Lalit Mehra , Meritnation Expert added an answer, on 7/3/12

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, 2*r* = 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 = 2*πR*

∴ 2*π R* = 14π cm

⇒ 2*R* = 14 cm

⇒ *R* = 7 cm

Slant height of the cone, *L* = 14 cm

⇒ 49 cm^{2} + *H * ^{2} = (14 cm)^{2} = 196 cm^{2}

⇒ *H * ^{2} = 196 cm^{2} – 49 cm^{2} = 147 cm^{2}

Capacity or volume of the conical cup =

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