The width of the rectangle of the paralepipid is 12 dm, it is 2 times less than the length and the height is 5 cm

univerkov | August 11, 2021 | education

The width of the rectangle of the paralepipid is 12 dm, it is 2 times less than the length and the height is 5 cm more than the sherren, find the surface area of the paralepipid and its volume

Let us find the length of the base of the rectangular parallelepiped, if it is known that it is 2 times greater than the width:
12 * 2 = 24 (dm).
Find the height of the parallelepiped if we know that it is 5 cm larger than the width:
12 dm + 5 cm = (12 * 10) cm + 5 cm = 120 cm + 5 cm = 125 cm.
The surface area of a parallelepiped is the sum of its two bases and four side faces.
Find the area of the base:
24 * 12 = 288 (dm).
Let’s find the area of the lateral surface:
2 * (12.5 * 12 + 12.5 * 24) = 2 * (150 + 300) = 2 * 450 = 900 (dm).
Let’s find the total surface area of the parallelepiped:
900 + 288 * 2 = 576 + 900 = 1476 (dm).
Let’s find the volume of the parallelepiped:
12 * 24 * 12.5 = 3600 (dm3).
ANSWER: total surface area = 1476 dm; volume = 3600 dm3.

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