Let ABC is a 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) = (l₁ - l/l₁) ⇒ r₂/r₁ = 1 - h/h₁ = 1 - l/l₁ ⇒ r₂/r₁ = 1 - l/l₁ ⇒ l/l₁ = 1 - r₂/r₁Read more
Let ABC is a 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) = (l₁ – l/l₁) ⇒ r₂/r₁ = 1 – h/h₁ = 1 – l/l₁
⇒ r₂/r₁ = 1 – l/l₁ ⇒ l/l₁ = 1 – r₂/r₁ = r₁ – r₂/r₁
⇒ l₁/l = (r₁/r₁ – r₂) ⇒ l₁ = r₁l/r₁ – r₂
Curved surface area of frustum = CSA of cone ABC – CSA of cone ADE = πr₁l₁ – πr₂(l₁ – l)
= πr₁(r₁l/r₁ – r₂) – πr₂((r₁l/r₁ – r₂) – l) = (πr₁²l/r₁ – r₂) – πr₂ (r₁l – r₁l + r₂l/r₁ – r₂)
= (πr₁²l/r₁ -r₂) – (πr₂²l/r₁ – r₂) = πl (r₁² – r₂²/r₁ – r₂) = π(r₁ + r₂)l
total surface area of frustum = Curved surface area + Area of upper circle + Area of lower circle.
= π(r₁ + r₂)l + πr₂² + πr₁²
In ΔAEG, EG/AG = tan 30° ⇒ EG/10 = 1/√3 ⇒ EG = 10/√3 In ΔABD, BD/AD = tan 30° ⇒ BD/20 = 1/√3 ⇒ BD = 20√3 Radius of upper part of frustum (r₁) = 10/√3 Radius of lower part of frustum (r₂) = 20/√3 Height of frustum (h) = 10 cm Volume of frustum = 1/3πh(r₁² + r₂² + r₁r₂) = 1/3π × 10 × [(10/√3)² + (20/√Read more
In ΔAEG,
EG/AG = tan 30°
⇒ EG/10 = 1/√3
⇒ EG = 10/√3
In ΔABD,
BD/AD = tan 30°
⇒ BD/20 = 1/√3
⇒ BD = 20√3
Radius of upper part of frustum (r₁) = 10/√3
Radius of lower part of frustum (r₂) = 20/√3
Height of frustum (h) = 10 cm
Volume of frustum = 1/3πh(r₁² + r₂² + r₁r₂)
= 1/3π × 10 × [(10/√3)² + (20/√3)² + 10/√3 × 20/√3]
= 1/3 × 22/7 × 10 × (100/3 + 400/3 + 200/3)
= 1/3 × 22/7 × 10 × (700/3) = 22000/9 cm³
Radius of wire (r) = 1/2 × 1/16 = 1/32 cm
Let the lenght of wire = l
Volume of wire = area of cross- section of wire × length of wire
= (πr²) (l)
= π(1/32)²× l
Volume of frustum = Volume of wire
⇒ 22000/9 = π(1/32)² × l
⇒ 22000/9 = 22 /7 × 1/1024 × l
⇒ l = 22000/9 × 7/22 × 1024
= 796444.44 cm
7964. 44 m
Hence, the length of wire is 7964.44 m.
It can be observed that 1 round of wire will cover 3 mm height of cylinder. Therefore, the number of round = Height of cylinder/Diameter of wire = 12/0.3 = 40 Lenght of wire for one round = radius of cylinder = 2πr = 2π × 5 = 10π Length of wire to wound 40 times = 40 × 10π = (400 × 22/7) = 8800/7 =Read more
It can be observed that 1 round of wire will cover 3 mm height of cylinder.
