A semi-circular sheet of metal of diameter 28 cm is bent into an open conical cup. The depth of the cup is approximately

Solution (By Examveda Team)

Radius of semi-circular sheet = r ⇒ $$\frac{{28}}{2}$$
r = 14 cm
Circumference of sheet = πr = 14π cm
Sheet is folded to form a cone
Let radius of cone = r₁

∴ The circumference of base of cone ⇒ Circumference of sheet
∴ 2πr₁ = 14π
r₁ = 7 cm
∴ Radius of cone = 7 cm
Slant height = Radius of semi-circular sheet
r = 14 cm
$$\eqalign{ & \therefore {\text{Height}} = \sqrt {{{\left( {14} \right)}^2} - {{\left( 7 \right)}^2}} \cr & = \sqrt {147} \cr & = 12{\text{ cm}}\,\,\,\left( {{\text{approx}}} \right) \cr} $$