A semicircular sheet of metal os diameter 28cm is bent into an open conical cup The depth of the cup - Maths - Surface Areas and Volumes

Question

A semicircular sheet of metal os diameter 28cm is bent into an
open conical cup The depth of the cup is

Lalit Mehra · Accepted 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 = 2.8 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 = πr

∴ 2πr = πr = 14π cm

⇒ 2R = 14 cm

⇒ R = 7 cm

Slant height of the cone, L = 14 cm

$∴ \sqrt{(7 cm)^{2} + H^{2}} = 14 cm [∵ L = \sqrt{R^{2} + H^{2}}]$

⇒ 49 cm² + H ² = (14 cm)² = 196 cm²

⇒ H ² = 196 cm² – 49 cm² = 147 cm²

$\Rightarrow H = \sqrt{147 {cm}^{2}} = 7 \sqrt{3} cm$

Thus, depth of the conical cup is $7 \sqrt{3} cm$ .

A semicircular sheet of metal os diameter 28cm is bent into an open conical cup.The depth of the cup is..........