triangle ABC is an isosceles triangle in which AB=AC. Each of the congruent sides is 3 cm more than the length of the base. If the perimeter of the triangle is 36 cm,line AM is perpendicular to line BC. Find the length of seg AM
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Given- ABC is an isosceles triangle with AB=AC.BC=12cm and the perimeter of ΔABC=44cm. A circle is inscribed in ΔABC such that it AB,BC & AC touch at P,R & Q respectively.
To find out- AP=?AQ=?
Solution- AB+AC=2AB or 2AC since AB=AC.
∴ the perimeter of ΔABC=AB+AC+BC=2AB+BC=2AB+12=44cm⟹AB=16cm=AC.......(i)
Now BR=CR=(
2
12
)cm=6cm since a circle, inscribed in an isosceles triangle, touches the base at mid point. But BR=BP=6cm and CR=CQ=6cm since tangents to a circle from an external point are equal in length. So AP=AQ=(16−6)cm=10cm (from i). ∴ The tangent segments AP=AQ=10cm.
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