The projector fully illuminates Screen A at a height of 50 cm, located 140 cm from the projector. What is the smallest distance (in centimeters) from the projector that you want to place screen B 260 cm high so that it is fully lit if the projector settings remain unchanged
Denote the triangles and their key points as shown.
Consider the triangles EGI and EFJ.
The straight line EH is perpendicular to both screens and passes through their center, therefore it is the middle perpendicular.
That is, FK = FJ / 2 = 50/2 = 25 and GH = GI / 2 = 260/2 = 130.
Consider the triangles EFK and EGH.
∠FEK is common to both triangles.
∠EKF = ∠EHG = 90 ° (because EH is the middle perpendicular).
Then, according to the first sign of similarity, these triangles are similar.
Therefore, we can write the aspect ratio:
EH / EK = GH / FK
EH / 140 = 130/25
EH = (130 * 140) / 25 = 728
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