Question

A circle with centre P is in the Triangle ABC .side AB ,side BC ,side AC touch the circle at points L,M,N respectively .Radius of the circle is r . prove that A (ABC )1/2(AB +BC + AC )×r

Answer 1

Answer:

Step-by-step explanation:

Let P be the point of intersection.

Area (Δ BPC) = 1/2 x base x height = 1/2 x PM x BC

Area (Δ APB ) = 1/2 x base x height = 1/2 x PL x AB

Area (Δ APC ) = 1/2 x base x height = 1/2 x PN x AC

Now, PM = PL = PN = r

Area (Δ ABC) = Area (Δ BPC) + Area (Δ APB ) + Area (Δ APC )

Area (Δ ABC) = 1/2 x r x BC + 1/2 x r x AB + 1/2 x r x AC

Area (Δ ABC) = 1/2 x r x ( AB + BC + AC)

Answer 2

Answer:

Proved

Step-by-step explanation:

A circle with centre P is in the Triangle ABC .side AB ,side BC ,side AC touch the circle at points L,M,N respectively .Radius of the circle is r . prove that A (ABC )1/2(AB +BC + AC )×r

prove that area of triangle ABC =1/2(AB+BC+AC)×r​

Let say center point = O

if we draw line from points A , B & C  at point O

we can Divide ΔABC into three triangle

ΔAOB  , ΔBOC & ΔCOA

Area of ΔAOB =  (1/2) * AB * OL  ( Base * Perpendicular)

OL = Radius = r

Area of ΔAOB =  (1/2) * AB * r

SImilarly

Area of ΔBOC =   (1/2) * BC * r

Area of ΔCOA =   (1/2) * AC * r

Area of ΔABC = Area of ΔAOB + Area of ΔBOC  + Area of ΔCOA

=> Area of ΔABC =  (1/2) * AB * r +  (1/2) * BC * r  +  (1/2) * AC * r

=> Area of ΔABC =  (1/2) * (AB + BC + AC) * r

QED