Radius of cone r = 24/2 = 12 cm and slant height l = 21 cm Total surface area of cone = πr(r + l) = 22/7 × 12 × (12 + 21) = 22/7 × 12 × 33 = 1244.57 m² Hence, the total surface area of cone is 1244.57 m².
Radius of cone r = 24/2 = 12 cm and slant height l = 21 cm
Total surface area of cone = πr(r + l)
= 22/7 × 12 × (12 + 21)
= 22/7 × 12 × 33
= 1244.57 m²
Hence, the total surface area of cone is 1244.57 m².
(i) curved surface area of cone = 308 cm² and slant height l = 14 cm. Let, the radius of base of cone = r cm Curved surface area of cone = πrl ⇒ 308 = 22/7 × r × 14 ⇒ 308 = 44r ⇒ r = 308/44 = 7 cm Hence, the radius of base of cone is 7 cm. (ii) Total surface area of cone = πr(r + l) = 22/7 × 7 × (7Read more
(i) curved surface area of cone = 308 cm² and slant height l = 14 cm.
Let, the radius of base of cone = r cm
Curved surface area of cone = πrl
⇒ 308 = 22/7 × r × 14
⇒ 308 = 44r
⇒ r = 308/44 = 7 cm
Hence, the radius of base of cone is 7 cm.
(ii) Total surface area of cone = πr(r + l)
= 22/7 × 7 × (7 + 14)
= 22 × 21
= 462 cm²
Hence, the total surface area of cone is 462 cm².
Radius of cone r = 24 m and height h = 10 m Let, the slant height = l m We know that, l² = r² + h² ⇒ l² = 24² + 10² = 576 + 100 = 676 ⇒ l = √676 = 26 m
Radius of cone r = 24 m and height h = 10 m
Let, the slant height = l m
We know that, l² = r² + h²
⇒ l² = 24² + 10² = 576 + 100 = 676
⇒ l = √676 = 26 m
Radius of cone r = 6 m and height h = 8 m let, the slant height = l m We know that, l² = r² + h² ⇒ l² = 6² + 8² = 36 + 64 = 100 ⇒ l = √100 = 10 m Area of tarpaulin to make the tent = πrl = 3.14 × 6 × 10 = 188.40 m² Let, the length of 3 m wide tarpaulin = L Therefore, the area of tarpaulin required =Read more
Radius of cone r = 6 m and height h = 8 m
let, the slant height = l m
We know that, l² = r² + h²
⇒ l² = 6² + 8² = 36 + 64 = 100
⇒ l = √100 = 10 m
Area of tarpaulin to make the tent = πrl
= 3.14 × 6 × 10 = 188.40 m²
Let, the length of 3 m wide tarpaulin = L
Therefore, the area of tarpaulin required = 3 × L
According to question,
3 × L = 188.40
⇒ L = 188.40/3 = 62.80 m
Extra tarpaulin for stitching margins and wastage = 20 cm = 0.20 m
Therefore, the total lenght of tarpaulin = 62.80 + 0.20 = 63 m
Hence, the length of 3 m wide tarpaulin is 63 m to make the tent.
Total surface area of cylindrical Petrol storage tank = 2πr(r + h) = 2 × 22/7 × 2.1 × (2.1 + 4.5) = 2 × 22/7 × 2.1 × 6.6 = 87. 12 m² Let, the area of steel used to make this cylindrical petrol storage tank = x m² Steel get wasted in preparation of petrol storage tank = 1/12x m² Therefore, the totalRead more
Total surface area of cylindrical Petrol storage tank = 2πr(r + h)
= 2 × 22/7 × 2.1 × (2.1 + 4.5) = 2 × 22/7 × 2.1 × 6.6 = 87. 12 m²
Let, the area of steel used to make this cylindrical petrol storage tank = x m²
Steel get wasted in preparation of petrol storage tank = 1/12x m²
Therefore, the total steel used in cylindrical petrol storage tank = x – 1/12x = 11/12x m²
According to the questions: 11/12x = 87.12 ⇒ x = 87.12 × 12/11 = 95.04 m²
Hence, 95.04 m² steel is required to make this cylindrical petrol storage tank.
