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 them so that it touches each of these spheres without disturbing them. How many cm above the table is the centre of the fifth sphere?a)10√6b)10(1+ √2)c)10(1+ √3)d)10(4-√2)Correct answer is option 'B'. Can you explain this answer?

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 them so that it touches each of these spheres without disturbing them. How many cm above the table is the centre of the fifth sphere?a)10&#8730;6b)10(1+ &#8730;2)c)10(1+ &#8730;3)d)10(4-&#8730;2)Correct answer is option 'B'. Can you explain this answer?

Answer

Step1: solve it like a prisum and find the hight of prisum
step2: to get total height add radius of bottom spear to high of the prisum

Answer

Cm
b)15 cm
c)20 cm
d)25 cm

Answer: b) 15 cm

Explanation:

Let's consider the square formed by the centres of the four spheres as the base of a pyramid. The height of this pyramid will be the distance between the base and the centre of the fifth sphere.

[asy]
import three;
triple A=(0,0,0),B=(0,0,20),C=(20,0,20),D=(20,0,0),E=(10,20,10);
draw(surface(A--B--C--D--cycle),white,nolight);
draw(B--C--D--B);
draw(A--B);
draw(A--D);
draw(C--D);
draw(A--E);
draw(B--E);
draw(C--E);
draw(D--E);
label('$A$',A,SW);
label('$B$',B,W);
label('$C$',C,N);
label('$D$',D,E);
label('$E$',E,NW);
draw(circle((5,0,5),10),linewidth(0.7));
draw(circle((5,0,15),10),linewidth(0.7));
draw(circle((15,0,15),10),linewidth(0.7));
draw(circle((15,0,5),10),linewidth(0.7));
draw(circle((10,20,10),10),linewidth(0.7));
[/asy]

We can see that the height of the pyramid will be equal to the radius of the fifth sphere, which is 10 cm.

Now, let's consider the cross-section of the pyramid passing through the centres of two adjacent spheres and the centre of the fifth sphere. This cross-section will be an isosceles triangle with two sides of length 20 cm (the distance between the centres of the adjacent spheres) and one side of length 20 cm + 20 cm + 20 cm - 20 cm (the distance between the centres of the four spheres).

[asy]
import three;
triple A=(0,0,0),B=(0,0,20),C=(20,0,20),D=(20,0,0),E=(10,20,10),F=(10,0,10),G=(0,0,10);
draw(surface(A--B--F--cycle),white,nolight);
draw(surface(B--C--F--cycle),white,nolight);
draw(surface(C--D--F--cycle),white,nolight);
draw(surface(D--A--F--cycle),white,nolight);
draw(B--C--D--B);
draw(A--B);
draw(A--D);
draw(C--D);
draw(A--E);
draw(B--E);
draw(C--E);
draw(D--E);
label('$A$',A,SW);
label('$B$',B,W);
label('$C$',C,N);
label('$D$',D,E);
label('$E$',E,NW);
label('$F$',F,S);
draw(circle((5,0,5),10),linewidth(0.7));
draw(circle((5,0,15),10),linewidth(0.7));
draw(circle((15,0,15),10),linewidth(0.7));
draw(circle((15,0,5),10),linewidth(0.7));
draw(circle

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