Question
Suresh is standing on vertex A of triangle ABC, with AB = 3, BC = 5, and CA = 4. Suresh walks according to the following plan: He moves along the altitude-to-the-hypotenuse until he reaches the hypotenuse. He has now cut the original triangle into two triangles; he now walks along the altitude to the hypotenuse of the larger one. He repeats this process forever. What is the total distance that Suresh walks?
Answer: Option C
:
C
Let M be the end point of the altitude on the hypotenuse. Since, we are dealing with right angle
triangles, ΔMAC ~ΔABC, so AM = 125. Let N be the endpoint he reaches on side AC. Δ
MAC ~Δ NAM, So, MNAM =45. This means that each altitude that he walks gets shorter
by a factor of 45. The total distance is thus an infinite G.P. =[a1−r] = 12
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:
C
Let M be the end point of the altitude on the hypotenuse. Since, we are dealing with right angle
triangles, ΔMAC ~ΔABC, so AM = 125. Let N be the endpoint he reaches on side AC. Δ
MAC ~Δ NAM, So, MNAM =45. This means that each altitude that he walks gets shorter
by a factor of 45. The total distance is thus an infinite G.P. =[a1−r] = 12
Was this answer helpful ?
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