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The doorsill of a campus building is h = 5 feet above ground level. To allow wheelchair access, the steps in front of the door are to be replaced by a straight ramp with constant slope 1/10, as shown in the figure. How long must the ramp be? [The answer is not 50 feet.]...help please :)
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|dw:1328596978457:dw| \[\frac{1}{10}=\frac{5}{50}\] \[x=\sqrt{50^2+5^2}=5\sqrt{101} ft \approx 50.2494 ft\]
thank you so much!!!
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