Q. 25.0( 1 Vote )
Draw the incircle of ΔABC in which AB = 6cm, AC = 7cm and ∠A = 40° Also find its inradius.
Draw a line segment AC with 7cm and take that as base. Take A as Centre and draw an arc with 6cm radius, taking A as Centre draw ray of 40o that should cut the previous arc. Name the intersection point as B and connect it to C. Thus, triangle is formed.
Construct the angular bisector for ∠ BAC
Taking A as Centre and with any radius draw an arc on the line segments AB, AC. From both the intersected points again draw arcs in the circle. Both the arcs should bisect each other.
Draw a line from A to the opposite side through the intersection point.
We need two angular bisectors to draw an incircle. So, we draw another angular bisector from ∠BCA by following the similar procedures. Where the two angular bisectors meet name the point as I. That is the Centre for the incircle.
Draw an perpendicular bisector with the external point I
For this we need to draw an arc with any radius and cut the line segment AC.
From the intersection points taking more than the half the distance between the points. Draw an arc that should bisector each other.
Connect the point with I. name the point that the line meeting with segment AC as D
With ID as radius and I as Centre draw a circle that will be inscribed in a circle.
Thus, incircle is drawn with the given dimensions.
From the figure we got the inradius as 1.50cm
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