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Standard X

Mathematics

Area of a Circle

In ABC AB =6 ...

Question

In ∆ABC AB=6 BC=8 AC=10 . A perpendicular dropped from B meets AC at D . A circle of radius BD (centre B) is drawn . In the circle cuts AB and BC at P and Q then find AP:QC

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Solution

Area of triangle = 0.5(ABBC)
= 24 which is also
= 0.5(ACBD)
=> BD = 24/5.
Now, AP = 6 - (24/5)
= 6/5 and
BQ = 8 - (24/5)
= 16/5. t

hus the ratio is 3 : 8

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In ∆ABC AB=6 BC=8 AC=10 . A perpendicular dropped from B meets AC at D . A circle of radius BD (centre B) is drawn . In the circle cuts AB and BC at P and Q then find AP:QC

Area of triangle = 0.5(AB*BC) = 24 which is also = 0.5(AC*BD) => BD = 24/5. Now, AP = 6 - (24/5) = 6/5 and BQ = 8 - (24/5) = 16/5. t hus the ratio is 3 : 8

Area of triangle = 0.5(ABBC)
= 24 which is also
= 0.5(ACBD)
=> BD = 24/5.
Now, AP = 6 - (24/5)
= 6/5 and
BQ = 8 - (24/5)
= 16/5. t

hus the ratio is 3 : 8