Three Dimensional Solids Class 10 OP Malhotra Exe-15C ICSE Maths Solutions Ch-15 questions as latest prescribe guideline for upcoming exam. In this article you would learn how to solve problems on Surface Area and Volume of Cone. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Three Dimensional Solids Class 10 OP Malhotra Exe-15C ICSE Maths Solutions Ch-15
Board | ICSE |
Publications | S Chand |
Subject | Maths |
Class | 10th |
Chapter-15 | Three Dimensional Solids |
Writer | OP Malhotra |
Exe-15C | Surface Area and Volume of Cone |
Edition | 2024-2025 |
Surface Area and Volume of Cone
The surface area of a cone is calculated as πr(l + r), where “r” is the radius of the base and “l” is the slant height, while the volume of a cone is (1/3)πr²h where “h” is the height of the cone. Curved Surface area of cone = πrl
Exercise- 15C
Three Dimensional Solids Class 10 OP Malhotra Exe-15C ICSE Maths Solutions Ch-15
Que-2: Find the volume of the cone given ; (i) height 8m, area of base 156m² (ii) Slant height 17cm, radius 8cm (iii) Height 8cm,slant height 10cm (iv) height 5cm,perimeter of base 8cm
Sol: (i) Given, height of cone (h) = 8 m
Area of base = πr2 = 156 m2
Now, volume of cone = (1/3)πr²h
= (1/3)×156×8
= 52 × 8
= 416 m3
(ii) Slant height (l) = 17 cm
Radius (r) = 8 cm
But, l² = r² + h²
⇒ h² = 1² – r²
⇒ h² = 17² – 8²
⇒ h² = 289 – 64 = 225 = (15)²
∴ h = 15
Now, volume of cone = (1/3)πr²h
= (1/3)×(22/7)×8×8×15 cm³
= 7040/7 cm³
= 1005.71 cm³.
(iii) Slant height (l) = 10 cm
Height (h) = 8 cm
But, l² = r² + h²
⇒ 10² = r² – 8²
⇒ r² = 10² – 8²
⇒ r² = 100 – 64 = 36 = (6)²
∴ r = 6
Now, volume of cone = (1/3)πr²h
= (1/3)×(22/7)×6×6×8 cm³
= 2112/7 cm³
= 301.71 cm³.
(iv) height = 5cm
perimeter of base = 8cm
Perimeter = 2πr
8 = 2×(22/7)×r
r = (8×7)/(22×2)
r = 14/11
volume of cone = (1/3)πr²h
= (1/3)×(22/7)×(14/11)×(14/11)×5 cm³
= 280/33 cm³
= 8.48 cm³.
Que-3: Find the area of curved surface of the circular cone, given ; (i) height 8m, slant height 10m (ii) Perimeter of the base 88 cm, slant height 2 dm (iii) Area of the base 154 cm², height 24 cm
Sol: (i) A right circular cone with height 8 m and slant height 10 m
Radius of cone = √[(l)2 – (h)2]
Radius = √[(10)² – (8)²]
⇒ R = √[100 – 64]
⇒ R = √36 = 6 m
Curve surface area of cone = πrl
⇒ CSA = (22/7) × 6 × 10
⇒ CSA = 188.6 m2
(ii) Circumference of the base of the cone = 88 cm
Slant Height of the cone = 2 dm =20 cm
Circumference of the base of the cone = 2πr, where r is the radius
88 = 2πr
r = 88/2π = 14 cm
Now, we can use the formula for the curved surface area of a cone:
The curved surface area of the cone = πrℓ
= (22\7) × (14)(20)
= 880 cm²
(iii) Area of base = 154 cm²
height = 54 cm
Area of base = πr²
154 = (22/7) r²
r² = (154×7)/22
r = √49 = 7cm
l = √{h²+r²}
l = √{24²+7²}
l = √(576+49)
l = √625 = 25 cm
Curved Surface area = πrl
= (22/7)×7×25
= 550 cm²
Que-4: Find the height of the cone whose base-radius is 5cm and volume 50π cm³.
Sol: Volume of cone = (1/3)×(πr²)×h
⇒ 50π = (1/3)×π×5×5×h
⇒ h = 150/25
⇒ h = 6 cm
Height of the cone = 6 cm.
Que-5: The curved surface area of a right circular cone of radius 11.3 cm is 710 cm². What is the slant height of the cone ?
Sol: Curved surface area = 710 cm2
Radius (r) of base = 11.3 cm
Let Slant height be l .
∴ πrl = 710
⇒ (22/7)×11.3×l = 710
⇒ l = (710×7)/(11.3×22)
⇒ l = 19.99 cm = 20cm
The slant height is 20 cm.