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.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Radius of cone r = 24/2 = 12 cm and slant height l = 21 cm Total surface area of cone = πr(r + l) = 22/7 × 12 × (12 + 21) = 22/7 × 12 × 33 = 1244.57 m² Hence, the total surface area of cone is 1244.57 m².
Radius of cone r = 24/2 = 12 cm and slant height l = 21 cm
See lessTotal surface area of cone = πr(r + l)
= 22/7 × 12 × (12 + 21)
= 22/7 × 12 × 33
= 1244.57 m²
Hence, the total surface area of cone is 1244.57 m².
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.
(i) curved surface area of cone = 308 cm² and slant height l = 14 cm. Let, the radius of base of cone = r cm Curved surface area of cone = πrl ⇒ 308 = 22/7 × r × 14 ⇒ 308 = 44r ⇒ r = 308/44 = 7 cm Hence, the radius of base of cone is 7 cm. (ii) Total surface area of cone = πr(r + l) = 22/7 × 7 × (7Read more
(i) curved surface area of cone = 308 cm² and slant height l = 14 cm.
Let, the radius of base of cone = r cm
Curved surface area of cone = πrl
⇒ 308 = 22/7 × r × 14
⇒ 308 = 44r
⇒ r = 308/44 = 7 cm
Hence, the radius of base of cone is 7 cm.
(ii) Total surface area of cone = πr(r + l)
See less= 22/7 × 7 × (7 + 14)
= 22 × 21
= 462 cm²
Hence, the total surface area of cone is 462 cm².
A conical tent is 10 m high and the radius of its base is 24 m. Find slant height of the tent.
Radius of cone r = 24 m and height h = 10 m Let, the slant height = l m We know that, l² = r² + h² ⇒ l² = 24² + 10² = 576 + 100 = 676 ⇒ l = √676 = 26 m
Radius of cone r = 24 m and height h = 10 m
See lessLet, the slant height = l m
We know that, l² = r² + h²
⇒ l² = 24² + 10² = 576 + 100 = 676
⇒ l = √676 = 26 m
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π= 3.14).
Radius of cone r = 6 m and height h = 8 m let, the slant height = l m We know that, l² = r² + h² ⇒ l² = 6² + 8² = 36 + 64 = 100 ⇒ l = √100 = 10 m Area of tarpaulin to make the tent = πrl = 3.14 × 6 × 10 = 188.40 m² Let, the length of 3 m wide tarpaulin = L Therefore, the area of tarpaulin required =Read more
Radius of cone r = 6 m and height h = 8 m
See lesslet, the slant height = l m
We know that, l² = r² + h²
⇒ l² = 6² + 8² = 36 + 64 = 100
⇒ l = √100 = 10 m
Area of tarpaulin to make the tent = πrl
= 3.14 × 6 × 10 = 188.40 m²
Let, the length of 3 m wide tarpaulin = L
Therefore, the area of tarpaulin required = 3 × L
According to question,
3 × L = 188.40
⇒ L = 188.40/3 = 62.80 m
Extra tarpaulin for stitching margins and wastage = 20 cm = 0.20 m
Therefore, the total lenght of tarpaulin = 62.80 + 0.20 = 63 m
Hence, the length of 3 m wide tarpaulin is 63 m to make the tent.
Find how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.
Total surface area of cylindrical Petrol storage tank = 2πr(r + h) = 2 × 22/7 × 2.1 × (2.1 + 4.5) = 2 × 22/7 × 2.1 × 6.6 = 87. 12 m² Let, the area of steel used to make this cylindrical petrol storage tank = x m² Steel get wasted in preparation of petrol storage tank = 1/12x m² Therefore, the totalRead more
Total surface area of cylindrical Petrol storage tank = 2πr(r + h)
See less= 2 × 22/7 × 2.1 × (2.1 + 4.5) = 2 × 22/7 × 2.1 × 6.6 = 87. 12 m²
Let, the area of steel used to make this cylindrical petrol storage tank = x m²
Steel get wasted in preparation of petrol storage tank = 1/12x m²
Therefore, the total steel used in cylindrical petrol storage tank = x – 1/12x = 11/12x m²
According to the questions: 11/12x = 87.12 ⇒ x = 87.12 × 12/11 = 95.04 m²
Hence, 95.04 m² steel is required to make this cylindrical petrol storage 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.