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प्रश्न
State the locus of a point in a rhombus ABCD, which is equidistant
- from AB and AD;
- from the vertices A and C.
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उत्तर
Steps of construction:
i. In rhombus ABCD, draw angle bisector of ∠A which meets in C.
ii. Join BD, which intersects AC at O.
O is the required locus.
iii. From O, draw OL ⊥ AB and OM ⊥ AD
In ΔAOL and ΔAOM
∠OLA = ∠OMA = 90°
∠OAL = ∠OAM ...(AC is bisector of angle A)
AO = OA ...(Common)
By Angle-Angle – side criterion of congruence,
ΔAOL ≅ ΔAOM ...(AAS Postulate)
The corresponding parts of the congruent triangles are congruent
`=>` OL = OM ...(C.P.C.T.)
Therefore, O is equidistant from AB and AD.
Diagonal AC and BD bisect each other at right angles at O.
Therefore, AO = OC
Hence, O is equidistant from A and C.
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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.
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
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 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 given figure, ABCD is a kite. AB = AD and BC =CD. Prove that the diagona AC is the perpendirular bisector of the diagonal BD.
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.
State and draw the locus of a point equidistant from two given parallel lines.
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.