Que-6: If the radius of the base of a circular cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone ?
Sol: Let radius of cone is r and height is h
Volume V1 = (1/3)πr²h
In another case,
Radius of cone = half of radius = r/2
Height =h
∴ Volume = (V2) = (1/3)π(1/2r)²h
= (1/3)π × (r²/4) × h
= (1/12) πr²h
∴ V1/V2 = {(1/12)πr²h}/{(1/3)πr²h} = 3/12 = 1/4
∴ Ratio will be (1 : 4).
Que-7: A conical tent requires 264 m² of canvas. If the slant height is 12m, find the vertical height.
Sol: Curved surface area of the tent = 264 m2
Slant height (I) = 12 m.
⇒ πrl = 264
⇒ (22/7)×r×12 = 264
⇒ r = (264×7)/(22×12)
⇒ r = 7cm
Let h be the vertical height.
We know,
l² = r²+h²
⇒ h = √{l²-r²}
⇒ h = √{12²-7²}
⇒ h = √(144-49) = √95
⇒ h = 9.75 m
Vertical height of cone = 9.75 m
Que-8: The vertical height of a right circular cone is twice its diameter and its volume is 36π cm³. Find the height.
Sol: height = 2 * 2r
height = 4r
volume = (1/3)πr²h
36π = (1/3)πr²4r
36 * (3/4) = r³
r = cube root (3 * 3 * 3)
r = 3 cm
height = 4 *r = 4 * 3 = 12 cm
Que-9: The radius and height of a cone are in the ratio 3:4. If its volume is 301.44 cm³, what is the radius? What is its slant height ?
Sol: Let the radius of cone be 3x cm and the height 4x cm, then
Volume of cone = (1/3)πr2h
⇒ (1/3) π (3x)2 (4x) = 301.44
⇒ x3 = 8
⇒ x = 2
Thus, radius of cone is 6 cm and height 8 cm.
Now,
slant height of cone = (6)²+(8)² = 10 cm.
Que-10: The radius and slant height of a cone are in the ratio of 4:7. If its curved surface area is 792 cm², find its radius.
Sol: Given that,
Curved surface area = πrl = 792
Let the radius (r) = 4x
Height (h) = 7x
Now, CSA = 792
4x(22/7)×4x×7x = 792
⇒ 88x² = 792
⇒ x² = 792/88 = 9
⇒ x = 3
∴ Radius = 4x = 4×3 = 12cm.
Que-11: The base radii of two right circular cones of the same height are i the ratio 3:5. Find the ratio of their volumes.
Sol: Let r1 = 3x; r2 = 5x; and h1 = h2 = h
Volume of 1st cylinder/Volume of 2nd cylinder = {πr²1h1}/{πr²1h2}
= {9x²h}/{25x²h} = 9/25
∴ Required ratio = 9 : 25
Que-12: The circumference of the base of a 10m high conical tent is 44m. Calculate the length of canvas used in making the tent if width of canvas is 2 cm.
Sol: Let r m be the radius of the base, h m be the height and l m be the slant height of the cone. Then,
Circumference = 44 metres
⇒ 2πr = 44
⇒ 2 x (22/7) x r = 44
⇒ r = 7 metres
It is given that h = 10 metres
∴ l2 = r2 + h2
⇒ l = √{r²+h²}
l = √(49+100) = √149 = 12.2m
Now, Surface area of the tent = πrl m2
= (22/7)×7×12.2 m2
= 268.4 m2
∴ Area of the canvas used = 268.4 m2
It is given that the width of the canvas is 2 m.
∴ Length of the canvas used = Area/Width
= 268.42
= 134.2 metres.
Que-13: How many metres of cloth 5m wide will be required to make a conical tent, the radius of whose base is 7m and whose height is 24m ?
Sol: r = 7m, h = 24m, l = ?
l² = r²+h² = 7²+24² = 49+576 = 625
∴ l = √625 = 25m
Area of canvas used = CSA of conical tent
∴ CSA of a cone = πrl = (22/7)×7m×25m = 550 m²
Area of canvas = 550 m²
Length× width = 550m²
∴ length = 550 m²/width = (550m²)/5 = 110m
∴ Length of canvas used = 110m.
Que-14: A conical tent of capacity 1232 m³ stands on a circular base of area 154 m². Find in m² the area of canvas.
Sol: Let l be the slant height of the conical tent, then
= l = √{h²+r²}
∴ l = √{h²+r²} = √{(24)²+(7)²} = √(576+49) = √625 = 25m
The area of the canvas required to make the tent = πrl m²
∴πrl = (22/7)×7×25 = 550 m²
Que-15: A conical tent is to accommodate 11 persons, each person must have 4 m² of space on the ground and 20 m³ of air to breathe. Find the height of the cone.
