Advertisements
Advertisements
प्रश्न
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
Advertisements
उत्तर
(i) The locus of points equidistant from three given points A, B & C is the straight line PQ, which bisects AB at right angles.
(ii) Similarly, the locus of points equidistant from B and C is the straight line RS which bisects BC at right angles.
Hence, the point common to PQ and RS must satisfy both conditions; that is to say, X the point of intersection of PQ and RS will be equidistant from A, B and C.
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.
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.
Ruler and compasses 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.
- Construct a ΔABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
- Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
- Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
- Mark the point Q, in your construction, which would make ΔQBC equal in area to ΔABC, and isosceles.
- Measure and record the length of CQ.
Construct a rhombus ABCD with sides of length 5 cm and diagonal AC of length 6 cm. Measure ∠ ABC. Find the point R on AD such that RB = RC. Measure the length of AR.
In Δ PQR, bisectors of ∠ PQR and ∠ PRQ meet at I. Prove that I is equidistant from the three sides of the triangle , and PI bisects ∠ QPR .
Draw and describe the locus in the following case:
The locus of a point in rhombus ABCD which is equidistant from AB and AD.
Using only ruler and compasses, construct a triangle ABC 1 with AB = 5 cm, BC = 3.5 cm and AC= 4 cm. Mark a point P, which is equidistant from AB, BC and also from Band C. Measure the length of PB.
Construct a Δ ABC, with AB = 6 cm, AC = BC = 9 cm; find a point 4 cm from A and equidistant from B and C.
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 compass 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 Δ ABC, in which BC = 8 cm, AB = 5 cm, ∠ ABC = 60°.
(ii) Construct the locus of point 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 as P, the point which is equidistant from AB, BC and also equidistant from B and C.
(v) Measure and record the length of PB.
