A semi circular sheet of diameter 35cm is bent into an open conical cup . Find the depth and capacity of cup.
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Dear Student,

Please find below the solution to the asked query :

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 = 35 cm

⇒ r = 17.5 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 = 17.5 cm

Circumference of base of cone = 2πR

∴ 2π R  = 17.5π cm

⇒ 2R = 17.5 cm

⇒ R =8.75 cm

Slant height of the cone, L = 17.5 cm
 

$$\begin{gathered} \sqrt{{\left(8.75\right)}^2+{\mathrm{H}}^2}= \left(17.5\right) \left[\mathrm{Using} \mathrm{thge} \mathrm{formula} \mathrm{L}\hspace{0.17em}= \sqrt{{\mathrm{R}}^2+{\mathrm{H}}^2}\right] \\ \mathrm{From} \mathrm{here} \mathrm{we} \mathrm{get} \\ \mathrm{H} = 15.15 \mathrm{cm} \\ \mathrm{Now}, \mathrm{capacity} \mathrm{or} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{conical} \mathrm{cup} \\ = \frac{1}{3}{\mathrm{\pi R}}^2\mathrm{H} \\ = \frac{1}{3}\times \frac{22}{7}\times 8.75\times 8.75\times 15.15 \\ =1215.15 {\mathrm{cm}}^3 \end{gathered}$$

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