylindrical pipe r = 5/2 cm = 2.5 cm = 0.025 m and lenght h = 28 m Total surface area of cylindrical pipe = 2πr (r + h) = 2 × 22/7 × 0.025 × (0.025 + 28) = 2 × 22/7 × 0.025 × 28.025 = 4.4 m² (approx) Hence, the total radiating surface in the system is 4.4 m².
Radis of cylindrical pipe r = 5/2 cm = 2.5 cm = 0.025 m and lenght h = 28 m
See lessTotal surface area of cylindrical pipe = 2πr (r + h)
= 2 × 22/7 × 0.025 × (0.025 + 28) = 2 × 22/7 × 0.025 × 28.025 = 4.4 m² (approx)
Hence, the total radiating surface in the system is 4.4 m².
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².
Radius of roller r = 84/2 = 42 cm = 0.42 m and length h = 120 cm = 1.2 m outer curved surface area of roller = 2πrh = 2 × 22/7 × 0.42 × 1.2 = 2 × 22 × 0.06 × 1.2 = 3.168 m² Area of ground levelled in on revolution = 3.168 m² Therefore, area of ground levelled in 500 revolutions = 500 × 3.168 = 1584Read more
Radius of roller r = 84/2 = 42 cm = 0.42 m and length h = 120 cm = 1.2 m
See lessouter curved surface area of roller = 2πrh
= 2 × 22/7 × 0.42 × 1.2 = 2 × 22 × 0.06 × 1.2 = 3.168 m²
Area of ground levelled in on revolution = 3.168 m²
Therefore, area of ground levelled in 500 revolutions = 500 × 3.168 = 1584 m²
Hence, the area of playground is 1584 m².
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12.50 per m².
Radius of cylindrical pillar r = 50/2 = 25 cm = 0.25 m and height h = 3.5 m Curved surface area of cylindrical pillar = 2πrh = 2 × 22/7 × 0.25 × 3.5 = 2 × 22 × 0.25 × 0.5 = 5.5 m² Cost of painting 1 m² area = Rs. 12.50 Therefore, cost of painting 5.5 m² area = Rs. 12.50 × 5.5 = Rs 68.75 Hence, the tRead more
Radius of cylindrical pillar r = 50/2 = 25 cm = 0.25 m and height h = 3.5 m
See lessCurved surface area of cylindrical pillar = 2πrh
= 2 × 22/7 × 0.25 × 3.5 = 2 × 22 × 0.25 × 0.5 = 5.5 m²
Cost of painting 1 m² area = Rs. 12.50
Therefore, cost of painting 5.5 m² area = Rs. 12.50 × 5.5 = Rs 68.75
Hence, the total cost of painting the cylindrical pillar is Rs 68.75.
Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.
Curved surface area of cylinder 4.4 m² and radius r = 0.7 m Let, the height of cylinder = h m Curved surface area of cylinder = 2πrh ⇒ 4.4 = 2 × 22/7 × 0.7 × h ⇒ 4.4 = 4.4h ⇒ h = 1 m Hence, the height of the cylinder is 1 m.
Curved surface area of cylinder 4.4 m² and radius r = 0.7 m
See lessLet, the height of cylinder = h m
Curved surface area of cylinder = 2πrh
⇒ 4.4 = 2 × 22/7 × 0.7 × h ⇒ 4.4 = 4.4h ⇒ h = 1 m
Hence, the height of the cylinder is 1 m.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.
Curved surface area of cylinder 88 cm² and height h = 14 cm Let, the radius of base of cylinder = r cm Curved surface area of cylinder = 2πrh ⇒ 88 = 2 × 22/7 × r × 14 ⇒ 88 = 88r ⇒ r = 1 cm Hence, the diameter of base of cylinder = 2r = 2 × 1 = 2 cm.
Curved surface area of cylinder 88 cm² and height h = 14 cm
See lessLet, the radius of base of cylinder = r cm
Curved surface area of cylinder = 2πrh
⇒ 88 = 2 × 22/7 × r × 14 ⇒ 88 = 88r ⇒ r = 1 cm
Hence, the diameter of base of cylinder = 2r = 2 × 1 = 2 cm.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?
Radius of base of cylinder r = 140/ 2 = 70 cm 0.7 m and height h = 1 m Total surface area of cylinder = 2πr(r + h) = 2 × 22/7 × 0.7(0.7 + 1) = 4.4 × 1.7 = 7.48 m² Hence, 7.48 square meter metal sheet is required to make the closed cylindrical tank.
Radius of base of cylinder r = 140/ 2 = 70 cm 0.7 m and height h = 1 m
See lessTotal surface area of cylinder = 2πr(r + h)
= 2 × 22/7 × 0.7(0.7 + 1) = 4.4 × 1.7 = 7.48 m²
Hence, 7.48 square meter metal sheet is required to make the closed cylindrical tank.