a right angle triangle, right angle at A. a circle is inscribed in it. the length of two sides containing angle A is 12cm and 5cm. find the radius
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in the given triangle we can conclude from Pythagoras theorem that BC=13 also let BP=BR=a [tangents from same external point are equal] so PA=12-a, AQ=PA=12-a, QC=13-a=RC. SO AB+AC+BC=5+12+13
from above we can say
a+13-a+13-a+a+12-a+12-a=30
-2*a+50=30
a=10
construction:from center O draw a line segment perpendicular to meet tangent AC at Q and a perpendicular line segment OP to meet tangent AB at P. since AQOP is a square [all angles are right angles] AQ=AP=PO=OQ=12-a=12-10=2
so radius of given circle is 2 cm.
from above we can say
a+13-a+13-a+a+12-a+12-a=30
-2*a+50=30
a=10
construction:from center O draw a line segment perpendicular to meet tangent AC at Q and a perpendicular line segment OP to meet tangent AB at P. since AQOP is a square [all angles are right angles] AQ=AP=PO=OQ=12-a=12-10=2
so radius of given circle is 2 cm.
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