Question

A tent is in form of a right circular cylinder surmounted by a cone.The diameter of the cylinder is 24 m.The height of the cylindrical portion is 11 m, while the vertex of the cone is 16 m above the ground.Find the area of the canvas required making for the tent.(Use $(\pi=\frac{22}{7})$)

Answer 1

Answer:

The area of the Canvas required for the tent is 1320 cm² .

Step-by-step explanation:

SOLUTION :  

Given :  

Diameter of a right  circular cylinder = 24 m  

Radius of a right circular cylinder, r = 24/2 = 12 cm

Height of the cylindrical portion , H = 11 m

Height of a tent = 16 m

Height of a cone, h  = 16 - 11 = 5 m

Radius of a cone = Radius of a cylinder = 12 cm

Slant height of a cone, l = √r² + h²

l = √12² + 5² = √144 + 25 = √169  

l =  √169

l = 13 cm  

Slant height of a cone, l = 13 cm  

Surface area of the tent , S = curved surface area of a cylinder + curved surface area of a cone

S = 2πrH + πrl

S = πr(2H + l)

S = 22/7 × 12(2 × 11 + 13)

S =  22/7 × 12(22 + 13)

S = 22/7 × 12 × 35

S = 22 × 12 × 5  

S = 1320 cm²

Hence, the area of the Canvas required for the tent is 1320 cm² .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

$hello \: mate$

Area required= Curved surface area of right circular cylinder+Curved surface area of cone

➡ Height of cone=16-11=5 m

➡ Height of cylinder=11 m

➡Radius=12 m

Slant height, l²=r²+h²

l²=(12)²+(5)²

l²=144+25

l²=169

l=13 m

Now, Area of canvas=CSA of cone+CSA of cylinder

=>πrl+2πrh

➡πr(l+rh)

➡22/7×12(13+2×11)

➡22/7×12(13+22)

➡22/7×12×35

➡22×12×5

➡1320 m²