Question

A circus tent is in the form of a right circular cylinder surmounted by a cone as shown in the figure. The radius of the cylinder is 120 m and has a height of 33 m. The vertex of the cone is 55 m above the ground. Find the cost of canvasing the tent at rs 10 per square metre​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

A circus tent is in the form of a right circular cylinder surmounted by a cone as shown in the figure. The radius of the cylinder is 120 m and has height of 33 m. The vertex of the cone is 55 m above the ground.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The cost of canvasing the tent at Rs.10/m².

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We know that formula of the area of canvas of the circus tent :

$\bf{\boxed{\bf{Area\:of\:canvas=C.S.A\:of\:cylinder+C.S.A\:of\:cone}}}}}$

A/q

$\underline{\underline{\red{\bf{Curved\:surface\:area\:of\:cylinder\::}}}}$

$\longrightarrow\sf{2\pi rh}\\\\\longrightarrow\sf{2\pi \times 120\times 33}\\\\\longrightarrow\sf{\green{7920\pi m^{2} }}$

$\underline{\underline{\red{\bf{Curved\:surface\:area\:of\:cone\::}}}}$

$\longrightarrow\sf{\pi rl}$

We know that formula of the slant height (l) of cone :

$\bf{\boxed{\bf{l=\sqrt{(r)^{2} +(h)^{2}}}}}}$

$\therefore\sf{l=\sqrt{(120)^{2}+(22)^{2} } }\\\\\sf{l=\sqrt{14400+484}}\\\\\sf{l=\sqrt{14884} m^{2} }\\\\\bf{l=122\:m}}$

Now;

$\longrightarrow\sf{\pi \times 120\times 122}\\\\\longrightarrow\sf{\green{14640\pi \:m^{2} }}$

Total area of canvas = C.S.A of cylinder + C.S.A of cone

Total area of canvas = 7920π + 14640π

Total area of canvas = 22560 π

Total area of canvas = 22560 × 22/7

Total area of canvas = (496320/7 )m²

Total area of canvas = 70902.85 m² .

$\dag\:\underline{\underline{\red{\bf{The\:cost\:of\:canvasing\:circus\:tent\::}}}}$

$\longrightarrow\sf{1m^{2} =Rs.10}\\\\\longrightarrow\sf{70902.85m^{2}=Rs.(70902.85\times 10)}\\ \\\longrightarrow\sf{\green{Rs.709028.5}}$

Thus;

Total cost of canvasing tent is Rs.709028.5 .

Answer 2

✰✰|| Given ||✰✰

• The radius of the cylinder is 120 m

• height of cylinder = 33 m

• length between vertex of cone and ground is 55m

• cost of canvasing the tent at rs 10 per square metre

✰✰|| To Find ||✰✰

• slanting height of cone

• surface area of cone

• surface area of cylinder

• area of canvas

• cost of canvasing

✪|| Solution ||✪

Let, the slanting height of cone be I

I = $\bold{\sqrt{ {r}^{2} + {h}^{2}}}$

I = $\bold{\sqrt{ 120m × 120m + (55 - 33m) × (55 - 33m)}}$

I = $\bold{\sqrt{ {14400m}^{2} + 22m × 22m}}$

I = $\bold{\sqrt{ {14400m}^{2} + {484m}^{2}}}$

I = $\bold{\sqrt{ {14884m}^{2}}}$

I = $\bold{\sqrt{ 122 × 122 × m × m}}$

I = 122m

$<font color = green>$

surface area of cone = πrl

→ surface area of cone = 22/7 × 120m × 122m

→ surface area of cone = 22/7 × 14,640m²

→ surface area of cone = 322080/7m²

→ surface area of cone = 46011.42m²

surface area of cylinder = 2πrh

⇒ surface area of cylinder = 2 × 22/7 × 120m × 33m

⇒ surface area of cylinder = 44/7 × 120m × 33m

⇒ surface area of cylinder = 44/7 × 3960m²

⇒ surface area of cylinder = 174240m²/7

⇒ surface area of Cylinder = 24891.42m²

Area of canvas = CSA of cone + CSA of cylinder

↬ Area of canvas = 46011.42m² + 24891.42m²

↬ Area of canvas = 46011.42m² + 24891.42m²

↬ Area of canvas = 70,902.84m²

cost of canvasing = 70,902.84 × rs 10

↪ cost of canvasing = 709028.4

Cost of canvasing is 709028.4