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प्रश्न
Use ruler and compass to answer this question. Construct ∠ABC = 90°, where AB = 6 cm, BC = 8 cm.
- Construct the locus of points equidistant from B and C.
- Construct the locus of points equidistant from A and B.
- Mark the point which satisfies both the conditions (a) and (b) as 0. Construct the locus of points keeping a fixed distance OA from the fixed point 0.
- Construct the locus of points which are equidistant from BA and BC.
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उत्तर
- The locus of points equidistant from B and C is on BC's perpendicular bisector.
- Similarly, the locus will be at the perpendicular bisector of AB.
- The locus will be the circle that touches all three points, A, B, and C.
- The point equidistant from BA and BC will be the angle bisector of ∠ABC.
संबंधित प्रश्न
Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
- Plot the points A(1, 1), B(5, 3) and C(2, 7).
- Construct the locus of points equidistant from A and B.
- Construct the locus of points equidistant from AB and AC.
- Locate the point P such that PA = PB and P is equidistant from AB and AC.
- Measure and record the length PA in cm.
Plot the points A(2, 9), B(–1, 3) and C(6, 3) on graph paper. On the same graph paper draw the locus of point A so that the area of ΔABC remains the same as A moves.
Construct a ti.PQR, in which PQ=S. 5 cm, QR=3. 2 cm and PR=4.8 cm. Draw the locus of a point which moves so that it is always 2.5 cm from Q.
In given figure 1 ABCD is an arrowhead. AB = AD and BC = CD. Prove th at AC produced bisects BD at right angles at the point M
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 .
Describe completely the locus of a point in the following case:
Centre of a ball, rolling along a straight line on a level floor.
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.
State and draw the locus of a point equidistant from two given parallel lines.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
Without using set squares or protractor.
(i) Construct a ΔABC, given BC = 4 cm, angle B = 75° and CA = 6 cm.
(ii) Find the point P such that PB = PC and P is equidistant from the side BC and BA. Measure AP.
