Construct ∆ PQR, in which angle Q = 70° , angle R = 80° and PQ+QR+PR = 9.5 cm
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SSC (English Medium) 9th Standard
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Diagram
Construct Δ PQR, in which ∠Q = 70°, ∠R = 80° and PQ + QR + PR = 9.5 cm.
SOLUTION
Steps of construction:
(1) Draw seg AB of 9.5 cm length.
(2) Draw a ray making angle of 35° at point A.
(3) Draw another ray making an angle of 40° at point B.
(4) Name the point of intersection of the two rays as P.
(5) Draw the perpendicular bisector of seg AP and seg BP.
Name the points as Q and R respectively where the perpendicular bisectors intersect line AB.
(6) Draw seg PQ and seg PR.
Therefore, △PQR is the required triangle.
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