Sol: Area of the base = 11 x 4 = 44 m2
and Volume of the cone = 11 x 20 = 220 m3
(1/3) x πR2h = 220 m3
Area of the base = πR2
∴ πR2 = 44
∴ R2 = 4422×7
∴ R2 = 14
∴ R = √14
By equation (i) and (ii), we get
(1/3)×(22/7)×√14×√14×h = 220
h = (220×3)/(22×2)
h = 30/2 = 15 cm.
Que-16: Find out whether the following statement is true or false :
The volume of the cone is one-half of the volume of cylinder of the same radius and height.
Sol: False, because the volume of a cone is one-third (1/3) of the volume of cylinder of the same radius and height.
Que-17: The volume of a cone is the same as that of a cylinder whose height is 9cm and diameter 40cm. Find the radius of the base of cone if its height is 108 cm.
Sol: Diameter of the cylinder = 40 cm.
Radius (r) = 40/2 = 20 cm.
Height (h) = 9 cm.
∴ Volume of cylinder = πr2h = π × 20 × 20 × 9 = 3600π cm3.
Height of cone (H) = 108 cm.
Let radius of cone = R.
Volume of cone = (1/3)πR²H
Given, volume of cone = volume of cylinder.
∴ (1/3)πR²H = πr²h
⇒ R² = (3×π×20×20×9)/(π×108)
⇒ R² = 10800/108
⇒ R² = 100
⇒ R = √100
⇒ R = 10 cm.
Hence, the radius of the cone is 10 cm.
Que-18: A canvas tent is in the shape of a cylinder surmounted by a conical roof. The common diameter of a cone and cylinder is 14m. The height of the cylindrical part is 8m and the height of conical roof is 4m. Find the area of the canvas used to make the tent. Give your answer in m² correct to one decimal place.
Sol: Height of the cylindrical part = H = 8 m
Height of the conical part = h = 4 m
Diameter = 14 m ⇒ radius = r = 7 m
Slant height of the cone = I
l = √(r²+h²)
l = √(7²+4²)
l = √65 = 8.06 m
Slant height of cone = 8.06 m
Area of the canvas used = Curved surface area of cylinder + curved surface area of cone
= 2πrH+πrl
= (2×22/7×7×8)+(22/7×7×8.06)
= 352 + 177.32
= 529.32 m2
Que-19: A circus tent is cylindrical to a height 3m and conical above it. If its diameter is 105m and slant height of the cone is 53m, calculate the total area of the canvas required.
Sol: Given diameter = 105 m
Radius = 105/2 m = 52.5 m
∴ Curved surface area of circus tent = πrl + 2πrh
= {(22/7)×52.5×53} + {2×52.5×3×(22/7)}
= 8745 + 990
= 9735 m²
Que-20: Find the volume of the largest circular cone that can be cut out of a cube whose edge is 9 cm.
Sol: The length of each side of the cube is 9 cm. We have to find the volume of the largest right circular cone contained in the cube.
The diameter of the base circle is same as the length of the side of the cube. Thus, the diameter of the base circle of the right circular cone is 9 cm. Therefore, the radius of the base of the right circular cone is r = 4.5cm.
From the right angled triangle ΔAOB we have
h = 9 cm
Therefore, the volume of the solid right circular cone is
V = (1/3) πr²h
= (1/3)×(22/7)×(4.5)²×9
= 190.93
Hence largest volume of cone is = 190.93 cm³.
Que-21: A tent is of the shape of a right circular cylinder up to a height of 3m and then becomes a right circular cone with a maximum height of 13.5 m above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs2 per square metre, if the radius of the base is 14 metres.
Sol: Let r m be the radius of cylindrical base of cylinder of height by m
r = 14 m and h1 = 3m
Curved surface area of cylinder
= 2πrh1 m²
= 25(22/7)×14×3 m²
= 264 m²
The radius of cylindrical box of cylinder is also equal to the radius of right circular cons.
Let h2 be the height of cone and l be the slant height of cone
and r = 14m and h2 = (13.5-3)
= 10.5m
l² = {r²+h²2}
l² = {(14)²+(10.5)²}
l = √{196+110.25}
l = √306.25 = 17.5m
Curved surface area of the cone πrl
= (22/7)×14×17.5
Curved surface of area of cone = πrl
= (22/7)×1417.5
= 770m²
Therefore,
Total area of tent which is to be painted
= curved surface area of cylinder + curved surface area of cone
= (264+770) m²
= 1034 m²
Now cost of painting 1 m2 of inner side of tent = Rs. 2
Cost of painting 1034 m2 inner side of tent
= 2×1034
Rs. 2068
Que-22: A girl fills a cylindrical bucket 32cm in height and 18cm in radius with sand. She empties the bucket on the ground and makes a conical heap of the sand. If the height of the conical heap is 24cm, find (i) the radius and (ii) the slant height of the heap. Give your answer correct to one decimal place.
