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 = $10 + 10 \sqrt{2} = 10 (1 + \sqrt{2}) c m$