Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Describe the locus of the centres of all circles passing through two fixed points.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
- AB and BC;
- B and D.
Describe the locus of points at distances greater than 4 cm from a given point.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
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.
In Fig. AB = AC, BD and CE are the bisectors of ∠ABC and ∠ACB respectively such that BD and CE intersect each other at O. AO produced meets BC at F. Prove that AF is the right bisector of BC.
