radius of heap of wheat r = 10.5/2 = 5.25 m and height h = 3 m Volume of heap of wheat = 1/3πr²h = 1/3 × 22/7 × 5.25 × 5.25 × 3 = 22 × 0.75 × 5.25 = 86.625 m³. Hence, the volume of heap of wheat is 86.625 m³. Let, the slant height of heap of wheat = l m We know that, l² = h² + r² ⇒ l² = 3² + (5.25)²Read more
radius of heap of wheat r = 10.5/2 = 5.25 m and height h = 3 m
Volume of heap of wheat = 1/3πr²h = 1/3 × 22/7 × 5.25 × 5.25 × 3 = 22 × 0.75 × 5.25 = 86.625 m³.
Hence, the volume of heap of wheat is 86.625 m³.
Let, the slant height of heap of wheat = l m
We know that, l² = h² + r² ⇒ l² = 3² + (5.25)² ⇒ l² = 9 + 27.5625 ⇒ l² = 36.5625
⇒ l = √36.5625 = 6.05 m (approx.)
Required area of canvas = πrl
= 22/7 × 5.25 × 6.05 = 22 × 0.75 × 6.05 = 99.825 m²
Hence, the required area of canvas to protect wheat is 99.825 m².
Volume of cone V = 1570 cm³ and height h = 15 cm Let, the radius of base of cone = r cm Volume of cone = 1/3πr²h ⇒ 1570 = 1/3 × 3.14 × r² × 15 ⇒ 1570 = 3.14 × r² 5 ⇒ r² = 1570/(3.14×5) = 100 ⇒ r = √100 = 10 cm Hence, the radius of base of cone is 10 cm.
Volume of cone V = 1570 cm³ and height h = 15 cm
Let, the radius of base of cone = r cm
Volume of cone = 1/3πr²h
⇒ 1570 = 1/3 × 3.14 × r² × 15 ⇒ 1570 = 3.14 × r² 5 ⇒ r² = 1570/(3.14×5) = 100
⇒ r = √100 = 10 cm
Hence, the radius of base of cone is 10 cm.
Volume of cone V = 48π cm³ and height h = 9 cm Let, the radius of base of cone = r cm Volume of cone = 1/3πr²h ⇒ 48π = 1/3 × π × r² × 9 ⇒ 48π = π × r² × 3 ⇒ r² = 48π/π×3 = 16 ⇒ r = √16 = 4 cm Therefore, the diameter of base = 2 × 4 = 8 cm Hence, the diameter of base of cone is 8 cm.
Volume of cone V = 48π cm³ and height h = 9 cm
Let, the radius of base of cone = r cm
Volume of cone = 1/3πr²h
⇒ 48π = 1/3 × π × r² × 9 ⇒ 48π = π × r² × 3 ⇒ r² = 48π/π×3 = 16 ⇒ r = √16 = 4 cm
Therefore, the diameter of base = 2 × 4 = 8 cm
Hence, the diameter of base of cone is 8 cm.
Radius of pit r = 3.5/2 = 1.75 m and height h = 12 m. Volume of pit = 1/3πr²h = 1/3 × 22/7 × 1.75 × 12 = 38.5 m³ = 38.5 kiloliters [∵ 1 m³ = 1 kilolitres] Hence, the capacity of pit is 38.5 kilolitres.
Radius of pit r = 3.5/2 = 1.75 m and height h = 12 m.
Volume of pit = 1/3πr²h
= 1/3 × 22/7 × 1.75 × 12 = 38.5 m³
= 38.5 kiloliters [∵ 1 m³ = 1 kilolitres]
Hence, the capacity of pit is 38.5 kilolitres.
(I) Volume of cone V = 9856 cm³ and radius r = 28/2 = 14 cm Let, the height of cone be h cm, therefore volume of cone = 1/3πr²h ⇒ 9856 = 1/3 × 22/7 × 14 × 14 × h ⇒ 9856 = 1/3 × 22 × 2 × 14 × h ⇒ h = 9856 × 3/22 × 2 × 14 ⇒ h = 48 cm Hence, the height of cone is 48 cm. (II) Height of cone h = 48 cm anRead more
(I) Volume of cone V = 9856 cm³ and radius r = 28/2 = 14 cm
Let, the height of cone be h cm, therefore volume of cone = 1/3πr²h
⇒ 9856 = 1/3 × 22/7 × 14 × 14 × h
⇒ 9856 = 1/3 × 22 × 2 × 14 × h
⇒ h = 9856 × 3/22 × 2 × 14 ⇒ h = 48 cm
Hence, the height of cone is 48 cm.
