. Construct a triangle PQR in which PQ = PR = 6cm and QR = 7cm. Draw all its lines
of symmetry.
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(i) Draw a line segment PQ of length 6 cm.
(ii) With P as centre, draw an arc of radius 4.5 cm.
(iii) With Q as centre, draw an arc of radius 7 cm which intersects the previous arc at R.
(iv) Join PR and QR. Then ∆PQR is the required triangle.
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