i must want to know answer of this question

see pretty triangel...www.munee.info\jp34prt3.htm me put longest side as base=along x-axis, & left end=(0,0)... triangel area=9.92157, rim-leng=15, base=6, hite=3.30719... top at (2.25, 3.3072)... alfa=55.77114 deg, beta=41.40963, gamma=82.81923 deg... inside serkel: radius=1.32288, senter=(2.5, 1.32288)... outside serkel: radius=3.023716, senter=(3,0.37797)...
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To solve this problem, use the facts that the center of a internal circle, which is inscribed in a triangle, is equidistant from the 3 sides of triangle, and the center of a external circle, which circumscribes a triangle, is equidistant from the 3 vertices of triangle.

First, we draw the triangle.  Draw a horizontal ray,  Mark point A and B on the ray.  Point B is 4cm away from A to the right.  Draw 2 arcs crossing above AB: one from A with radius 6cm, the other from B with radius 5cm, and label the crossing of 2 arcs C.  So, AB=4cm, BC=5cm and CA=6cm.

Second, we draw the internal circle.  Bisect ∠A and ∠B, and label the crossing of 2 angle bisectors above AB, O.  From O, draw 3 perpendiculars to AB, BC and CA, crossing AB at G, BC at H and CA at I.  Draw a circle around the center O with radius OG.  Circle O is the required internal circle that only touches the triangle at points G, H and I.  Because △AOG≡△AOI, △BOG≡△BOH and △COH≡△COI, so OG=OH=OI.

Third, we draw the external circle.  Draw 2 perpendicular bisectors of segment AB and BC, and label the crossing of 2 bisectors above AB, P.  Draw a circle around the center P with radius PA.  The circle P is the required external circle that passes thru 3 points, A, B and C.  Because △PAB,△PBC and △PCA are isoscelese triangle, so PA=PB=PC.

Therfore, circle O is the required internal circle with radius approx. 1.32cm, and circle P is the required external circle with radius approx. 3.02cm.

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