The figure shows the cross section of a tunnel which was dug through a mountain. Rectangle MNPQ is 40 feet by
20 feet. Arc NOP is a semicircle.
O
N
-P
M
Q
If the length of the tunnel is 1050 feet, find the volume of
dirt and rock that was dug out to create the tunnel. (Round
answer to the nearest 100.000 cubic feet.)

This is the same as 1.5 million cubic feet

Work Shown:

area of rectangle = length*width

area of rectangle = 40*20

area of rectangle = 800 square feet

area of semicircle = 0.5*(area of circle)

area of semicircle = 0.5*pi*r^2

area of semicircle = 0.5*pi*20^2

area of semicircle = 200*pi exact area

total area = (rectangle area) + (semicircle area)

total area = 800 + 200pi

total area = 1428.31853 approximately

Multiply this with the length of the tunnel to get the volume

volume of tunnel = (cross sectional area)*(length of tunnel)

volume of tunnel =  1428.31853*1050

volume of tunnel = 1,499,734.4565

Rounding to the nearest 100,000 cubic feet, we get roughly 1,500,000 cubic feet as the approximate volume.

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