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प्रश्न
Draw and describe the locus in the following case:
The locus of points inside a circle and equidistant from two fixed points on the circle.
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उत्तर
The locus of the points inside the circle that are equidistant from the fixed points on the circle will be the diameter, which is the perpendicular bisector of the line joining the two fixed points on the circle.
संबंधित प्रश्न
Use ruler and compasses only for this question:
I. Construct ABC, where AB = 3.5 cm, BC = 6 cm and ABC = 60o.
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. Measure and records the length of PB.
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distantce from the given line is always equal to 3 units
Describe the locus of a point in space, which is always at a distance of 4 cm from a fixed point.
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Two straight roads AB and CD cross each other at Pat an angle of 75° . X is a stone on the road AB, 800m from P towards B. BY taking an appropriate scale draw a figure to locate the position of a pole, which is equidistant from P and X, and is also equidistant from the roads.
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.
In Δ ABC, B and Care fixed points. Find the locus of point A which moves such that the area of Δ ABC remains the same.
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.
Ruler and compasses only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.
(i) Construct a ΔABC, in which BC = 6 cm, AB = 9 cm and ∠ABC = 60°.
(ii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to ΔABC.
(iii) Mark the point Q, in your construction, which would make ΔQBC equal in area to ΔABC, and isosceles.
(iv) Measure and record the length of CQ.
