In a right angle triangle ABC, right angled at B, BC=5cm and AB=12 CM. the circle is touching the sides of triangle ABC .Find the radius of the circle. (pls explain with diagram.)
Answers
Step-by-step explanation:
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 cm^2
By Pythagoras theorem,
BC^2 = AC^2 + AB^2
⇒ BC^2 = 12^2 + 5^2
⇒ 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.
Given:
AB = 5 cm, BC = 12 cm
Using Pythagoras theorem,
AC
2
=AB
2
+BC
2
= 5
2
+12
2
= 25+144
= 169
AC=13.
We know that two tangents drawn to a circle from the same point that is exterior to the circle are of equal lengths.
So, AM=AQ=a
Similarly MB=BP=b and PC=CQ=c
We know
AB=a+b=5
BC=b+c=12 and
AC=a+c=13
Solving simultaneously we get a=3,b=2 and c=10
We also know that the tangent is perpendicular to the radius
Thus OMBP is a square with side b.
Hence the length of the radius of the circle inscribed in the right angled triangle is 2cm.