Draw an equilateral triangle of side 6.5 cm and locate its incentre.Also draw the incircle
Answers
In equilateral ΔABC,
A B =B C=C A=6.5 cm
Draw angle bisector of ∠A, ∠B and ∠C. The point where these angle bisector meet is the incentre of Δ ABC.
Point P, Q, R are points on sides B C, CA and AB respectively where the angle bisector from A,B and C meet.
⇒∠ CAP=30° , ∠ C =60° , ∠ C PA=90°→ [by angle sum property of triangle in Δ CAP]
In Δ ABC , AP ⊥ BC, BQ⊥AC, CR⊥AB,
Δ APC≅ΔAPB [SAS, As, AB=AC, AP is common,and ∠BAP=∠CAP]
∴ AP, BQ, CR are perpendicular bisector of sides, as well as angles of ΔABC.
Let these bisectors meet at point O.
In Δ OPB, ∠OPB=90°, OP=r, BP=BC/2=6.25/2=3.25 cm
Tan(∠ OBP)=
Tan(30°) = r/3.25
1/√3=r/3.25
⇒ r=3.25/1.732=1.876 cm →(approx)
In Δ APC
Tan 60°=AP/AC
⇒ AP=3.25√3=5.629 cm
So, incenter lies (5.629- 1.876=3.753 cm) from any vertex and at a distance of 1.876 from the point where the angle bisector meets the sides.