Advertisements
Advertisements
Question
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that:
- P is equidistant from B and C.
- P is equidistant from AB and BC.
- Measure and record the length of PB.
Advertisements
Solution
Steps of construction:
-
- Draw a line segment AB = 7 cm.
- Draw angle ∠ABC = 60° with the help of a compass.
- Cut off BC = 8 cm.
- Join A and C.
- The triangle ABC so formed is the required triangle.
- Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
- Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.
The point that fulfills the condition required in i. and ii. is the intersection point of the bisector of line BC and the angular bisector of ∠ABC.
P is the required point, which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.
APPEARS IN
RELATED QUESTIONS
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Draw an ∠ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC; and also equidistant from A and B.
In the figure given below, find a point P on CD equidistant from points A and B.
Describe the locus for questions 1 to 13 given below:
1. The locus of a point at a distant 3 cm from a fixed point.
Describe the locus of points at a distance 2 cm from a fixed line.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
In Fig. ABCD is a quadrilateral in which AB = BC. E is the point of intersection of the right bisectors of AD and CD. Prove that BE bisects ∠ABC.
ΔPBC and ΔQBC are two isosceles triangles on the same base. Show that the line PQ is bisector of BC and is perpendicular to BC.
ΔPBC and ΔQBC are two isosceles triangles on the same base BC but on the opposite sides of line BC. Show that PQ bisects BC at right angles.
