Question

A semicircular sheet of metal diameter 28cm is bent into an open conical cup. Find the depth and capacity of the cup.

Answer 1

Answer:

Yes, you use the formula for a cone volume to find the capacity.

depth = √147≈12.1cm

capacity ≈622cm3

Explanation:

C=2⋅π⋅r and V=13⋅π⋅r22⋅h

Diameter=28cm,→r1=14cm

From the semicircular piece of metal we first find the circumference of the base of the cone, which is the same as ½ of the full circle,
C=2⋅π⋅r12
C=2⋅π⋅142=14π≈44cm

Now find our cone radius from the cone circumference.

C=2⋅π⋅r2→ r2=(C2⋅π)

r2=14π2⋅π=7

From Pythagoras, the equation for a right triangle

r21=r22+h2 we obtain:

h=√r21–r22 → h=√196–49

h=√147(≈12.1cm  this is the depth of the cone cup)

V=13⋅π⋅r22⋅h

V=13⋅π⋅49⋅√147

V=622cm3 volume capacity

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Answer 2

 diameter =d = 28 cm
Radius= r = 14 cm
Circumference of semi circle = πr
= (22/7) x 14 
= 44 cm

so the circumference of base of cone
2πR = 44
R = 44 x (7/44)
R = 7 cm

The radius of semi circular sheet = slant height of conical cup
which is 
 l = 7 cm
As,
 r2 + h2 = l2

142 + h2 = 72

196 – 49 = h2 

h2 = 147

hence; h = 7√3 cm

depth of the conical cup= 7√3 cm

Capacity of cup = (1/3) πr2h

= (1/3) x (22/7) x 72 x 7√3

= 622.37 cu cm approx