A sphere of radius r is cut by a plane at a distance of h from its cen

Question

A sphere of radius r is cut by a plane at a distance of h from its center, thereby breaking this sphere into two different pieces. If the cumulative surface area of these two pieces is 25% more than that of the sphere, what is the value of h ? (The surface area of a sphere = 4πr^2)

A. r/√2
B. r/√3
C. r/√5
D. r/√6
E. r/√6

A. r/√2
B. r/√3
C. r/√5
D. r/√6
E. r/√6

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GMATWhizTeam · Answer

Solution 
 • Before the cut, the total surface area of the sphere  = 4*\pi*r^2 
• Now, as shown in the below figure, the sphere is at cut by a plane at a distance h.
 https://www.dropbox.com/s/dfzezcxsmjhl7b7/sphere_cut.png?dl=1 
• After the cut, two extra identical circular faces will come into picture ( top view of the identical circular faces is shown below)
 https://www.dropbox.com/s/8l6kqffnxo98a2i/sphere_top_view.png?dl=1 
 o Radius of each of these circular faces  = \sqrt {r^2 - h^2} 
o Thus, the cumulative surface area of the two pieces = total surface area of the sphere + 2* area of the new circular face = 4*\pi*r^2 + 2*\pi*(r^2 – h^2)  
• It is given that the cumulative surface area is 25% more than that of the original sphere.
 o This means,  2*\pi*(r^2 – h^2) = \frac{1}{4}*4*\pi*r^2 
 Or,  2r^2 – 2h^2 = r^2 
 Or,  h^2 =\frac{r^2}{2} 
 Or,  h = \frac{r}{\sqrt{2}}   
Thus, the correct answer is  Option A.

GMATinsight · Answer

Please find the step-by-step video explanation of the problem here

Answer: Option A

https://www.youtube.com/watch?v=JAc_VOAwSs0

ScottTargetTestPrep · Answer

When a sphere with radius r is cut into two different pieces by a plane at a distance of h from its center, the total surface area of the two pieces exceeds the surface area of the sphere by the total area of the two circular cross sections of the two pieces. Furthermore, since the total surface area of the two pieces is 25% more than that of the sphere, the total area of the two circular cross sections must be 25% of the surface area of the sphere.

Since the two cross sections have the same area and they are circles, we can let the radius of each be R. Using the Pythagorean theorem, we can determine R, in terms of r and h, as √(r^2 - h^2). Thus the total area of the two circular cross sections is 2πR^2 = 2π ^2 = 2π(r^2 - h^2). Since this must be 25% of the surface area of the sphere, we can create the equation:

2π(r^2 - h^2) = ¼(4πr^2)

2π(r^2 - h^2) = πr^2

2r^2 - 2h^2 = r^2

r^2 = 2h^2

r^2/2 = h^2

√(r^2/2) = h

r/√2 = h

Answer: A

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A sphere of radius r is cut by a plane at a distance of h from its cen

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A sphere of radius r is cut by a plane at a distance of h from its cen

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