Answer:
A. 2ab
Explanation:
For solid P with height 2c and a base of width a and length b,
Total surface area = 2(2ca + 2cb + ab)
For the two other rectangular solids, Q and R, each have height c and bases of width a and length b
Total surface area of Q = 2(ca + cb + ab)
Total surface area of R = 2(ca + cb + ab)
Total surface area of the two solids = 2(ca + cb + ab) + 2(ca + cb + ab)
= 4(ca + cb + ab)
Difference between sum of the surface areas of Q and R and the surface area of P
= 4(ca + cb + ab) - 2(2ca + 2cb + ab)
= 4ca + 4cb + 4ab - 4ca - 4cb - 2ab
= 4ca - 4ca + 4cb - 4cb + 4ab - 2ab
= 2ab
The sum of the surface areas of Q and R exceeds the surface areas of P by 2ab. Option A