Advertisements
Advertisements
प्रश्न
In given figure, ABCD is a kite. AB = AD and BC =CD. Prove that the diagona AC is the perpendirular bisector of the diagonal BD.
Advertisements
उत्तर
A is equidistant from B and D. Therefore, A lies on perpendicular bisector of BD.
C is equidistant from Band D. Therefore, C lies on perpendicular bisector ofBD.
A and C both lie on perpendicular bisector of BD.
Hence, AC is perpendicular bi sector of BD.
APPEARS IN
संबंधित प्रश्न
Describe the locus of vertices of all isosceles triangles having a common base.
Describe the locus of a point in space, which is always at a distance of 4 cm from a fixed point.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
- Complete the rectangle ABCD such that:
- P is equidistant from AB and BC.
- P is equidistant from C and D.
- Measure and record the length of AB.
Construct a Δ XYZ in which XY= 4 cm, YZ = 5 cm and ∠ Y = 1200. Locate a point T such that ∠ YXT is a right angle and Tis equidistant from Y and Z. Measure TZ.
Draw and describe the lorus in the following cases:
The locus of points at a distance of 4 cm from a fixed line.
Draw and describe the locus in the following case:
The locus of a point in rhombus ABCD which is equidistant from AB and AD.
Describe completely the locus of a point in the following case:
Centre of a circle of varying radius and touching the two arms of ∠ ABC.
Use ruler and compass only for the following question. All construction lines and arcs must be clearly shown.
- Construct a ΔABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
- Construct the locus of points at a distance of 3.5 cm from A.
- Construct the locus of points equidistant from AC and BC.
- Mark 2 points X and Y which are at a distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Using only a ruler and compass construct ∠ABC = 120°, where AB = BC = 5 cm.
(i) Mark two points D and E which satisfy the condition that they are equidistant from both ABA and BC.
(ii) In the above figure, join AD, DC, AE and EC. Describe the figures:
(a) AECB, (b) ABD, (c) ABE.
