What is the Area of a Right Circular Cone

What is the Area of a Right Circular Cone

If r, h and ℓ denote respectively the radius of base, height and slant height of a right circular cone, then-

1. ℓ ² = r² + h²
2. Area of base = πr²
3. Curved (lateral) surface area = πrℓ
4. Total surface area = πr (ℓ + r)
5. Volume = \(\frac{1}{3}\) πr²h

Read more about Radius of a Right Circular Cylinder

Right Circular Cone Example Problems with Solutions

Example 1:     The height of a cone is 48 cm and the radius of its base is 36 cm. Find the curved surface
area and the total surface area of the cone.
(Take π = 3.14).
Solution:     Given : h = 48 cm and r = 36 cm.
∴    ℓ² = h² + r²
⇒   ℓ² = 48² + 36² = 2304 + 1296 = 3600
⇒   ℓ = \(\sqrt {3600} \,\,cm\) = 60 cm
∴    The curved surface area = πrℓ
= 3.14 × 36 × 60 cm² = 6782.4 cm²
And, the total surface area of the cone
= πrℓ  + πr² = πr (ℓ + r)
= 3.14 × 36 × (60 + 36)cm²
= 10851.84 cm²

Example 2:     Curved surface area of a cone is 2200 cm².
It its slant height is 50 cm, find :
(i)   radius of the base.
(ii)  total surface area.
(iii) height of the cone.
Solution:     (i)   Given : πrℓ = 2200 cm² and ℓ = 50 cm
⇒  \(\frac{{22}}{7} \times r \times 50\) = 2200
i.e., r = \(\frac{{2200 \times 7}}{{22 \times 50}}cm\) = 14 cm
(ii)  Total surface area = πrℓ (ℓ + r)
= \(\frac{{22}}{7}cm \times 14\) × (50 + 14) cm²
= 2816 cm²       Ans.
(iii)  ℓ²  = h² + r²   ⇒   h² = ℓ² – r²
= 50² – 14² = 2500 – 196 = 2304
∴    h = \(\sqrt {2304}\) cm = 48 cm

Example 3:     A conical tent is 10 m high and radius of its base is 24 m. Find
(i)   slant height of the  tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs. 70.
Solution:     (i)   Given : h = 10 m and r = 24 m
∴    ℓ² = h² + r²
⇒   ℓ² = 10² + 24²  = 100 + 576 = 676
⇒   ℓ = \(\sqrt {676}\) m = 26 m
(ii) Area of canvas required
= Curved surface area of the tent
= πrℓ  = \(\frac{{22}}{7} \times 24 \times 26m\) = \(\frac{{13728}}{7}{m^2}\)
∵    Cost of 1 m² canvas is Rs. 70
∴    Total cost of canvas required
= \(\frac{{13728}}{7} \times Rs.70\)
= Rs. 1,37,280

Example 4:     What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? 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).
Solution:     For the tent : h = 8 m and r = 6 m
ℓ²  = h² + r²  ⇒   ℓ² = 8² + 6²
= 64 + 36 = 100 and   ℓ = \(\sqrt {100} \,\,m\) = 10m
Curved surface area of the tent = πrℓ
= 3.14 × 6 × 10 m²  = 188.4 m²
⇒  Area of tarpaulin used = 188.4 m²
⇒  Length of tarpaulin × its width = 188.4 m²
⇒  Length of tarpaulin × 3 m = 188.4 m²
⇒  Length of tarpaulin = \(\frac{{188.4}}{3}m = 62.8m\)
∵    Extra length of tarpaulin required
= 20 cm = 0.2 m
∴    Total length of tarpaulin required
= 62.8 m + 0.2 m  = 63 m

Example 5:     A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m², what will be the cost of painting all these cones?
(Use π = 3.14 and take \(\sqrt {1.04}\) = 1.02)
Solution:     For each cone :
r = \(\frac{{40}}{2}\) cm = 20 cm = 0.2 m and h = 1 m
∴    ℓ² = h² + r² ⇒   ℓ² = (1)² + (0.2)²
= 1 + 0.04 = 1.04
⇒   ℓ = \(\sqrt {1.04}\)m = 1.02 m
C.S.A of each cone = πrℓ
= 3.14 × 0.2 × 1.02 m² = 0.64056 m²
⇒   C.S.A of 50 cones = 50 × 0.64056 m²
= 30.028 m² = Area to be painted
∵ The cost of painting is Rs. 12 per m²
∴ The total cost of painting outer sides of 50 cones
= 30.028 × Rs. 12 = Rs. 384.34

Example 6:     The radius and the slant height of a cone are in the ratio 3 : 5. If its curved surface area is 2310 cm2, find its height.
Solution:     Given : r : ℓ = 3 : 5
⇒   if r = 3x cm, ℓ = 5x cm
C.S.A. = πrℓ   ⇒  \(\frac{22}{7}\) × 3x × 5x = 2310
⇒   x² = \(\frac{{2310 \times 7}}{{22 \times 3 \times 5}} = 49\) ⇒    x = 7
∴    r = 3x = 3 × 7 cm = 21 cm
and ℓ = 5x = 5 × 7 cm = 35 cm,
ℓ² = h² + r² ⇒  h² = 35² –21²
= 1225 – 441 = 784
∴    Height (h) = \(\sqrt {784} cm\) = 28 cm

