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Question

In an acute angled triangle ABC, AP is the altitude. Circle drawn with AP as its diameter cuts the sides AB and AC at D and E, respectively, then length DE is equal to
  1. Δ2R
  2. Δ3R
  3. Δ4R
  4. ΔR

A
Δ3R
B
ΔR
C
Δ2R
D
Δ4R
Solution
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AC=b, AB=c, BC=a and R is the radius of the circle

AP=c sinB

In ADE ,

DEsinA=2R=AP=c sinB

DE=c sinB sinA=c b2RsinA

Consider,
DE=c b2RsinA

But, =12cbsinA, area of triangle

DE=R

Option D

938376_297075_ans_53ef1ba1e7b5420cbcad0401bff185c9.JPG

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