Question

Construct a $\Delta ABC$ in which AB = 4 cm, BC = 5 cm and $\angle ABC={120}^{\circ }$. Locate the point P, such that $\angle BAP={90}^{\circ }$ and BP = CP. Then the length of BP is

• A. 5.1 cm
• B. 5.6 cm
• C. 6.1 cm
• D. 6.6 cm

Answer 1

The correct option is C

6.1 cm

(I) Draw a line segment AB = 4 cm.
(ii) Draw $\angle B={120}^{\circ }$ and BC = 5 cm.

(iii) Join A to C to make $\Delta ABC$.
(iv) Draw $\perp$ bisector of BC.

(v) Draw $\angle A={90}^{\circ }$ which intersects $\perp$ bisector at P.

(Given, P must be such that $\angle BAP={90}^{\circ }$ and BP=CP. i.e, point P must be equidistant from B and C. We know that 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. So, we draw the perpendicular bisectors of BC so as to get the locus of the point which is equidistant from B and C).
From constructions, length of BP = 6.1 cm.