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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.
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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
State the locus of a point in a rhombus ABCD, which is equidistant
- from AB and AD;
- from the vertices A and C.
Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
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.
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.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
Use ruler and compasses only for the following questions:
Construct triangle BCP, when CB = 5 cm, BP = 4 cm, ∠PBC = 45°.
Complete the rectangle ABCD such that :
(i) P is equidistant from AB and BC and
(ii) P is equidistant from C and D. Measure and write down the length of AB.
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.
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.
