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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.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Construct a rhombus ABCD with sides of length 5 cm and diagonal AC of length 6 cm. Measure ∠ ABC. Find the point R on AD such that RB = RC. Measure the length of AR.
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.
Draw and describe the lorus in the following cases:
The Iocus of the mid-points of all parallel chords of a circle.
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.
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 triangle ABC, such that AB= 6 cm, BC= 7.3 cm and CA= 5.2 cm. Locate a point which is equidistant from A, B and C.
Using a ruler and compass only:
(i) Construct a triangle ABC with BC = 6 cm, ∠ABC = 120° and AB = 3.5 cm.
(ii) In the above figure, draw a circle with BC as diameter. Find a point 'P' on the circumference of the circle which is equidistant from Ab and BC.
Measure ∠BCP.
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.
