Let ABC is cone, which is converted into a frustum by cutting the top. Let r₁ and r₂ be the radius of two ends and h be its height. In ΔABG and ΔADF, DF∥BG ∴ ΔABG ∼ ΔADF DF/BG = AF/AG = AD/AB ⇒ r²/r₁ = h₁ - h/h₁ ⇒ r²/r₁ = 1 - h/h₁ = 1 - 1/l₁ ⇒ r₂/r₁ = 1 - h/h₁ ⇒ h/h₁ = 1 - r₂/r₁ = (r₁ - r₂/r₁) ⇒ h₁/Read more
Let ABC is cone, which is converted into a frustum by cutting the top. Let r₁ and r₂ be the radius of two ends and h be its height.
In ΔABG and ΔADF, DF∥BG ∴ ΔABG ∼ ΔADF
DF/BG = AF/AG = AD/AB
⇒ r²/r₁ = h₁ – h/h₁ ⇒ r²/r₁ = 1 – h/h₁ = 1 – 1/l₁
⇒ r₂/r₁ = 1 – h/h₁ ⇒ h/h₁ = 1 – r₂/r₁ = (r₁ – r₂/r₁)
⇒ h₁/h = (r₁/r₁ – r₂) ⇒ h₁ = r₁h/r₁ – r₂
Volume of frustum = Volume of cone ABC- Volume of cone ADE
= (1/3)πr₁²h₁ – (1/3)πr₂²(h₁ – h) = 1/3π[r₁²h₁ – r₂²(h₁ – h)]
= (1/3)π[r₁²(r₁h/r₁ – r₂) – r₂²(r₁h/r₁ – r₂) – h)] = (1/3)π[r₁³h/r₁ – r₂) – r₂²(r₁h – r₁h + r₂h/r₁ – r₂)]
= (1/3)π [(r₁³h/r₁ – r₂) – (r₂³h/r₁ -r₂)] = (1/3)πh[r₁³ – r₂³/r₁ – r₂] = (1/3)πh [(r₁ – r₂)(r₁² + r₁r₂ + r₂²/r₁ – r₂] = (1/3)πh(r₁² + r₁r₂ + r₂².
Two formed cones (which are formed by revolving the right angle triangle) are given. Hypotenuse AC = √(3² + 4²) = √25 = 5 cm Area of right angled triangle ABC = 1/2 × AB × AC ⇒ 1/2 × AC × OB = 1/2 × 4 × 3 ⇒ 1/2 × 5 × OB = 6 ⇒ OB = 12/5 = 2.4 cm Volume of double cone = Volume of cone 1 + Volume of coRead more
Two formed cones (which are formed by revolving the right angle triangle) are given.
Hypotenuse AC = √(3² + 4²) = √25 = 5 cm
Area of right angled triangle ABC = 1/2 × AB × AC
⇒ 1/2 × AC × OB = 1/2 × 4 × 3
⇒ 1/2 × 5 × OB = 6
⇒ OB = 12/5 = 2.4 cm
Volume of double cone = Volume of cone 1 + Volume of cone 2
= 1/3πr²h₁ + 1/3πr²h₂
= 1/3πr²(h₁ + h₂) = 1/3πr²(OA + OC)
1/3 × 3.14 × (2.4)² (5)
= 30.14 cm³
Curved surface area of double cone = Curved surface area of cone 1 + curved surface area of cone 2
= πrl₁ + πrl₂ = πr[4 + 3]
= 3.14 × 2.4 × 7 = 52.75 cm².
Volume of cistern = 150 × 120 × 110 = 1980000 cm³ Volume of water in cistern = 129600 cm³ Volume of bricks to be filled in cistern = 1980000 - 129600 = 1850400 cm³ Let, the number of bricks = n Therefore, Volume of n bricks = n × 22.5 × 7.5 × 6.5 = 1096.875 n As each brick absorbs 1/17 of its volumeRead more
Volume of cistern = 150 × 120 × 110 = 1980000 cm³
Volume of water in cistern = 129600 cm³
Volume of bricks to be filled in cistern = 1980000 – 129600 = 1850400 cm³
Let, the number of bricks = n
Therefore, Volume of n bricks = n × 22.5 × 7.5 × 6.5 = 1096.875 n
As each brick absorbs 1/17 of its volume, therefore, the volume of water obsorbed by n bricks = n/17 × 1096.875
According to question,
18504000 + n/17 × 1096.875 = 1096.875n
⇒ 18504000 = 16n/17 × 1096.875n ⇒ n = 1792.41
Hence, 1792 bricks can be put into the cistern without overflow the water.
