In triangle ABC with different rib inscribed circular with radius r=4 which touches AB at point M and M divided in two parts with length AM=8cm and MB=6cm.Find area of triangle ABC?!

I used the same......
we can consider 'r' be the radii of incircle and we know (according to solutions of triangle) that
r=(area of triangle/semiperimeter of triangle) ,,which can be furthr written as

r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.

lets consider the circle touches triangle at pts M,N,P on sides AB,AC,BC respctivly
since when we draw tangents from a point on circle they r equal in length thrfore
AM=AN=8
MB=BP=6
PC=CN= x

NOW USING THIS FORMULA

r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}

where r is radii=4

s = \frac{a+b+c}{2}

a,b,c are sides opp. to A,B,C resp.
i.e. a=6+x ; b=8+x ; c=8+6=14

we can find x
and use
r= \frac{area of triangle}{s}

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TopNewestIs the answer 84 sq.units......

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Can you explain ? I think we should use property of tangents and incircle - perimeter of triangle relation

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SORRY i got late

I used the same...... we can consider 'r' be the radii of incircle and we know (according to solutions of triangle) that

r=(area of triangle/semiperimeter of triangle) ,,which can be furthr written as

lets consider the circle touches triangle at pts M,N,P on sides AB,AC,BC respctivly since when we draw tangents from a point on circle they r equal in length thrfore AM=AN=8 MB=BP=6 PC=CN= x

NOW USING THIS FORMULA

r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}

where r is radii=4

s = \frac{a+b+c}{2}

a,b,c are sides opp. to A,B,C resp. i.e. a=6+x ; b=8+x ; c=8+6=14

an finally the answer i got is 84...

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