(II) Height of cone h = 48 cm and radius r = 14 cm
Let, the slant height of cone = l cm
We, know that, l² = h² + r²
⇒ l² = 48² + 14² ⇒ l² = 2304 + 196
⇒ l² = 2500 ⇒ l = √2500 = 50 cm
Hence, the slant height of cone is 50 cm.
(III) Slant height of cone l = 50 cm and radius r = 14 cm
Curved surface area of cone = πrl
= 22/7 × 14 × 50 = 22 × 2 × 50 = 2200 cm²
Hence, the curved surface area of cone is 2200 cm².
If the triangle is revolved about 12 cm side, a cone will be formed. Therefore, the dadius of cones r = 5 cm height h = 12 cm and slant height l = 13 cm. Volume of solid (cone) = 1/3πr²h = 1/3 × π × 5 × 5 × 12 = 100π cm³ Hence, the volume of solid is 100π cm³.
If the triangle is revolved about 12 cm side, a cone will be formed. Therefore, the dadius of cones r = 5 cm height h = 12 cm and slant height l = 13 cm.
Volume of solid (cone) = 1/3πr²h = 1/3 × π × 5 × 5 × 12 = 100π cm³
Hence, the volume of solid is 100π cm³.
If the triangle is revolved about 5 cm side, a cone will be formed with radius r = 12 cm, height h = 5 cm slant height l = 13 cm. Volume of solid = 1/3 πr²h = 1/3 × π × 12 × 12 × 5 = 240π cm³ Hence, the volume of solid is 240π cm³.
If the triangle is revolved about 5 cm side, a cone will be formed with radius
r = 12 cm, height h = 5 cm slant height l = 13 cm.
Volume of solid = 1/3 πr²h = 1/3 × π × 12 × 12 × 5 = 240π cm³
Hence, the volume of solid is 240π cm³.
(I) Radius of conicon vessel r = 7 cm and slant height l = 25 cm. Let, the height of conical vessel = h cm We know that, l² = r² + h² ⇒ 25² = 7² + h² ⇒ 625 = 49 + h² ⇒ h² = 625 - 49 = 576 ⇒ h = √576 = 24 cm Capacity of conical vessel = 1/3πr²h = 1/3 × 22/7 × 7 × 24 = 1232 cm³ = 1232/1000 = 1.232 litRead more
(I) Radius of conicon vessel r = 7 cm and slant height l = 25 cm.
Let, the height of conical vessel = h cm
We know that, l² = r² + h²
⇒ 25² = 7² + h²
⇒ 625 = 49 + h² ⇒ h² = 625 – 49 = 576
⇒ h = √576 = 24 cm
Capacity of conical vessel = 1/3πr²h
= 1/3 × 22/7 × 7 × 24 = 1232 cm³ = 1232/1000 = 1.232 litres
Hence, the capacity of conical vessel is 1.232 litres.
(II) Height of conical vessel h = 12 cm and slant height l = 13 cm
Let, the radius of conical vessel = r cm
We know that, l² = h² + r²
⇒ 13² = 12² + r²
⇒ 169 = 144 + r²
⇒ r² = 169 – 144 = 25 ⇒
r = √25 = 5 cm
Capacity of conical vessel = 1/3πr²h
= 1/3 × 22/7 × 5 × 5 × 12 = 2200/7 cm³
= 2200/(7×1000) = 11/35 litres
Hence, the capacity of conical vessel is 11/35 litres.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Radius of hemispherical bowl r = 10.5/2 = 5.25 cm Therefore, the volume of hemispherical bowl = 2/3πr³ =2/3 × 22/7 × 5.25 × 5.25 × = 2 × 22 × 0.25 × 5.25 × 5.25 = 303 cm³ (Approx.) = 303/1000Litre [∵ 1 cm³ = 1/1000 litre] = 0.303 litre Hence, the hemispherical bowl holds 0.303 litres of milk.
Radius of hemispherical bowl r = 10.5/2 = 5.25 cm
See lessTherefore, the volume of hemispherical bowl = 2/3πr³
=2/3 × 22/7 × 5.25 × 5.25 × = 2 × 22 × 0.25 × 5.25 × 5.25 = 303 cm³ (Approx.)
