# 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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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