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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.
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उत्तर
Steps of constructions:
1) Draw a line BC = 6 cm and CBX = 60°. Cut off AB = 3.5 cm. Join AC, ΔABC is the required triangle.
2) Draw perpendicular bisector of BC and bisector of ∠B.
3) The bisector of ∠B meets bisector of BC at P, therefore BP is the required length,
Where BP = 3.5 cm
4) P is the point which is equidistant from BA and BC, also equidistant from B 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.
Without using set squares or protractor, construct a quadrilateral ABCD in which ∠ BAD = 45°, AD = AB = 6 cm, BC= 3.6 cm and CD=5 cm. Locate the point P on BD which is equidistant from BC and CD.
Construct a Δ XYZ in which XY= 4 cm, YZ = 5 cm and ∠ Y = 1200. Locate a point T such that ∠ YXT is a right angle and Tis equidistant from Y and Z. Measure TZ.
In Δ PQR, s is a point on PR such that ∠ PQS = ∠ RQS . Prove thats is equidistant from PQ and QR.
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
In Δ PQR, bisectors of ∠ PQR and ∠ PRQ meet at I. Prove that I is equidistant from the three sides of the triangle , and PI bisects ∠ QPR .
Describe completely the locus of a point in the following case:
Centre of a circle of varying radius and touching the two arms of ∠ ABC.
Without using set squares or protractor construct a triangle ABC in which AB = 4 cm, BC = 5 cm and ∠ABC = 120°.
(i) Locate the point P such that ∠BAp = 90° and BP = CP.
(ii) Measure the length of BP.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
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.