= 303/1000Litre [∵ 1 cm³ = 1/1000 litre]
= 0.303 litre
Hence, the hemispherical bowl holds 0.303 litres of milk.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
radius of heap of wheat r = 10.5/2 = 5.25 m and height h = 3 m Volume of heap of wheat = 1/3πr²h = 1/3 × 22/7 × 5.25 × 5.25 × 3 = 22 × 0.75 × 5.25 = 86.625 m³. Hence, the volume of heap of wheat is 86.625 m³. Let, the slant height of heap of wheat = l m We know that, l² = h² + r² ⇒ l² = 3² + (5.25)²Read more
radius of heap of wheat r = 10.5/2 = 5.25 m and height h = 3 m
See lessVolume of heap of wheat = 1/3πr²h = 1/3 × 22/7 × 5.25 × 5.25 × 3 = 22 × 0.75 × 5.25 = 86.625 m³.
Hence, the volume of heap of wheat is 86.625 m³.
Let, the slant height of heap of wheat = l m
We know that, l² = h² + r² ⇒ l² = 3² + (5.25)² ⇒ l² = 9 + 27.5625 ⇒ l² = 36.5625
⇒ l = √36.5625 = 6.05 m (approx.)
Required area of canvas = πrl
= 22/7 × 5.25 × 6.05 = 22 × 0.75 × 6.05 = 99.825 m²
Hence, the required area of canvas to protect wheat is 99.825 m².
Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m
Radius of sphere r = 7 cm Therefore, volume of sphere = 4/3πr³ = 4/3 × 22/7 × 7 × 7 × 7 = 4/3 × 22 × 7 × 7 = 1437(1/3) cm³ Hence, the volume of sphere = 4/3πr³ = 4/3 × 22/7 × 0.63 × 0.63 × 0.63 = 4 × 22 × 0.03 × 0.63 × 0.63 = 1.05 m³(approx.) Hence, the volume of sphere is 1.05 m³.
Radius of sphere r = 7 cm
See lessTherefore, volume of sphere = 4/3πr³
= 4/3 × 22/7 × 7 × 7 × 7 = 4/3 × 22 × 7 × 7 = 1437(1/3) cm³
Hence, the volume of sphere = 4/3πr³
= 4/3 × 22/7 × 0.63 × 0.63 × 0.63 = 4 × 22 × 0.03 × 0.63 × 0.63 = 1.05 m³(approx.)
Hence, the volume of sphere is 1.05 m³.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base.
Volume of cone V = 1570 cm³ and height h = 15 cm Let, the radius of base of cone = r cm Volume of cone = 1/3πr²h ⇒ 1570 = 1/3 × 3.14 × r² × 15 ⇒ 1570 = 3.14 × r² 5 ⇒ r² = 1570/(3.14×5) = 100 ⇒ r = √100 = 10 cm Hence, the radius of base of cone is 10 cm.
Volume of cone V = 1570 cm³ and height h = 15 cm
See lessLet, the radius of base of cone = r cm
Volume of cone = 1/3πr²h
⇒ 1570 = 1/3 × 3.14 × r² × 15 ⇒ 1570 = 3.14 × r² 5 ⇒ r² = 1570/(3.14×5) = 100
⇒ r = √100 = 10 cm
Hence, the radius of base of cone is 10 cm.
If the volume of a right circular cone of height 9 cm is 48 π cm³, find the diameter of its base.
Volume of cone V = 48π cm³ and height h = 9 cm Let, the radius of base of cone = r cm Volume of cone = 1/3πr²h ⇒ 48π = 1/3 × π × r² × 9 ⇒ 48π = π × r² × 3 ⇒ r² = 48π/π×3 = 16 ⇒ r = √16 = 4 cm Therefore, the diameter of base = 2 × 4 = 8 cm Hence, the diameter of base of cone is 8 cm.
Volume of cone V = 48π cm³ and height h = 9 cm
See lessLet, the radius of base of cone = r cm
Volume of cone = 1/3πr²h
⇒ 48π = 1/3 × π × r² × 9 ⇒ 48π = π × r² × 3 ⇒ r² = 48π/π×3 = 16 ⇒ r = √16 = 4 cm
Therefore, the diameter of base = 2 × 4 = 8 cm
Hence, the diameter of base of cone is 8 cm.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Radius of pit r = 3.5/2 = 1.75 m and height h = 12 m. Volume of pit = 1/3πr²h = 1/3 × 22/7 × 1.75 × 12 = 38.5 m³ = 38.5 kiloliters [∵ 1 m³ = 1 kilolitres] Hence, the capacity of pit is 38.5 kilolitres.
Radius of pit r = 3.5/2 = 1.75 m and height h = 12 m.
See lessVolume of pit = 1/3πr²h
= 1/3 × 22/7 × 1.75 × 12 = 38.5 m³
= 38.5 kiloliters [∵ 1 m³ = 1 kilolitres]
Hence, the capacity of pit is 38.5 kilolitres.
