Question

Use ruler and compass only for this question
(i) Construct $\Delta ABC$, where AB = 3.5 cm, BC = 6 cm and $\angle ABC={60}^{\circ }$
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C.

Then the length of PB is

• A. 2 cm
• B. 2.5 cm
• C. 3 cm
• D. 3.5 cm

Answer 1

The correct option is D

3.5 cm

(i) $\Delta ABC$ is the required triangle in which AB = 3.5 cm, $\angle B={60}^{\circ }$ and BC = 6 cm.
(ii) Draw BD, the angle bisector of $\angle B$ which is the locus of points inside the triangle which are equidistant from BA and BC. (Because the locus of a point which is equidistant from two intersecting straight lines is a pair of straight lines which bisect the angles between the given lines).
(iii) Draw EF, the perpendicular bisector of BC which is the locus of a point inside the triangle equidistant from B and C. (Because the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points.).
On measuring, we get, PB = 3.5 cm