a tent is in the form of cylinder of diameter 4.3 metre had height 3.8 m surmounted by a cone whose Vertical angle is a right angle find the surface area and volume of the tent
Answers
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
HOPE IT HELPS!!!
PLS MARK IT AS THE BRAINLIEST