Given that a circle touches all the three sides of a right angled ΔABC.
We have to prove that radius of a circle =
Let radius be r and the centre of the circle touching all the sides of a triangle be I i.e. incentre.
In the figure,
ID = IE = IF = r
Since ΔABC is a right angled triangle, ∠B = 90°.
Also ID ⊥ BC and AB ⊥ BC.
∴ ID || AB and ID || FB
Similarly, IF || BD
∴ IFBD is a parallelogram.
∴ FB = ID = r and BD = IF = r … (1)
∴ Parallelogram IFBD is a rhombus.
Since ∠B = 90°, parallelogram IFBD is a square.
Now AE = AF
⇒ AE = AB – FB
= AB – r [From (1)] … (2)
And CE = CD
⇒ CE = BC – BD
= BC – r [From (1)] … (3)
Now, AC = AE + CE,
⇒ AC = AB – r + BC – r
⇒ AC = AB + BC – 2r
⇒ 2r = AB + BC – AC
⇒ r =
∴ The radius of a circle is .
Rate this question :
A chord ofGujarat Board Mathematics
Two concentric ciGujarat Board Mathematics
In example 4, ifGujarat Board Mathematics
⨀<spGujarat Board Mathematics
<img width="33" hGujarat Board Mathematics
In ΔABC, AB = 3,Gujarat Board Mathematics
∠<spGujarat Board Mathematics