Sol: Height of the cylindrical bucket, h₁ = 32 cm
Radius of the cylindrical bucket, r₁ = 18 cm
Height the of conical heap, h = 24 cm
(i) volume the conical heap of sand = volume of sand in the cylindrical bucket
1/3 πr2h = πr₁2h₁
r2 = 3r₁2h₁ / h
r2 = [3 × (18 cm)2 × 32 cm] /24 cm
r2 = (18 cm)2 × 4
r = 18 cm × 2 = 36 cm
(ii) Slant height, l = √r2 + h2
= √(36 cm)2 + (24 cm)2
= √1296 cm2 + 576 cm2
= √1872 cm2
= 12√13 cm
= 43.3 cm.
Que-23: From a solid cylinder whose height is 8cm and radius is 6cm, a conical cavity of height 8cm and of base radius 6cm is hollowed out. Find the volume of the remaining solid correct to 4 places of decimals.
Sol: Volume of the solid left = Volume of cylinder – Volume of cone
= πr²h – (1/3) πr²h
= (2/3)×(22/7)×8×6×6 = 603.428 cm³
The slant length of the cone ,l = √{r²+h²} = √(36+64) = 10cm
Total surface area of final solid = Area of base circle + Curved surface area of cylinder + curved area of once
= πrl²+2πrh+πrl = πr(r+2h+l)
= (22/7)×6×(6+16+10)
= 603.42 cm².
Que-24: A metallic cylinder has radius 3cm and height 5cm. It is made of a metal A. To reduce its weight, a conical hole is drilled in the cylinder as shown in figure and it is completely filled with a lighter metal B. The conical hole has a radius of (3/2)cm and is depth is (8/9)cm. Calculate the ratio of the volume of the metal A to the volume of the metal B in the solid.
Sol: Given:
Radius of the cylinder, R = 3 cm
Height of the cylinder, H = 5 cm
∴ Volume of the cylinder = πR²H
= π×(3)²×(5)
= 45π cm³
Radius of the cone, r = 3/2 cm
Height of the cone, h = 8/9 cm
∴ Volume of the cone removed from the cylinder = (1/3)πr²h
= (1/3)×π×(3/2)²×(8/9)
= 2π/3 cm³
According to the question,
Volume of the metal left in the cylinder = Volume of the cylinder − Volume of the cone
= 45π − (2π/3)
=(45−2/3)π
Volume of metal left in the cylinder/Volume of metal taken out in conical shape
= {(133π/3)/{(2π/3)
= 133/2
Thus, the ratio of the volume of the metal left in the cylinder to the volume of the metal taken out in conical shape is 133 : 2
Que-25: An open cylindrical vessel of internal diameter 7cm and height 8cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is (7/2)cm and height 8cm. Find the volume of the water required to fill the vessel. If the cone is replaced by another cone, whose height is {1*(3/4)}cm and the radius of whose base is 2cm, find the drop in the water vessel.
Sol: Diameter of the base of the cylinder = 7 cm
Therefore, radius of the cylinder = 7/2 cm
Volume of the cylinder = πr2h
= (22/7)×(7/2)×(7/2)×8
= 308 cm3
Diameter of the base of the cone = 7/2cm
Therefore, radius of the cone = 7/4cm
Volume of the cone = (1/3) πr²h
= (1/3)×(22/7)×(7/4)×(7/4)×8
= 773 cm³
On placing the cone into the cylindrical vessel, the volume of the remaining portion where the water is to be filled
= 308 – (77/3)
= (924-77)/3
= 847/3
= 282.33 cm3
Height of new cone = 1*(3/4) = 7/4 cm
Radius = 2 cm
Therefore, volume of new cone
= (1/3) πr²h
= (1/3)×(22/7)×2×2×(7/4)
= 223 cm³
Volume of water which comes down = (77/3) – (22/3) cm3 = 55/3 cm3 …(1)
Let h be the height of water which is dropped down.
Radius = 7/2cm
∴ Volume = πr²h
V = (22/7)×(7/2)×(7/2)×h …(2)
From (1) and (2)
(77/2) h = 55/3
⇒ h = (55/3) × (2/77)
⇒ h = 10/21
Drop in water level = 10/21 cm
–: End of Three Dimensional Solids Class 10 OP Malhotra Exe-15C ICSE Maths Ch-15 :–
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