Question

Construct $△$PQR such that PQ = 9.2 cm , $\angle$PRQ = 112$°$ ,RK is an altitude,RK = 2.4cm.

Answer 1

Steps of construction:
(i) Draw a line segment PQ = 9.2 cm.
(ii) Through P, draw PY such that $\angle$QPY= 112$°$.
(iii) Draw PX$\perp$PY.
(iv) Draw the perpendicular bisector of PQ intersecting PX at O and PQ at H.
(v) With O as the centre and OP as the radius, draw a circle.
(vi) The minor arc PWQ of the circle contains the vertical 112$°$.
(vii) With H as the centre, cut off HS = 2.4 cm.
(viii) Draw a line RR' parallel to PQ passing through S.
(ix) With R as the centre, draw a line parallel to SH intersecting PQ at K.
(x) Join RQ and RP

Thus, $△$PQR or $△$PQR' is the required triangle.