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प्रश्न
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.
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उत्तर
Steps of Constructions:
(i) (1) Mark a horizontal line XY on your paper and take BC = 6 cm on it.
(2) Construct ∠ABC = 60° with arm AB = 9 cm.
(3) Join A and C to get the required ΔABC.
(ii) (1) Draw AD ⊥ BC.
(2) Construct a line X'Y', perpendicular to AD, parallel to XY and passing through A.
(3) X'Y', is the required locus of the vertices of Δs with base BC and area to ΔABC.
[∵ Δs having same base and height an equal in area]
(iii) (1) Draw right bisector PQ of BC, meeting X'Y', in Q.
(2) Then Q is the point such that ΔQBC is an isosceles triangle and area (ΔQBC) = area (ΔABC).
(iv) On measuring, we find CQ = 8·4 cm.
संबंधित प्रश्न
Describe the locus of a point P, so that:
AB2 = AP2 + BP2,
where A and B are two fixed points.
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 = 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.
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.
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 the given figure ABC is a triangle. CP bisects angle ACB and MN is perpendicular bisector of BC. MN cuts CP at Q. Prove Q is equidistant from B and C, and also that Q is equidistant from BC and AC.

Describe completely the locus of a point in the following case:
Centre of a ball, rolling along a straight line on a level floor.
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 ruler and compasses construct:
(i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm.
(ii) the locus of point equidistant from A and C.
(iii) a circle touching AB at A and passing through C.
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.