Rainfall in the velley = 10 cm = 0.1 m. Area of valley = 97280 km² = 97280000000 m² Volume of rain water = 97280000000 × 0.1 = 9728000000 m³ Volume of one river = 1072000 × 75 × 3 = 241200000 m³ Volume of there rivers = 3 × 241200000 = 723600000 m³ Hence, 723600000 m³ ≈ 9728000000 m³
Rainfall in the velley = 10 cm = 0.1 m. Area of valley = 97280 km² = 97280000000 m²
Volume of rain water = 97280000000 × 0.1 = 9728000000 m³
Volume of one river = 1072000 × 75 × 3 = 241200000 m³
Volume of there rivers = 3 × 241200000 = 723600000 m³
Hence, 723600000 m³ ≈ 9728000000 m³
Upper radius of frustum (r₁) = 9 cm and lower radius of frustum (r₂) = 4 cm Therefore, the radius of cylindrical part = 4 cm Height of frustum (h₁) = 22 - 10 = 12 cm Height of cylindrical part (h₂) = 10 cm Slant height of frustum (l) = √(r₁ - r₂)² + h² = √(9 - 4)² + 12² = 13 cm Area of required tinRead more
Upper radius of frustum (r₁) = 9 cm and lower radius of frustum (r₂) = 4 cm
Therefore, the radius of cylindrical part = 4 cm
Height of frustum (h₁) = 22 – 10 = 12 cm
Height of cylindrical part (h₂) = 10 cm
Slant height of frustum (l) = √(r₁ – r₂)² + h² = √(9 – 4)² + 12² = 13 cm
Area of required tin = Curved surface area of frustum + curved surface area of cylindrical part = π(r₂ + r₂)l + 2πr₂h₂
= 22/7 × (9 + 4) × 13 + 2 × 22/7 × 4 × 10
= 22/7 [169 + 80] = (22 × 249/7) = 782(4/7) cm²
Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Let ABC is cone, which is converted into a frustum by cutting the top. Let r₁ and r₂ be the radius of two ends and h be its height. In ΔABG and ΔADF, DF∥BG ∴ ΔABG ∼ ΔADF DF/BG = AF/AG = AD/AB ⇒ r²/r₁ = h₁ - h/h₁ ⇒ r²/r₁ = 1 - h/h₁ = 1 - 1/l₁ ⇒ r₂/r₁ = 1 - h/h₁ ⇒ h/h₁ = 1 - r₂/r₁ = (r₁ - r₂/r₁) ⇒ h₁/Read more
Let ABC is cone, which is converted into a frustum by cutting the top. Let r₁ and r₂ be the radius of two ends and h be its height.
See lessIn ΔABG and ΔADF, DF∥BG ∴ ΔABG ∼ ΔADF
DF/BG = AF/AG = AD/AB
⇒ r²/r₁ = h₁ – h/h₁ ⇒ r²/r₁ = 1 – h/h₁ = 1 – 1/l₁
⇒ r₂/r₁ = 1 – h/h₁ ⇒ h/h₁ = 1 – r₂/r₁ = (r₁ – r₂/r₁)
⇒ h₁/h = (r₁/r₁ – r₂) ⇒ h₁ = r₁h/r₁ – r₂
Volume of frustum = Volume of cone ABC- Volume of cone ADE
= (1/3)πr₁²h₁ – (1/3)πr₂²(h₁ – h) = 1/3π[r₁²h₁ – r₂²(h₁ – h)]
= (1/3)π[r₁²(r₁h/r₁ – r₂) – r₂²(r₁h/r₁ – r₂) – h)] = (1/3)π[r₁³h/r₁ – r₂) – r₂²(r₁h – r₁h + r₂h/r₁ – r₂)]
= (1/3)π [(r₁³h/r₁ – r₂) – (r₂³h/r₁ -r₂)] = (1/3)πh[r₁³ – r₂³/r₁ – r₂] = (1/3)πh [(r₁ – r₂)(r₁² + r₁r₂ + r₂²/r₁ – r₂] = (1/3)πh(r₁² + r₁r₂ + r₂².
