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A Triangle ABC is drawn to circumscribe a circle of radius 10 cm such that the segments BP and PC into which BC is divided by the point of contact P , are of lenghts 15 cm and 20 cm respectively . If the area of triangle ABC=525 cm sq , then find the lenghts of sides AB and AC

 http://www.youtube.com/watch?v=6J5Lk4W4dhE

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sry..i did a mistake

Let O be the incentre of ΔABC having radius 10 cm.

Also, BP = 15 cm and PC = 62 cm

⇒ BR = 15 cm and CQ = 62 cm (Length of the tangents from an external point are equal)

Let AQ = AR = x cm

So, AB = AR + RB = (x + 15) cm

BC = BP + PC = (15 + 62) cm = 77 cm

AC = AQ + QC = (x + 62) cm

Also, area of ΔABC = area of ΔOBC + area of ΔOCA + area of ΔAOB

It is given that area of ΔABC = 525 cm2

⇒ 10 (x + 77) = 525

⇒ x + 77 = 52.5

⇒ x = 52.5 – 77

⇒ x = –24.5, which is not possible

So, this means the given data is incorrect.

(Such a triangle and incircle is not possible)

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i think so this answer is wrong where did 62 came from?????

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Yes, where did 62 come from ?

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I think this is right u may get calaculations right but i think it is right way

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Hope that this would help you...

  • 18
32.5 nd 37.5
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Hey its not 62 its 20
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AB=50cm AC=55cm Since x=35cm
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Sorry x=17.5cm
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Answer in pic is correct
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Ab is equal to X + 15 BC is equal to 15 + 20 is equal to 35 AC is equal to X + 20

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Hope this is helpful

  • 3
Answer is 32.5
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Triangle is drawn to circumscribe a circle of radius 10 cm such that the segment BP and CP into which BC is divided by point of contact of length 15cm and 20cm respectively if the area of triangle ABC is 525 CM square then find the length of the side a b and ac
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Answer

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