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Question
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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Solution
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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