a circle is drawn touching sides of triangle ABC at P,Q,R. If AP+BC=13 cm,then perimeter of triangle ABC is?
Answers
Given : circle is drawn touching sides of triangle ABC at P,Q,R.
AP+BC=13
To Find : perimeter of triangle ABC
Solution:
Perimeter of Triangle ABC
= AB + BC + AC
= AP + PB + BQ + CQ + CR + AR
AP = AR Equal Tangent
BP = BQ Equal Tangent
CQ = CR Equal Tangent
= AP + BQ + BQ + QC + CQ + AP
= 2( AP + BQ + QC)
= 2 ( AP + BC)
= 2 (13)
= 26
perimeter of Δ ABC = 26 cm
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Answer: circle is drawn touching sides of triangle ABC at P,Q,R.
AP+BC=13
Step-by-step explanation:
In the figure shown,
Let the sides,
AB=c
AC=b
BC=a
PC=x
PB=a−x
Now, AQ and AR are two pair of tangents drawn from the same extrernal point A. Also, PC and CR are tangents from point C and PB and QB are tangents from point B.
Using the property that lenght of tangents drawn to a circle from a given external point is always the same, we can say that
PC=CR=x
BP=QB=a−x
AQ=AR
=>AB+BQ=AC+CR
=> c+a−x=b+x
=> x=2a−b+c
Now,
AQ=AR=b+x
=> b+2a−b+c
=> 2a+b+c