Therefore, the number of round = Height of cylinder/Diameter of wire = 12/0.3 = 40
Lenght of wire for one round = radius of cylinder
= 2πr = 2π × 5 = 10π
Length of wire to wound 40 times = 40 × 10π
= (400 × 22/7) = 8800/7
= 1257.14 cm = 12.57 m
Radius of wire = 0.15 cm
Volume of wire = Area of cross- section of wire × length
= π(0.15)² × 1257.14 = 88.898 cm³
Mass = Volume × Density = 88.898 × 8.88 = 789.41 gm
Circumference of upper part of frustum = 18 cm ⇒ 2πr₁ = 18 ⇒r₁ = 9/π Circumference of lower part of frustum = 6 cm ⇒ 2πr₂ = 6 ⇒r₂ = 3/π Height of frustum = 4 cm Curved surface area of the frustum = π (r₁ + r₂)l = π (9/π+3/π)4 = 12 × 4 = 48 cm² Hence, the curved area of the frustum is 48 cm².
Circumference of upper part of frustum = 18 cm
⇒ 2πr₁ = 18 ⇒r₁ = 9/π
Circumference of lower part of frustum = 6 cm
⇒ 2πr₂ = 6 ⇒r₂ = 3/π
Height of frustum = 4 cm
Curved surface area of the frustum = π (r₁ + r₂)l
= π (9/π+3/π)4 = 12 × 4 = 48 cm²
Hence, the curved area of the frustum is 48 cm².
Radius of lower part of cap (r₁) = 10 cm Radius of upper part of cap (r₂) = 4 cm Slant height of cap = 15 cm Area of material used for making it = CSA of frustum + Area of upper part = π(r₁ + r₂)l + πr₂ = π(10 + 4) × 15 + π × 4² = 210π + 16π = 226π = 226 × 22/7 = 710(2/7) cm² Hence, the area of mateRead more
Radius of lower part of cap (r₁) = 10 cm
Radius of upper part of cap (r₂) = 4 cm
Slant height of cap = 15 cm
Area of material used for making it = CSA of frustum + Area of upper part
= π(r₁ + r₂)l + πr₂
= π(10 + 4) × 15 + π × 4²
= 210π + 16π = 226π
= 226 × 22/7 = 710(2/7) cm²
Hence, the area of material used for making is 710(2/7) cm².
Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Let ABC is a 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) = (l₁ - l/l₁) ⇒ r₂/r₁ = 1 - h/h₁ = 1 - l/l₁ ⇒ r₂/r₁ = 1 - l/l₁ ⇒ l/l₁ = 1 - r₂/r₁Read more
Let ABC is a 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) = (l₁ – l/l₁) ⇒ r₂/r₁ = 1 – h/h₁ = 1 – l/l₁
⇒ r₂/r₁ = 1 – l/l₁ ⇒ l/l₁ = 1 – r₂/r₁ = r₁ – r₂/r₁
⇒ l₁/l = (r₁/r₁ – r₂) ⇒ l₁ = r₁l/r₁ – r₂
Curved surface area of frustum = CSA of cone ABC – CSA of cone ADE = πr₁l₁ – πr₂(l₁ – l)
= πr₁(r₁l/r₁ – r₂) – πr₂((r₁l/r₁ – r₂) – l) = (πr₁²l/r₁ – r₂) – πr₂ (r₁l – r₁l + r₂l/r₁ – r₂)
= (πr₁²l/r₁ -r₂) – (πr₂²l/r₁ – r₂) = πl (r₁² – r₂²/r₁ – r₂) = π(r₁ + r₂)l
total surface area of frustum = Curved surface area + Area of upper circle + Area of lower circle.
= π(r₁ + r₂)l + πr₂² + πr₁²
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/6 cm find the length of the wire.
