5 identical spheres are put together in the form of pyra.mid such that four place touching the ground as well as two adjacent spares and 50 ball is placed on top of structure formed by the 4 balls the radius of its where is 5 units what will be the distance of centre of the topmost fares from the ground
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consider pqrs are centeres of the spheres places to form the base square formed by 4 spheres.. each side is 2*r
so diagonal length 2r√2
let A be center of 5th spheres
then triangle apq is equilatrel of side 2r
drop a perpendicular from A to square PQRS
then height of A = perpendicular length AA' + r
perpendicular length = √(4r^2-2r^2)=√2r
(√2+1)5
so diagonal length 2r√2
let A be center of 5th spheres
then triangle apq is equilatrel of side 2r
drop a perpendicular from A to square PQRS
then height of A = perpendicular length AA' + r
perpendicular length = √(4r^2-2r^2)=√2r
(√2+1)5
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