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38 votes
For a project in his Geometry class, Yusuf uses a mirror on the ground to measure the

height of his school building. He walks a distance of 10.25 meters from the building,
then places a mirror flat on the ground, marked with an X at the center. He then
walks 1.65 more meters past the mirror, so that when he turns around and looks
down at the mirror, he can see the top of the school clearly marked in the X. His
partner measures the distance from his eyes to the ground to be 1.55 meters. How tall
is the school? Round your answer to the nearest hundredth of a meter.

User Ade
by
2.9k points

1 Answer

11 votes
11 votes

Explanation:

sadly we don't know anything about the size of the mirror on the ground, or if e.g. the 1.65 m from the mirror mean from the edge of the mirror or from the x at the center of the mirror.

so, we must assume the mirror as a dot without any significant lengths in the calculation.

we have 2 right-angled triangles.

1. the small one : from Yusuf's eyes to the ground at his feet to the x on the mirror on the ground.

2. the large one : from the top of the building to the bottom of the building on the ground to the x on the mirror on the ground.

the mirroring creates a dilation from the large triangle to the small triangle through the point x on the mirror.

in other words, all the angles stay the same, and there is one constant scale factor for all the lengths of one triangle to the other, making both triangles similar.

that means that the scale factor "f" of the ground distances from the small to the large triangle is

10.25 / 1.65 = 1025 / 165 = 6 35/165 = 6 7/33 =

= 6.212121212...

the scaling from Yusuf's height (well, his eye level) to the height of the school must follow the same ratio :

school height / Yusuf's eye level =

= h/1.55 = 10.25/1.65

h = 10.25 × 1.55 / 1.65 = 1025 × 155 / 16500 =

= 9.628787879... m ≈ 9.63 m

the school is approximately 9.63 m tall.

FYI : this is the same principle as with a camera. the lens (and the mirror inside if applicable) create a dilated "projection" onto a film or an array of digital light sensors, so that the captured image retains all the angles and all the length ratios of the original.

but to know from the picture e.g. how tall the original was, we need to know also the distance from which the picture was taken. otherwise this can't be answered.

User Tao Wang
by
3.4k points
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