Incircle radius of a Right Angled Triangle.
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| Incircle radius of a right angled triangle |
From fig In ∆ABC,
'O' is in centre, AB= p, BC=h, AC=b,
Inradius r=OD=OE=OF ; OD⊥AC, OE⊥BC, OF⊥AB ;
⇒ar(ΔABC) = ¹/₂ ×p×b
⇒ar∆(ABC) = ar∆(AOB)+ar(∆AOC)+ar(∆BOC)
⇒ ¹/₂ ×p×b= (¹/₂ ×b×r )+( ¹/₂ ×p×r)+ (¹/₂ ×h×r)
⇒ ¹/₂ ×p×b= ¹/₂ ×r×(b+p+h)
⇒ r = (b×p) / (b+p+h)
⇒ Inradius =r= Area / semiperimeter
r = (b×p) / (b+p+h) {∵area=b×p/2, semi perimeter= (b+p+h) / 2}
multiply and divide with (p+b-h)
⇒ r = pb×(p+b-h) / (b+p+h)(p+b - h).
⇒ r = pb×(p+b-h) / [(p+b)² - h²]
⇒ r = pb×(p+b-h) / [p²+b²+2pb -h²]
⇒ r = pb×(p+b-h)/ h²+2pb - h² {∵p²+b² =h²}
⇒ r = pb[p+b - h] / 2pb
⇒ r = (p+b - h) / 2
∴r = {height+base - hypotenuse} / 2
