Question

Using ruler and compasses only, construct a triangle POR such that $\angle P={120}^{\circ }$, PO = 5 cm PR = 6 cm.In the same figure, find a point which is equidistant from its sides. Name this point With this point as centre draw a circle touching all the sides of the triangle.

• A. Circumcentre
• B. Incentre
• C. Mid point
• D. Data insufficient

Answer 1

Answer: (B)

Incentre

(I) We draw a triangle POR with $\angle RPO={120}^O.$

(II) OI & OR are the angular bisectors of $\angle RPO\&\angle ROP$

The bisectors intersect at I.

(i) IM & IN are drawn perpendiculars from i to PR & PO respectively.

(iii) The circle which touches the sides of the triangle POR.

has the radius IM=IN.

Justification-

Between $\Delta IPM\&\Delta IPN,\angle IMP={90}^o=\angle INP,\angle IPM=\angle IPN$

So the third angles $\angle PIN=\angle PIM$

Also the side IP is common.

So, by ASA rule, $\Delta IPM\equiv \Delta IPN.$

i.e IM=IN.

Similarly, by considering the triangles INO & ISO it can be shown that

IN=IS.

So IM=IN=IS.

i.e the circle with centre I touches the sides of the given triangle.

So I is the INCENTRE of the triangle POR.

Ans- Option B.