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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
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°
Hence:
1) Construct the locus of points equidistant from BA and BC
2) Construct the locus of points equidistant from B and C.
3) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
State the locus of a point in a rhombus ABCD, which is equidistant
- from AB and AD;
- from the vertices A and C.
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.
Draw a straight line AB of 9 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
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.
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.
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.
Without using set squares or a protractor, construct:
- Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
- Draw the locus of a point which moves so that it is always 2.5 cm from B.
- Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
- Mark the point of intersection of the loci with the letter P and measure PC.
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.
