हिंदी

Construct a Triangle Abc with Ab = 5.5 Cm, Ac = 6 Cm and ∠Bac = 105° Construct the Locus of Points Equidistant from Ba and Bc Construct the Locus of Points Equidistant from B and C. Mark the Point Which Satisfies the Above Two Loci as P. Measure and Write the Length of Pc.

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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.

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उत्तर

1) Draw a line segment AB of length 5.5 cm.

2) Make an angle m∠BAX = 105° using a protractor

3) Draw an arc AC with radius AC = 6 cm on AX with centre at A.

4) Join BC.

Thus ΔABC is the required triangle.

a) Draw BR, the bisector of ∠ABC, which is the locus of points equidistant from BA and BC.

b) Draw MN, the perpendicular bisector of BC, which is the locus of points equidistant from B and C

c) The angle bisector of ∠ABC and the perpendicular bisector of BC meet at point P. Thus, P satisfies the above two loci.

Length of PC = 4.8 cm

 

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2014-2015 (March)

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