C = 2*pi*r " and "V = 1/3 * pi * r_2^2* hC=2⋅π⋅r and V=13⋅π⋅r22⋅h
Diameter = 28 cm , rarr r_1 = 14cm=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 * pi * r_1)/2C=2⋅π⋅r12
C = (2 * pi * 14)/2 = 14pi ~~ 44cmC=2⋅π⋅142=14π≈44cm
Now find our cone radius from the cone circumference.
C = 2 * pi * r_2 rarr" " r_2 = (C/(2*pi)) C=2⋅π⋅r2→ r2=(C2⋅π)
r_2 = ( 14 pi)/(2*pi) = 7r2=14π2⋅π=7
From Pythagoras, the equation for a right triangle
r_1^2 = r_2^2 + h^2r21=r22+h2 we obtain:
h = sqrt(r_1^2 – r_2^2) " "rarr" " h = sqrt(196 – 49)
h= sqrt 147( ~~ 12.1 cm" " this is the depth of the cone cup)
V = 1/3 * pi * r_2^2 * h
V = 1/3 * pi * 49 * sqrt147
V= 622 cm^3 volume capacity
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