Axial section of the cone

The axial section of the cone is an isosceles triangle in which the ratio of cone diameter to cone side is 2: 3. Calculate its volume if you know its area is 314 cm square.

Result

V =  366.43 cm3

Solution:

D:s=2:3 S=312 cm2  2r:s=2:3 r:s=1:3  3r=s  S=πr(r+s) S=πr(r+3r) S=4 πr2  r=S4π=3124 3.14164.9828 cm D=2 r=2 4.98289.9656 cm s=3 r=3 4.982814.9484 cm   Correctness test:  k=D/s=9.9656/14.9484=230.6667  h=s2r2=14.948424.9828214.0935 cm  S1=π r2=3.1416 4.98282=78 cm2  V=13 S1 h=13 78 14.0935366.4297=366.43 cm3D:s = 2:3 \ \\ S = 312 \ cm^2 \ \\ \ \\ 2r:s = 2:3 \ \\ r:s = 1:3 \ \\ \ \\ 3r = s \ \\ \ \\ S = \pi r(r+s) \ \\ S = \pi r(r+3r) \ \\ S = 4 \ \pi r^2 \ \\ \ \\ r = \sqrt{ \dfrac{ S }{ 4 \pi } } = \sqrt{ \dfrac{ 312 }{ 4 \cdot \ 3.1416 } } \doteq 4.9828 \ cm \ \\ D = 2 \cdot \ r = 2 \cdot \ 4.9828 \doteq 9.9656 \ cm \ \\ s = 3 \cdot \ r = 3 \cdot \ 4.9828 \doteq 14.9484 \ cm \ \\ \ \\ \text{ Correctness test: } \ \\ k = D/s = 9.9656/14.9484 = \dfrac{ 2 }{ 3 } \doteq 0.6667 \ \\ \ \\ h = \sqrt{ s^2 - r^2 } = \sqrt{ 14.9484^2 - 4.9828^2 } \doteq 14.0935 \ cm \ \\ \ \\ S_{ 1 } = \pi \cdot \ r^2 = 3.1416 \cdot \ 4.9828^2 = 78 \ cm^2 \ \\ \ \\ V = \dfrac{ 1 }{ 3 } \cdot \ S_{ 1 } \cdot \ h = \dfrac{ 1 }{ 3 } \cdot \ 78 \cdot \ 14.0935 \doteq 366.4297 = 366.43 \ cm^3



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