1. Determine the length and perimeter of Laura's property.
(5)

9514 1404 393

Step-by-step explanation:

The outer dimensions of the property are 20 squares (length) by 14 squares. Each square is 2 m on a side, so the overall length of the property is ...

  (20 squares)×(2 m/square) = 40 m . . . . length

The overall width of the property is ...

  (14 squares)×(2 m/square) = 28 m . . . . width

Then the perimeter is ...

  P = 2(length + width) = 2(40 +28) m = 136 m . . . . perimeter

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Additional comment

We have computed the perimeter as though the plot were a rectangle. If you carefully consider the total length of horizontal edges and the total length of vertical edges, you see that those totals are the same as they would be for a 20×14 square (40×28 m) rectangle.

The area of open ground is perhaps most easily computed by finding the area that must be subtracted from the overall 20×14 square rectangle. Those exclusions will be (in dimensions of squares) ...

The total area of exclusions is 24+6+16+63+8 = 117 squares. The bounding rectangle is 20 by 14 = 280 squares, so the open ground is ...

  280 squares - 117 squares = 163 squares

At (2 m)(2 m) = 4 m² per grid square, that's an area of ...

  (163 squares)(4 m²/square) = 652 m²

The surface area of the open ground is 652 m².

The volume is given by the formula ...

  V = Bh

where B is the base area and h is the height.

1 cm = 1/100 m = 0.01 m thickness. That means the total volume is ...

  V = (652 m²)(0.01 m) = 6.52 m³

6.52 cubic meters of topsoil are needed to cover the open ground to a depth of 1 cm.

mean, 40.6

step-by-step explanation:

95.03 is the answer. hope this !