Math, asked by CherryF356, 1 year ago

Draw an equilateral triangle of side 6.5 cm and locate its incentre.Also draw the incircle

Answers

Answered by CarlynBronk
6

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)=\frac{OP}{BP}=\frac{r}{3.25}

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.

Attachments:
Similar questions