The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find
(I) Volume of cone V = 9856 cm³ and radius r = 28/2 = 14 cm Let, the height of cone be h cm, therefore volume of cone = 1/3πr²h ⇒ 9856 = 1/3 × 22/7 × 14 × 14 × h ⇒ 9856 = 1/3 × 22 × 2 × 14 × h ⇒ h = 9856 × 3/22 × 2 × 14 ⇒ h = 48 cm Hence, the height of cone is 48 cm. (II) Height of cone h = 48 cm anRead more
(I) Volume of cone V = 9856 cm³ and radius r = 28/2 = 14 cm
Let, the height of cone be h cm, therefore volume of cone = 1/3πr²h
⇒ 9856 = 1/3 × 22/7 × 14 × 14 × h
⇒ 9856 = 1/3 × 22 × 2 × 14 × h
⇒ h = 9856 × 3/22 × 2 × 14 ⇒ h = 48 cm
Hence, the height of cone is 48 cm.
(II) Height of cone h = 48 cm and radius r = 14 cm
Let, the slant height of cone = l cm
We, know that, l² = h² + r²
⇒ l² = 48² + 14² ⇒ l² = 2304 + 196
⇒ l² = 2500 ⇒ l = √2500 = 50 cm
Hence, the slant height of cone is 50 cm.
(III) Slant height of cone l = 50 cm and radius r = 14 cm
See lessCurved surface area of cone = πrl
= 22/7 × 14 × 50 = 22 × 2 × 50 = 2200 cm²
Hence, the curved surface area of cone is 2200 cm².
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
If the triangle is revolved about 12 cm side, a cone will be formed. Therefore, the dadius of cones r = 5 cm height h = 12 cm and slant height l = 13 cm. Volume of solid (cone) = 1/3πr²h = 1/3 × π × 5 × 5 × 12 = 100π cm³ Hence, the volume of solid is 100π cm³.
If the triangle is revolved about 12 cm side, a cone will be formed. Therefore, the dadius of cones r = 5 cm height h = 12 cm and slant height l = 13 cm.
See lessVolume of solid (cone) = 1/3πr²h = 1/3 × π × 5 × 5 × 12 = 100π cm³
Hence, the volume of solid is 100π cm³.
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
If the triangle is revolved about 5 cm side, a cone will be formed with radius r = 12 cm, height h = 5 cm slant height l = 13 cm. Volume of solid = 1/3 πr²h = 1/3 × π × 12 × 12 × 5 = 240π cm³ Hence, the volume of solid is 240π cm³.
If the triangle is revolved about 5 cm side, a cone will be formed with radius
See lessr = 12 cm, height h = 5 cm slant height l = 13 cm.
Volume of solid = 1/3 πr²h = 1/3 × π × 12 × 12 × 5 = 240π cm³
Hence, the volume of solid is 240π cm³.
Find the capacity in litres of a conical vessel with radius 7 cm, slant height 25 cm.
(I) Radius of conicon vessel r = 7 cm and slant height l = 25 cm. Let, the height of conical vessel = h cm We know that, l² = r² + h² ⇒ 25² = 7² + h² ⇒ 625 = 49 + h² ⇒ h² = 625 - 49 = 576 ⇒ h = √576 = 24 cm Capacity of conical vessel = 1/3πr²h = 1/3 × 22/7 × 7 × 24 = 1232 cm³ = 1232/1000 = 1.232 litRead more
(I) Radius of conicon vessel r = 7 cm and slant height l = 25 cm.
Let, the height of conical vessel = h cm
We know that, l² = r² + h²
⇒ 25² = 7² + h²
⇒ 625 = 49 + h² ⇒ h² = 625 – 49 = 576
⇒ h = √576 = 24 cm
Capacity of conical vessel = 1/3πr²h
= 1/3 × 22/7 × 7 × 24 = 1232 cm³ = 1232/1000 = 1.232 litres
Hence, the capacity of conical vessel is 1.232 litres.
(II) Height of conical vessel h = 12 cm and slant height l = 13 cm
See lessLet, the radius of conical vessel = r cm
We know that, l² = h² + r²
⇒ 13² = 12² + r²
⇒ 169 = 144 + r²
⇒ r² = 169 – 144 = 25 ⇒
r = √25 = 5 cm
Capacity of conical vessel = 1/3πr²h
= 1/3 × 22/7 × 5 × 5 × 12 = 2200/7 cm³
= 2200/(7×1000) = 11/35 litres
Hence, the capacity of conical vessel is 11/35 litres.