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प्रश्न
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.
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उत्तर
(i) Steps of Construction:
1. Draw a line segment BC = 8 cm.
2. Make ∠CBX = 60°
3. Set off BA = 5 cm, along BX.
4. Join CA.
Then, ΔABC is the required triangle.
(ii) We know that the locus of point equidistant from two intersecting straight lines consist of a pair of straight lines that bisect the angles between the given straight lines.
Therefore in this case is the angle bisector of angle B, It is shown in the adjoining figure.
(iii) We know that the locus of a point equidistant from two fixed points is the right bisector of the straight line joining the two fixed points.
Therefore, in this case the right bisector of side BC of ΔABC. It is shown in the given figure.
(iv) The point P, is the point in intersecting of angle bisector of ∠ABC and the right bisector of BC.
It is shown in the following figure.
(v) On measuring we find the length of PB = 3 cm.
संबंधित प्रश्न
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 vertices of all isosceles triangles having a common base.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respectively. Draw and describe the locus of a point which is:
- equidistant from BA and BC.
- 4 cm from M.
- 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
State the locus of a point in a rhombus ABCD, which is equidistant
- from AB and AD;
- from the vertices A and C.
Draw two intersecting lines to include an angle of 30°. Use ruler and compasses to locate points which are equidistant from these Iines and also 2 cm away from their point of intersection. How many such points exist?
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
Describe completely the locus of a point in the following case:
Centre of a circle of radius 2 cm and touching a fixed circle of radius 3 cm with centre O.
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.
