Abc is right angled triangle at b then find the diameter of incircle of triangle
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Raghunath answered 3 year(s) ago
ABC is a right angled triangle, right angled at A. a circle is inscribed in it. the length of two sodes containing angle A is 12 cm and 5 cm. find the radius.
Class-IX Maths
person
Asked by Anoop
Feb 18
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Raghunath , SubjectMatterExpert
Member since Apr 11 2014

Consider ABC be the right angled triangle such that ∠A = 90° and AB = 5cm, AC = 12 cm.
And O be the centre and r be the radius of the incircle.
AB, BC and CA are tangents to the circle at P, N and M.
∴ OP = ON = OM = r (radius of the circle)
Area of ΔABC = ½ × 5 × 12 = 30 cm2
By Pythagoras theorem,
BC2 = AC2 + AB2
⇒ BC2 = 122 + 52
⇒ BC2 = 169
⇒ BC = 13 cm
Area of ∆ABC = Area ∆OAB + Area ∆OBC + Area ∆OCA
30 = 1 2 r × AB + 1 2 r × BC + 1 2 r × CA
30 = 1 2 r(AB+BC+CA)
⇒ r = 2 × 30 (AB+BC+CA)
⇒ r = 60 5+13+12
⇒ r = 60/30 = 2 cm.
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Raghunath answered 3 year(s) ago
ABC is a right angled triangle, right angled at A. a circle is inscribed in it. the length of two sodes containing angle A is 12 cm and 5 cm. find the radius.
Class-IX Maths
person
Asked by Anoop
Feb 18
0 Like
11419 views
editAnswer
Like Follow
2 Answers
Top Recommend
| Recent
person
Raghunath , SubjectMatterExpert
Member since Apr 11 2014

Consider ABC be the right angled triangle such that ∠A = 90° and AB = 5cm, AC = 12 cm.
And O be the centre and r be the radius of the incircle.
AB, BC and CA are tangents to the circle at P, N and M.
∴ OP = ON = OM = r (radius of the circle)
Area of ΔABC = ½ × 5 × 12 = 30 cm2
By Pythagoras theorem,
BC2 = AC2 + AB2
⇒ BC2 = 122 + 52
⇒ BC2 = 169
⇒ BC = 13 cm
Area of ∆ABC = Area ∆OAB + Area ∆OBC + Area ∆OCA
30 = 1 2 r × AB + 1 2 r × BC + 1 2 r × CA
30 = 1 2 r(AB+BC+CA)
⇒ r = 2 × 30 (AB+BC+CA)
⇒ r = 60 5+13+12
⇒ r = 60/30 = 2 cm.
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