A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)
Two formed cones (which are formed by revolving the right angle triangle) are given. Hypotenuse AC = √(3² + 4²) = √25 = 5 cm Area of right angled triangle ABC = 1/2 × AB × AC ⇒ 1/2 × AC × OB = 1/2 × 4 × 3 ⇒ 1/2 × 5 × OB = 6 ⇒ OB = 12/5 = 2.4 cm Volume of double cone = Volume of cone 1 + Volume of coRead more
Two formed cones (which are formed by revolving the right angle triangle) are given.
See lessHypotenuse AC = √(3² + 4²) = √25 = 5 cm
Area of right angled triangle ABC = 1/2 × AB × AC
⇒ 1/2 × AC × OB = 1/2 × 4 × 3
⇒ 1/2 × 5 × OB = 6
⇒ OB = 12/5 = 2.4 cm
Volume of double cone = Volume of cone 1 + Volume of cone 2
= 1/3πr²h₁ + 1/3πr²h₂
= 1/3πr²(h₁ + h₂) = 1/3πr²(OA + OC)
1/3 × 3.14 × (2.4)² (5)
= 30.14 cm³
Curved surface area of double cone = Curved surface area of cone 1 + curved surface area of cone 2
= πrl₁ + πrl₂ = πr[4 + 3]
= 3.14 × 2.4 × 7 = 52.75 cm².
A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Volume of cistern = 150 × 120 × 110 = 1980000 cm³ Volume of water in cistern = 129600 cm³ Volume of bricks to be filled in cistern = 1980000 - 129600 = 1850400 cm³ Let, the number of bricks = n Therefore, Volume of n bricks = n × 22.5 × 7.5 × 6.5 = 1096.875 n As each brick absorbs 1/17 of its volumeRead more
Volume of cistern = 150 × 120 × 110 = 1980000 cm³
See lessVolume of water in cistern = 129600 cm³
Volume of bricks to be filled in cistern = 1980000 – 129600 = 1850400 cm³
Let, the number of bricks = n
Therefore, Volume of n bricks = n × 22.5 × 7.5 × 6.5 = 1096.875 n
As each brick absorbs 1/17 of its volume, therefore, the volume of water obsorbed by n bricks = n/17 × 1096.875
According to question,
18504000 + n/17 × 1096.875 = 1096.875n
⇒ 18504000 = 16n/17 × 1096.875n ⇒ n = 1792.41
Hence, 1792 bricks can be put into the cistern without overflow the water.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Rainfall in the velley = 10 cm = 0.1 m. Area of valley = 97280 km² = 97280000000 m² Volume of rain water = 97280000000 × 0.1 = 9728000000 m³ Volume of one river = 1072000 × 75 × 3 = 241200000 m³ Volume of there rivers = 3 × 241200000 = 723600000 m³ Hence, 723600000 m³ ≈ 9728000000 m³
Rainfall in the velley = 10 cm = 0.1 m. Area of valley = 97280 km² = 97280000000 m²
See lessVolume of rain water = 97280000000 × 0.1 = 9728000000 m³
Volume of one river = 1072000 × 75 × 3 = 241200000 m³
Volume of there rivers = 3 × 241200000 = 723600000 m³
Hence, 723600000 m³ ≈ 9728000000 m³
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.
Upper radius of frustum (r₁) = 9 cm and lower radius of frustum (r₂) = 4 cm Therefore, the radius of cylindrical part = 4 cm Height of frustum (h₁) = 22 - 10 = 12 cm Height of cylindrical part (h₂) = 10 cm Slant height of frustum (l) = √(r₁ - r₂)² + h² = √(9 - 4)² + 12² = 13 cm Area of required tinRead more
Upper radius of frustum (r₁) = 9 cm and lower radius of frustum (r₂) = 4 cm
See lessTherefore, the radius of cylindrical part = 4 cm
Height of frustum (h₁) = 22 – 10 = 12 cm
Height of cylindrical part (h₂) = 10 cm
Slant height of frustum (l) = √(r₁ – r₂)² + h² = √(9 – 4)² + 12² = 13 cm
Area of required tin = Curved surface area of frustum + curved surface area of cylindrical part = π(r₂ + r₂)l + 2πr₂h₂
= 22/7 × (9 + 4) × 13 + 2 × 22/7 × 4 × 10
= 22/7 [169 + 80] = (22 × 249/7) = 782(4/7) cm²