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Question
Construct a ti.PQR, in which PQ=S. 5 cm, QR=3. 2 cm and PR=4.8 cm. Draw the locus of a point which moves so that it is always 2.5 cm from Q.
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Solution
Steps of oonstruction:
(i) Draw PQ = 5.5 cm
(ii) With P as centre and radius 4.8 cm draw an arc.
(iii) With Q as centre and radius 3.2 cm cut another arc which meets the first arc at R. Join PR and QR. PQR is the required triangle.
(iv) Draw perpendicular bisector of PR.
(v) Q as centre and radius as 2.5 cm, draw an arc which intersects the perpendicular bisector of PR at O.
O is the required point which is at a distance of 2.5 cm from Q.
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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.
