The width of the rectangle of the parallelepiped is 60% of the length, the height is 20% more than the width

| January 13, 2021 | education

The width of the rectangle of the parallelepiped is 60% of the length, the height is 20% more than the width, and the sum of its three dimensions is 5.8 dm. find the volume and area of the side surface of this parallelepiped.

Let the length of the side of the parallelepiped be X cm, AD = X dm.
Then its width is X * 0.6 cm, AB = 0.6 * X dm.
The height of the parallelepiped will be: AA1 = AB + 0.2 * AB = 0.6 * AB + 0.2 * 0.6 * AB = 0.72 * AB dm.
By condition, the sum of the lengths of the three dimensions of the parallelepiped is 5.8 dm, then:
AD + AB + AA1 = 5.8 dm.
X + 0.6 * X + 0.72 * X = 5.8 dm.
2.32 * X = 5.8 dm.
X = 5.8 / 2.32 = 2.5 dm.
BP = 2.5 dm.
AB = 2.5 * 0.6 = 1.5 dm.
AA1 = 2.5 * 0.72 = 1.8 dm.
Determine the area of ​​the base of the parallelepiped.
Sosn = AB * AD = 1.5 * 2.5 = 3.75 dm2.
S side = P * AA1 = 2 * (AB + AD) * AA1 = 2 * 4 * 1.8 = 14.4 dm2.
V = Sbase * AA1 = 3.75 * 1.8 = 6.75 dm3.
Answer: The volume of the parallelepiped is 6.75 cm3, the side area is 14.4 dm2.



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