A commercial greenhouse is constructed

Mathematics

1 answer:

Llana [10]3 years ago

4 0

9514 1404 393

Answer:

155.52 m³

Step-by-step explanation:

The volume will be the product of the end area and the length of the greenhouse.

The end area consists of 4 triangles and 6 rectangles. We can do a little manipulation of the shapes to simplify the area calculation a bit.

We can divide half of the greenhouse face along its vertical lines to make two trapezoids. Their "bases" will be the lengths of the vertical lines, and their heights will be the dimensions 1.8 m and 0.6 m. Since there are two of each, we can find the end area to be ...

A = 2(1/2)(b1 +b2)(h1) +2(1/2)(b2 +b3)(h2)

A = (3.3 +2.7)(1.8) +(2.7 +0.9)(0.6) = 10.8 +2.16 = 12.96 . . . square meters

The volume is then ...

V = Bh

V = (12.96 m²)(12 m) = 155.52 m³ . . . . greenhouse volume

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