1 . The surface area of a sphere of radius r is
S=4πr2
then,
r=4πS
Given S=144π cm2
r=4π144πcm2=212=6 cm
2 . For a regular tetrahedron of edge length a :
Altitude h=32a
a=2r6 =126
h=32×126=24
3 . The diameter of the sphere would be equal to the side of the square .
d=12cm
then,
r=6cm
So,
V=34×πr3
V=34×π×216
V=904.32
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