**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-

- ℓ
^{ 2}= r^{2}+ h^{2} - Area of base = πr
^{2} - Curved (lateral) surface area = πrℓ
- Total surface area = πr (ℓ + r)
- Volume = πr
^{2}h

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.

∴ ℓ^{2} = h^{2} + r^{2}

⇒ ℓ^{2} = 48^{2} + 36^{2} = 2304 + 1296 = 3600

⇒ ℓ = = 60 cm

∴ The curved surface area = πrℓ

= 3.14 × 36 × 60 cm^{2} = 6782.4 cm^{2}

And, the total surface area of the cone

= πrℓ + πr^{2} = πr (ℓ + r)

= 3.14 × 36 × (60 + 36)cm^{2}

= 10851.84 cm^{2}

**Example 2: **Curved surface area of a cone is 2200 cm^{2}.

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 cm^{2} and ℓ = 50 cm

⇒ = 2200

i.e., r = = 14 cm

(ii) Total surface area = πrℓ (ℓ + r)

= × (50 + 14) cm^{2}

= 2816 cm^{2} Ans.

(iii) ℓ^{2} = h^{2} + r^{2} ⇒ h^{2} = ℓ^{2} – r^{2}

= 50^{2} – 14^{2} = 2500 – 196 = 2304

∴ h = 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 m^{2} canvas is Rs. 70.

**Solution: **(i) Given : h = 10 m and r = 24 m

∴ ℓ^{2} = h^{2} + r^{2}

⇒ ℓ^{2} = 10^{2} + 24^{2} = 100 + 576 = 676

⇒ ℓ = m = 26 m

(ii) Area of canvas required

= Curved surface area of the tent

= πrℓ = =

∵ Cost of 1 m^{2} canvas is Rs. 70

∴ Total cost of canvas required

=

= 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

ℓ^{2} = h^{2} + r^{2} ⇒ ℓ^{2} = 8^{2} + 6^{2}

= 64 + 36 = 100 and ℓ = = 10m

Curved surface area of the tent = πrℓ

= 3.14 × 6 × 10 m^{2} = 188.4 m^{2}

⇒ Area of tarpaulin used = 188.4 m^{2}

⇒ Length of tarpaulin × its width = 188.4 m^{2}

⇒ Length of tarpaulin × 3 m = 188.4 m^{2}

⇒ Length of tarpaulin =

∵ 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 m^{2}, what will be the cost of painting all these cones?

(Use π = 3.14 and take = 1.02)

**Solution: **For each cone :

r = cm = 20 cm = 0.2 m and h = 1 m

∴ ℓ^{2} = h^{2} + r^{2} ⇒ ℓ^{2} = (1)^{2} + (0.2)^{2}

= 1 + 0.04 = 1.04

⇒ ℓ = m = 1.02 m

C.S.A of each cone = πrℓ

= 3.14 × 0.2 × 1.02 m^{2} = 0.64056 m^{2}

⇒ C.S.A of 50 cones = 50 × 0.64056 m^{2}

= 30.028 m^{2} = Area to be painted

∵ The cost of painting is Rs. 12 per m^{2}

∴ 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 ⇒ × 3x × 5x = 2310

⇒ x^{2} = ⇒ x = 7

∴ r = 3x = 3 × 7 cm = 21 cm

and ℓ = 5x = 5 × 7 cm = 35 cm,

ℓ^{2} = h^{2} + r^{2} ⇒ h^{2} = 35^{2} –21^{2}

= 1225 – 441 = 784

∴ Height (h) = = 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 m^{2}.

**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 × × 28 × 8 m^{2} = 1408 m^{2}

For conical portion :

r = 28 m and h = 13 m – 8 m = 5 m

∴ ℓ^{2} = h^{2} + r^{2} ⇒ ℓ^{2} = 5^{2} + 28^{2} = 809

⇒ ℓ = m = 28.4 m

∴ Curved surface area = πrℓ

= × 28 × 28.4 m^{2} = 2499.2 m^{2}

∴ Total inner curved surface area of the tent.

= C.S.A. of cylindrical portion + C.S.A. of the conical portion

1408 m^{2} + 2499.2 m^{2} = 3907.2 m^{2}

(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 cm^{3}. Taking π = 3.14, find :

(i) radius of the base.

(ii) area of the base.

**Solution: **(i) Given : h = 30 cm and volume = 3140 cm^{3}

Volume =

⇒ 3140 = × r^{2} × 30

⇒ r^{2} = and r = 10cm

(ii) Area of the base = πr^{2}

= 3.14 × 10^{2} cm^{2} = 314 cm^{2}

Alternative Method :

× area of base × height = volume

⇒ × area of base × 30 = 3140

⇒ Area of base =

= 314 cm^{2}

**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: **∵ 5^{2} + 12^{2} = 13^{2} ⇒ 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 =

= × × 5 × 5 × 12 cm^{3}

= 314.29 cm^{3}

(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 =

= × × 12 × 12 × 5 cm^{3} = 754.29 cm^{3}

(iii) Required difference

= 754.29 cm^{3} – 314.29 cm^{3} = 440 cm^{3}

**Example 10: **The volume of a right circular cone is 9856 cm^{3}. 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 cm^{3}

and radius (r) = cm = 14cm

(i) Volume = πr^{2}h ⇒ 9856 = × × 14 × 14 × h

⇒ h = = 48 cm

(ii) ℓ^{2} = h^{2} + r^{2 }⇒ ℓ^{2} = 48^{2} + 14^{2}

= 2304 + 196 = 2500

⇒ ℓ = cm = 50 cm

∴ Slant height = 50 cm

(iii) Curved surface area = πrℓ

= × 14 × 50 cm^{2} = 2200 cm^{2}

**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) = = 5.25 m

and height (h) = 3m.

∴ Volume = πr^{2}h

= × 3 m^{3} = 86.625 m^{3}

Now, ℓ^{2} = h^{2} + r^{2} ⇒ ℓ^{2} = (3)^{2} + (5.25)^{2}

= 9 + 27.5625 = 36.5625

⇒ ℓ = m = 6.047 m

∴ Area of canvas required

= curved surface area of the conical heap

= πrℓ = = 99.7755 m^{2}

**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 h_{1} and h_{2} respectively.

Clearly, volume of the cylinder = πr^{2}h_{1}

and, volume of the cone = πr^{2}h_{2}

Given :

Volume of cylinder = 2 × volume of cone

⇒ πr^{2}h_{1} = 2 × πr^{2}h_{2}

⇒ h_{1} = h_{2}

⇒

i.e., h_{1} : h_{2} = 2 : 3

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