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a student 6 ft tall is standing 20 feet away from a 35 ft tall flagpole. The flagpole is 60 ft away from the building. From the student point of view, the top of the flag pole is lined up perfectly with the top of the building, how talls the building

User Gep
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1 Answer

9 votes
9 votes

Explanation:

between the student and the flagpole we create a trapezoid with the line of sight to the top of the flagpole and the ground connection between the student's feet and the flagpole being the other 2 sides.

and between the flagpole and the building we create a similar trapezoid.

the line of sight to the top of the building and the ground connection between the flagpole and the building are the other 2 sides.

since both shapes are created from the same projection (the "projector" is somewhere behind the student on the ground), they are truly and mathematically similar.

that means that the angles of one trapezoid are the same as in the other trapezoid.

and - there must be one constant scale factor between all the sides (or for any other lengths in or on the trapezoids) of one trapezoid to the sides of the other trapezoid.

we know the ground distances and how they relate :

20 ft to 60 ft

that means 20/60 = 1/3

so, any length of the first (smaller) trapezoid is 1/3 of the corresponding length of the second (larger) trapezoid.

so, we know the height of the flagpole (35 ft).

the height of the building must follow the same ratio (scaling factor) as the ground distances.

flagpole height = building height × 1/3

3× flagpole height = building height = 3×35 = 105 ft

the building is 105 ft tall.

User Mumfordwiz
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