Consider a cuboid of whose all sides are integers, and base is a square. Suppose sum of all it's edges is numerically equal to the sum of areas of all it's six faces, then the.sum of all it's edges is…
A)12 B)18 C)24 D)36
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Let the dimensions of the cuboid be a, a and b
Thus according to question
Sum of the edges = 8a + 4b
Sum of the areas of 6 faces = 2a^2 + 4ab
Thus according to question
8a + 4b = 2a^2 + 4ab
Thus
2a(4 - a) = 4b (a-1)
a(4-a) = 2b(a-1)
4a + 2b = 36
Thus according to question
Sum of the edges = 8a + 4b
Sum of the areas of 6 faces = 2a^2 + 4ab
Thus according to question
8a + 4b = 2a^2 + 4ab
Thus
2a(4 - a) = 4b (a-1)
a(4-a) = 2b(a-1)
4a + 2b = 36
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1
Answer:
the answer is 24
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