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
Δ2R
Δ3R
Δ4R
ΔR
A
Δ3R
B
ΔR
C
Δ2R
D
Δ4R
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Solution
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AC=b,AB=c,BC=a and R is the radius of the circle
AP=csinB
In △ADE ,
DEsinA=2R=AP=csinB
DE=csinBsinA=cb2RsinA
Consider,
DE=cb2RsinA
But, △=12cbsinA, area of triangle
∴DE=△R
Option D
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