Draw a line QR=3cm.
Construct an angle of 45° at Q.
With Q as center and 2cm as radius cut an arc on QX at S.
Join SR and bisect it.
Name the intersecting point of QX and the bisector P.
PQR is the required Triangle.
QR and ∠PQR =45°.
P lies on the perpendicular bisector of SR.
PS = PR
QS = PQ – PS
= PQ -PR (as PS=PR)
Thus The construction is justified.
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