In ΔAEG, EG/AG = tan 30° ⇒ EG/10 = 1/√3 ⇒ EG = 10/√3 In ΔABD, BD/AD = tan 30° ⇒ BD/20 = 1/√3 ⇒ BD = 20√3 Radius of upper part of frustum (r₁) = 10/√3 Radius of lower part of frustum (r₂) = 20/√3 Height of frustum (h) = 10 cm Volume of frustum = 1/3πh(r₁² + r₂² + r₁r₂) = 1/3π × 10 × [(10/√3)² + (20/√Read more
In ΔAEG,
See lessEG/AG = tan 30°
⇒ EG/10 = 1/√3
⇒ EG = 10/√3
In ΔABD,
BD/AD = tan 30°
⇒ BD/20 = 1/√3
⇒ BD = 20√3
Radius of upper part of frustum (r₁) = 10/√3
Radius of lower part of frustum (r₂) = 20/√3
Height of frustum (h) = 10 cm
Volume of frustum = 1/3πh(r₁² + r₂² + r₁r₂)
= 1/3π × 10 × [(10/√3)² + (20/√3)² + 10/√3 × 20/√3]
= 1/3 × 22/7 × 10 × (100/3 + 400/3 + 200/3)
= 1/3 × 22/7 × 10 × (700/3) = 22000/9 cm³
Radius of wire (r) = 1/2 × 1/16 = 1/32 cm
Let the lenght of wire = l
Volume of wire = area of cross- section of wire × length of wire
= (πr²) (l)
= π(1/32)²× l
Volume of frustum = Volume of wire
⇒ 22000/9 = π(1/32)² × l
⇒ 22000/9 = 22 /7 × 1/1024 × l
⇒ l = 22000/9 × 7/22 × 1024
= 796444.44 cm
7964. 44 m
Hence, the length of wire is 7964.44 m.
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
It can be observed that 1 round of wire will cover 3 mm height of cylinder. Therefore, the number of round = Height of cylinder/Diameter of wire = 12/0.3 = 40 Lenght of wire for one round = radius of cylinder = 2πr = 2π × 5 = 10π Length of wire to wound 40 times = 40 × 10π = (400 × 22/7) = 8800/7 =Read more
It can be observed that 1 round of wire will cover 3 mm height of cylinder.
See lessTherefore, the number of round = Height of cylinder/Diameter of wire = 12/0.3 = 40
Lenght of wire for one round = radius of cylinder
= 2πr = 2π × 5 = 10π
Length of wire to wound 40 times = 40 × 10π
= (400 × 22/7) = 8800/7
= 1257.14 cm = 12.57 m
Radius of wire = 0.15 cm
Volume of wire = Area of cross- section of wire × length
= π(0.15)² × 1257.14 = 88.898 cm³
Mass = Volume × Density = 88.898 × 8.88 = 789.41 gm
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Circumference of upper part of frustum = 18 cm ⇒ 2πr₁ = 18 ⇒r₁ = 9/π Circumference of lower part of frustum = 6 cm ⇒ 2πr₂ = 6 ⇒r₂ = 3/π Height of frustum = 4 cm Curved surface area of the frustum = π (r₁ + r₂)l = π (9/π+3/π)4 = 12 × 4 = 48 cm² Hence, the curved area of the frustum is 48 cm².
Circumference of upper part of frustum = 18 cm
See less⇒ 2πr₁ = 18 ⇒r₁ = 9/π
Circumference of lower part of frustum = 6 cm
⇒ 2πr₂ = 6 ⇒r₂ = 3/π
Height of frustum = 4 cm
Curved surface area of the frustum = π (r₁ + r₂)l
= π (9/π+3/π)4 = 12 × 4 = 48 cm²
Hence, the curved area of the frustum is 48 cm².
A fez, the cap used by the Turks, is shaped like the frustum of a cone If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
Radius of lower part of cap (r₁) = 10 cm Radius of upper part of cap (r₂) = 4 cm Slant height of cap = 15 cm Area of material used for making it = CSA of frustum + Area of upper part = π(r₁ + r₂)l + πr₂ = π(10 + 4) × 15 + π × 4² = 210π + 16π = 226π = 226 × 22/7 = 710(2/7) cm² Hence, the area of mateRead more
Radius of lower part of cap (r₁) = 10 cm
See lessRadius of upper part of cap (r₂) = 4 cm
Slant height of cap = 15 cm
Area of material used for making it = CSA of frustum + Area of upper part
= π(r₁ + r₂)l + πr₂
= π(10 + 4) × 15 + π × 4²
= 210π + 16π = 226π
= 226 × 22/7 = 710(2/7) cm²
Hence, the area of material used for making is 710(2/7) cm².