4. Given three cubes with integer side lengths, if the sum of the surface areas of the three cubes is
498 sq. cm, then the sum of the volumes of the cubes in all possible solutions is
A) 731
B) 495
C) 1226
D) none of these
Answers
Answered by
3
Answer:
Surface area of cube= 6*edge*edge
Sum of square of cubes = 498
6*edge*edge+6*edge*edge+6*edge*edge = 498
18*edge*edge=498
edge*edge= 27.67
edge=5.25
Volume of cube = edge*edge*edge
Volume of 3 cubes = 3*5.25*5.25*5.25=434.10
Answer is D
Step-by-step explanation:
Answered by
3
Answer:
6(x2 + y2 + z2
) = 498
x
2 + y2 + z2 = 83
for x, y, z to be integer
x = 49 , y = 25 , z = 9
x = 7, y = 5, z = 3
Sum of volumes = 73 + 53 + 33
343 + 125 + 27 = 495
for x, y, z to be integer
x = 81 , y = 11 , z = 1
x = 9, y = 1, z = 1
Sum of volumes = 93 + 13 + 13
= 729 + 1 + 1 = 731.
So, total sum = 495 + 731 = 1226
Option (C).
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