Question

In ΔABC, point L touches side AC, point N touches side AB and point M touches side BC, then prove that A(ΔABC) = 12 * (Perimeter of ABC) * (radius)

Answer 1

Answer:

explanation in attachment

Answer 2

Given : In ΔABC, point L touches side AC, point N touches side AB and point M touches side BC, then

To Find : prove that A(ΔABC) = (1/2) * (Perimeter of ABC) * (radius)

Solution:

Let say O is the center of circle which touches s side AC at L , side AB at N and side BC at M.

OL = OM = ON  = Radius

ON ⊥  AB ,   OL   ⊥  AC ,    OM  ⊥  BC

Join AO , BO , CO

Area of ΔABC  = Area of ΔAOB +  Area of ΔBOC + Area of ΔAOC

Area of ΔAOB = (1/2) * AB * ON  = (1/2) * AB *  Radius

Area of ΔAOC = (1/2) * AC * OL = (1/2) *AC *  Radius  

Area of ΔBOC = (1/2) * BC * OM = (1/2) * BC *  Radius

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

= (1/2) ( AB + AC + BC ) * Radius

= (1/2) (Perimeter of ΔABC )* Radius

QED

Hence proved

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