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)
Answers
Answer:
explanation in attachment
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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