The length of a rectangular parallelepiped is 40% longer than its width, and its width is 5 times less than its height. What is the total surface area of a parallelepiped if its volume is 56 dm3?
Let’s designate the width of the parallelepiped as “x” dm, then taking into account the condition of the problem, its length will be “x + 0.4x” dm, and the height will be equal to “5x” dm.
Let’s make an equation taking into account that the volume of this parallelepiped is 56 dm³.
x * 1.4x * 5x = 56;
1.4x² * 5x = 56;
7x³ = 56;
x³ = 56/7;
x³ = 8;
x = ³√8 = 2 dm width.
1.4 * 2 = 2.8 dm length.
5 * 2 = 10 dm height.
Let’s calculate the total surface area:
S = 2 * (a * b + b * c + a * c) = 2 * (2.8 * 2 + 2 * 10 + 2.8 * 10) = 2 * (5.6 + 20 + 28) = 107.2 dm².
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