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a semi circular sheet of metal of diameter 28 cm is bent to form a cone. find the capacity of the cone.

Asked by Sakshi Singh(student) , on 6/3/12


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EXPERT ANSWER

Let r cm and R cm be the radius of the semi-circular sheet and base of the conical cup respectively.

Suppose the depth of the conical cup is H cm.

Given, 2r = 28 cm

r = 14 cm

When the semi-circular sheet of metal is bent into an open conical cup, then 

Slant height of the cone, L = Radius of the semi-circular sheet = 14 cm

Circumference of base of cone = 2πR

∴ 2π R  = 14π cm

⇒ 2R = 14 cm

R = 7 cm

Slant height of the cone, L = 14 cm

⇒ 49 cm2 + H 2 = (14 cm)2 = 196 cm2

H 2 = 196 cm2 – 49 cm2 = 147 cm2

Capacity or volume of the conical cup =

Posted by Lalit Mehra(MeritNation Expert)on 7/3/12

This conversation is already closed by Expert

More Answers

 Let   r   cm and   R   cm be the radius of the semi-circular sheet and base of the conical cup respectively.

Suppose the depth of the conical cup is H cm.

Given, 2r = 2.8 cm

⇒ = 14 cm

When the semi-circular sheet of metal is bent into an open conical cup, then

Slant height of the cone, L = Radius of the semi-circular sheet = 14 cm

Circumference of base of cone = πr

∴ 2πr = πr = 14π cm

⇒ 2R = 14 cm

⇒ R = 7 cm

Slant height of the cone, L = 14 cm

⇒ 49 cm2 +  2 = (14 cm)2 = 196 cm2

⇒  2 = 196 cm2 – 49 cm2 = 147 cm2

Capacity or volume of the conical cup = 

 

:)

Posted by Crazzy(student)on 6/3/12

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