Question

Construct a triangle ∆PQR given that QR=3cm ∠Q=45 degree and PQ+QR=6cm.
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Answer 1

$\underline{\underline{\maltese\: \: \textbf{\textsf{Answer}}}}$

steps of constructions:

1. draw a line segment QR = 3cm

2. at Q draw an angle MQR= 45 deg

3. join MR.

4. now draw the perpendicular bisector of MR. let it intersect MQ at P.

5. join PR.

thus PQR is the required triangle.

now constructing the triangle P'QR' ~ triangle PQR with scale factor 3/2

steps of construction:

1. now extend QR to QR' such that QR': QR = 3:2

2. from R' draw a line parallel to PR let it intersect MQ at P'.

thus P'QR' is the required similar triangle.