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Question 1199155: There is a loading dock at Henry's grocery store. Attached to the loading dock, there is a ramp, whose run is 12 feet and whose rise is 39 inches. Henry is wondering whether a long rectangular box can be stored underneath the ramp, if the box is 2 feet tall and 5 feet long. Can the box be stored underneath the ramp?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
39 inches = 39/12 = 3.25 ft
This value 3.25 is the y coordinate of point C shown below.
Diagram
Points:
A = (0, 0)
B = (12, 0)
C = (0, 3.25)
D = (4.6153846154, 2)
E = (4.6153846154, 0)
The decimal values of points D and E are approximate.
Points B and C make up the ramp.
The equation of line BC is  which is of the form  (technically not standard form but somewhat close enough). It has slope 
Draw a horizontal line through y = 2 which represents the height of the box.
This horizontal line intersects line BC at the point D(4.6153846154, 2).
Just below it is E(4.6153846154, 0)
The horizontal width from A(0,0) to E(4.6153846154, 0) is roughly 4.6153846154 feet.
This is unfortunately smaller than the 5 ft width of the box Henry wants to store.
Another approach would be to draw a vertical line through x = 5.
This vertical line intersects line BC at an (x,y) point such that y < 2, which means there isn't enough height for the box at this x location.
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Answer: No, there isn't enough room.
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