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Out of 4 identical balls of radius r, 3 balls are placed on a plane such that each ball touches the other two balls. The 4th ball is placed on them such that this ball touches all the three balls. What is the distance of centre of 4th ball from the plane?

Out of 4 identical balls of radius r, 3 balls are placed on a plane such that each ball touches the other two balls. The 4th ball is placed on them such that this ball touches all the three balls. What is the distance of centre of 4th ball from the plane? Correct Answer <span class="math-tex">\(\frac{{\sqrt 3 + 2\sqrt 2 }}{{\sqrt 3 }}r\)</span> unit

Concept Used:

If the centres of the three balls placed on the plane are joined, they will form an equilateral triangle with side equal to ‘2r’ units, as shown below

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Height of triangular pyramid = √(2/3) × Base side

Calculation:

And, if the centre of the 4th ball is joined to the three centres, a triangular pyramid will be formed with a base of an equilateral triangle

⇒ Base side of triangular pyramid = 2r units Height of triangular pyramid = √(2/3) × Base side

⇒ Height of triangular pyramid = 2√2r/√3 units

This is equal to the perpendicular distance of the centre of 4th ball to the line joining the centres of the other balls

Thus, the distance of centre of 4th ball from plane can be computed by adding the radius of the ball to this distance

∴ Distance of centre of 4th ball from plane = 2√2r/√3 + r = (2√2 + √3)r/√3 units

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