At a certain time of the day, a man 6 feet tall, casts his shadow 8 feet long. Find the length of the shadow cast by a building 45 feet high, at the same time which is next to the man.
Explanation
Let the length of the shadow of the building =x.
In △ABC, AB || DE (both are standing vertically on the ground)
\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EC}}\)
\(\frac{45}{6}=\frac{x}{8} \Rightarrow x=\frac{45 \times 8}{6}=60 \mathrm{ft}\)
Length of shadow cast by the building is 60 ft.
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