Question

the interior of a building is in the form of cylinder of a diameter 4.3m and 3.8m high surmounted by cone having same base whose Vertical angle is right angle find the area of surface and volume of a building

Answer 1

Solution :-

Diameter of the cylindrical portion = 4.3 m

Radius = 4.3/2 = 2.15 m

Height = 3.8 m

Lateral Surface Area of cylindrical portion = 2πeh

⇒ 2*22/7*2.15*3.8

= 359.48/7

= 51.3543 m²

It is mentioned that vertical angle of the cone is a right angle.

Let BAC be the triangle.

In Δ BAC,

Slant height 'l' = AB = AC

⇒ l² + l² = (BC)²  (BC = diameter of the common base of cone and cylinder)

⇒ 2l² = (4.3)²

⇒ 2l² = 18.49

⇒ l² = 18.49/2

⇒ l² = 9.245

⇒ l = 3.04 m

So, slant height is 3.04 m

l² = r² + h²
 
(3.04)² = √(2.15)² + h²

9.2416 = 4.6225 + h²

h² = 9.2416 - 4.6225

h² = 4.6191

h = √4.6191

h = 2.149  or 2.15 m (Approx) 

Lateral surface area of conical portion = πrl

22/7*2.15*3.04

143.792/7

= 20.5417 m²

Total surface area of the building = Surface area of cylindrical portion + Surface area of the conical portion

⇒ 51.3543 + 20.5417

= 71.896 m² 

Volume of Cylindrical portion = πr²h

22/7*2.15*2.15*3.8

386.441/7

= 55.2059 m³

Volume of the conical portion = 1/3πr²h

1/3*22/7*2.15*2.15*2.15

218.64425/21

= 10.4116 m³

Total volume of the building = Volume of the cylindrical portion + Volume of the conical portion

55.2059 + 10.4116 

= 65.6175 m³

Answer.

Answer 2

Answer..

CSA -61.776

VOLUME- 52.381

THE EXPLANATION IS IN ATTACHMENT