Question

Construct a triangle $ABC$ with $AB=5.5cm,AC=6cm$ and $\angle BAX={105}^o$
Hence :
Construct the locus of points equidistant from $BA$ and $BC$ .

Answer 1

1) Draw a line segment AB of length 5.5 cm.

2) Make an angle MZBAX = 105° using a

protractor

3) Draw an arc AC with radius AC = 6 cm on AX with centre at A.

4) Join BC.

Thus ABC is the required triangle.

a) Draw BR, the bisector of ABC, which is the locus of points equidistant from BA and BC.

b) Draw MN, the perpendicular bisector of BC, which is the locus of points

equidistant from B and C c) The angle bisector of ABC and the perpendicular bisector of BC meet at

point P. Thus, P satisfies the above two

loci.

Length of PC = 4.8 cm