Question

A circle with centre P is inscribed in the AABC. Side AB, side
BC and side AC touch the circle at points L, M and N
respectively. Radius of the circle is r.
Prove that:
A(AABC) = (AB + BC + AC) x r​

Answer 1

Step-by-step explanation:

Answer:

Answer:Step-by-step explanation:

Consider P is the centre of the circle.

Radius=r

T. P= ar(∆ABC)= 1/2(AB+BC+AC)×r

Draw lines between PA PB and PC.

AB is tangent on the circle, touch the circle at the points L and PL is the radius.

According to tangent radius theory

PL_I_ AB

PN_I_ AC

PM_I_ BC

We know that

PL=PM=PN=r(Radius of the circle)

∆APL,∆APN,∆BPL,∆BPM,∆CPMand ∆CPN are right angle triangle

ar(∆ABC)= ar(∆APL)+ar(∆APN)+ar(∆BPL)+ ar∆(BPM)+ar(∆CPM)+ar(∆CPN)

1/2×PL×AL+1/2×PN×AN+1/2×PL×BL+1//2×PM×BM+1/2×PM×CM+1/2×PN×CN

1/2×r(AL+AN+BL+BM+CM+CN)

1/2×r(AL+BL+BM+CM+CN+AN)

1/2×r(AB+BC+AC)

PLEASE MARK AS BRAINLIST

Answer 2

hence.... it's proved!!!!!!!!! ✌