Example 7:     A circus tent is in the shape of a cylinder, upto a height of 8 m, surmounted by a cone of the same radius 28 m. If the total height of the tent is 13 m, find:
(i)   total inner curved surface area of the tent.
(ii)  cost of painting its inner surface at the rate of Rs. 3.50 per m².
Solution:     According to the given statement, the rough sketch of the circus tent will be as shown:
(i)   For the cylindrical portion :
r = 28 and h = 8 m
∴    Curved surface area = 2πrh
= 2 × \(\frac{22}{7}\) × 28 × 8 m² = 1408 m²
For conical portion :
r = 28 m and h = 13 m – 8 m = 5 m
∴    ℓ² = h² + r² ⇒   ℓ² = 5² + 28² = 809
⇒   ℓ = \(\sqrt {809}\)m = 28.4 m
∴    Curved surface area = πrℓ
=  \(\frac{22}{7}\) × 28 × 28.4 m²  = 2499.2 m²
∴    Total inner curved surface area of the tent.
=    C.S.A. of cylindrical portion + C.S.A. of the conical portion
1408 m² + 2499.2 m² = 3907.2 m²
(ii)  Cost of painting the inner surface
= 3907.2 × Rs. 3.50 = Rs. 13675.20

Example 8:     The height of a cone is 30 cm and its volume is 3140 cm³. Taking π = 3.14, find :
(i) radius of the base.
(ii) area of the base.
Solution:     (i)   Given : h = 30 cm and volume = 3140 cm³
Volume = \(\frac{1}{3}\pi {r^2}h\)
⇒   3140 = \(\frac{1}{3} \times 3.14\) × r² × 30
⇒   r² = \(\frac{{3140 \times 3}}{{3.14 \times 30}} = 100\) and r = 10cm
(ii)  Area of the base = πr²
= 3.14 × 10² cm² = 314 cm²
Alternative Method :
\(\frac{1}{3}\) × area of base × height = volume
⇒   \(\frac{1}{3}\) × area of base × 30 = 3140
⇒  Area of base = \(\frac{{3140 \times 3}}{{30}}c{m^2}\)
= 314 cm²

Example 9:     A right triangle ABC has sides 5 cm, 12 cm and 13 cm. Find the :
(i)   volume of solid obtained by revolving ∆ ABC about the side 12 cm.
(ii)  volume of solid obtained by revolving ∆ ABC about side 5 cm.
(iii) difference between the volumes of the solids obtained in step (i) and step (ii).
Solution:     ∵    5² + 12² = 13² ⇒  Angle opp. to 13 cm is right angle
(i)   When the ∆ is revolved about the side of
12 cm, for the cone formed :
h = 12 cm and r = 5.
∴    Volume of solid obtained = \(\frac{1}{3}\pi {r^2}h\)
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5 × 5 × 12 cm³
= 314.29 cm³
(ii)  When the ∆ is revolved about the side of 5 cm, for the cone formed :
h = 5 cm, and r = 12 cm.
∴    Volume of solid obtained = \(\frac{1}{3}\pi {r^2}h\)
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 12 × 12 × 5 cm³ = 754.29 cm³
(iii) Required difference
= 754.29 cm³ – 314.29 cm³ = 440 cm³

Example 10:     The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find:
(i) height of the cone.
(ii) slant height of the cone
(iii) curved surface area of the cone.
Solution:     Given : volume of cone = 9856 cm³
and radius (r) = \(\frac{28}{2}\) cm = 14cm
(i)   Volume = \(\frac{1}{3}\)πr²h   ⇒  9856 = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 14 × 14 × h
⇒     h = \(\frac{{9856 \times 3 \times 7}}{{22 \times 14 \times 14}}cm\) = 48 cm
(ii)   ℓ²  = h² + r²  ⇒   ℓ² = 48² + 14²
= 2304 + 196 = 2500
⇒   ℓ = \(\sqrt {2500}\)cm = 50 cm
∴    Slant height = 50 cm
(iii) Curved surface area = πrℓ
= \(\frac{22}{7}\) × 14 × 50 cm² = 2200 cm²

Example 11:     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.
Solution:     For the conical heap :
Radius (r) = \(\frac{{10.5}}{2}m\) = 5.25 m
and height (h) = 3m.
∴    Volume = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{{22}}{7} \times 5.25\) × 3 m³ = 86.625 m³
Now, ℓ² = h² + r²  ⇒  ℓ² = (3)² + (5.25)²
= 9 + 27.5625 = 36.5625
⇒   ℓ = \(\sqrt {36.5625}\)m = 6.047 m
∴    Area of canvas required
= curved surface area of the conical heap
= πrℓ = \(\frac{{22}}{7} \times 5.25 \times 6.047{m^2}\) = 99.7755 m²

Example 12:     A cylinder and a cone have same base area. But the volume of cylinder is twice the volume of cone. Find the ratio between their heights.
Solution:     Since, the base areas of the cylinder and the cone are the same.
⇒   their radius are equal (same).
Let the radius of their base be r and their heights be h₁ and h₂ respectively.
Clearly, volume of the cylinder = \(\frac{1}{3}\)πr²h₁
and, volume of the cone = \(\frac{1}{3}\)πr²h₂
Given :
Volume of cylinder = 2 × \(\frac{1}{3}\) volume of cone
⇒   πr²h₁ = 2 × πr²h₂
⇒   h₁ = \(\frac{2}{3}\)h₂
⇒ \(\frac{{{h_1}}}{{{h_2}}} = \frac{2}{3}\)
i.e., h₁ : h₂ = 2 : 3