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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.
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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(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.
