Three identical balls of radius 2 , cm each a | vedclass

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Question

(A) Let the radius of each ball be $r = 2 \, cm$. The centers of the three bottom balls form an equilateral triangle $ABC$ with side length $2r = 4 \,

Vedclass · Accepted Answer

$4 \left( \sqrt{\frac{2}{3}} + 1 \right)$

Vedclass · Answer

(A) Let the radius of each ball be $r = 2 \, cm$. The centers of the three bottom balls form an equilateral triangle $ABC$ with side length $2r = 4 \, cm$.Let $G$ be the centroid of $\Delta ABC$. The distance from a vertex to the centroid is $AG = \frac{2}{3} 	imes (2r \sin 60^{\circ}) = \frac{2}{3} 	imes (2r 	imes \frac{\sqrt{3}}{2}) = \frac{2r}{\sqrt{3}}$.The center of the fourth ball,$D$,forms a regular tetrahedron with the centers $A, B, C$. The vertical height $h$ of $D$ above the plane of $ABC$ is given by $h^2 + AG^2 = (2r)^2$.$h^2 = 4r^2 - \frac{4r^2}{3} = \frac{8r^2}{3} \implies h = 2r \sqrt{\frac{2}{3}}$.The total height of the highest point of the fourth ball above the table is $H = h + r + r = h + 2r$.$H = 2r \sqrt{\frac{2}{3}} + 2r = 2r \left( \sqrt{\frac{2}{3}} + 1 ight)$.Substituting $r = 2 \, cm$,we get $H = 2(2) \left( \sqrt{\frac{2}{3}} + 1 ight) = 4 \left( \sqrt{\frac{2}{3}} + 1 ight)$.

Three identical balls of radius $2 \, cm$ each are placed on a table such that they touch each other as well as the table. Now a fourth ball of the same radius is placed above these three balls. The height of the highest point on the fourth ball above the table is -

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