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.
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.