Four spheres each of radius 10 cm lie on a horizontal table so that the centres of the spheres form - Maths - Surface Areas and Volumes

Question

Four spheres each of radius 10 cm lie on a horizontal table so
that the centres of the spheres form a square of side 20 cm A fifth
sphere also of radius 10 cm is placed on then so that it touches
each of these spheres without disturbing them Find that how many
cms above the table is the centre of the fifth sphere ?

Ajanta Trivedi · Accepted Answer

in this case if we consider the relationship between the centres of the spheres, the problem reduces to one involving a pyramid first consider three spheres, each of radius 10 cm, touching each other as shown:

if the centers of the spheres are P, Q, R and S be the mid-point of QR in the right angled triangle PSR, $P S^{2} = P R^{2} - S R^{2} P S^{2} = 20^{2} - 10^{2} = 400 - 100 P S^{2} = 300 P S = 10 \sqrt{3} c m$ in the 2nd diagram P, Q, R, T and U represent the centres of the spheres while S represent the mid-point of QR and let V be the centre of the squareQRTU

using pythagoras' theorem in triangle PVS, $P V^{2} = P S^{2} - S V^{2} P V^{2} = 300 - 100 = 200 P V = 10 \sqrt{2} c m$ the height of the centre V of the square QRTU from the ground = 10 cm therefore the height of the centre of the fifth sphere = $10 + 10 \sqrt{2} = 10 (1 + \sqrt{2}) c m$