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प्रश्न
In Δ ABC, B and Care fixed points. Find the locus of point A which moves such that the area of Δ ABC remains the same.
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उत्तर
Steps of Construction:
(i) ABC is the required triangle.
(ii) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from Band C.
(iii) Through A, draw a line m 11 BC.
(iv) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(v) Then triangle QBC is equal in area to triangle ABC. mis the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.
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संबंधित प्रश्न
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.
State the locus of a point in a rhombus ABCD, which is equidistant
- from AB and AD;
- from the vertices A and C.
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 triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
- Complete the rectangle ABCD such that:
- P is equidistant from AB and BC.
- P is equidistant from C and D.
- Measure and record the length of AB.
Construct a rhombus ABCD whose diagonals AC and BD are 8 cm and 6 cm respectively. Find by construction a point P equidistant from AB and AD and also from C and D.
Construct a triangle BPC given BC = 5 cm, BP = 4 cm and .
i) complete the rectangle ABCD such that:
a) P is equidistant from AB and BCV
b) P is equidistant from C and D.
ii) Measure and record the length of AB.
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.
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.
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.
