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a student wants to know how far above the ground the top of a leaning f…

Question

a student wants to know how far above the ground the top of a leaning flagpole is. at high noon, when the sun is almost directly overhead, the shadow cast by the pole is 8 ft long. the student holds a plumb bob with a string 3 ft long up to the flagpole and determines that the point of the plumb bob touches the ground 13 in from the base of the flagpole. how far above the ground is the top of the pole? the top of the flag pole is about □ in above the ground (round to the nearest inch as needed)

Explanation:

Step1: Set up the proportion

Since the triangles are similar (by the AA similarity criterion, as the angles are the same due to the sun's rays), we can set up the proportion. Let \(h\) be the height of the flag - pole in inches. The length of the shadow of the flag - pole is \(8\) ft \(=8\times12 = 96\) inches. The length of the string of the plumb - bob is \(3\) ft \(=3\times12=36\) inches and its shadow is \(13\) inches.
The proportion is \(\frac{h}{96}=\frac{36}{13}\).

Step2: Solve for \(h\)

Cross - multiply the proportion: \(13h = 36\times96\).
First, calculate \(36\times96=(30 + 6)\times96=30\times96+6\times96=2880+576 = 3456\).
Then, \(h=\frac{3456}{13}\approx266\)

Answer:

\(266\)