Question

a semi circle of metal sheet with diameter 28cm is bend to form an open cone .find he cone's depth and volume?

Answer 1

When the semi-circular sheet is bent into an open cone, the radius of the sheet becomes slant height of the cone  and circumference of the sheet becomes circumference of the base of the cone.therefore, l = slant height of the cone = 14 cm.Let r cm be the radius and h cm the height(depth) of the cone. Then,circumference of the base of the cone = circumference of the sheet

 => 2πr = π 14 
r= 7 
now , l² = r²+h²
h=√l² - r²
h = √ 14² - 7²
h = 7√3 cm 
h= 12.12 cm = depth of cone
now vol / capacity of cone = 1/3πr²h cm³
1/3 × 22/7 × 7 × 7 × 12.12
 = 622.26 cm³

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Answer 2

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
_______
\/(7)2 +h2 =14 cm

⇒ 49 cm2 + H 2 = (14 cm)2 = 196 cm2

⇒ H 2 = 196 cm2 – 49 cm2 = 147 cm2
_______ __
H = \/ 147 cm2 = 7 \/ 3

Capacity or volume of the conical cup =
1/3 × 22/7 ×R2 ×H _
1/3 ×22/7 × ( 7)2 × 7 \/3
_
1078 \|3/3