Four spheres each of radius 10 cm lie on a horizontal table so that th...
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