Question

Construct triangle PQR such that QR=3.5,LQ=40 and LR=60'.
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Answer 1

Answer: Hope this answer helps

Step-by-step explanation:  [ fig. is in the attachment]

Given:  

In ∆PQR, QR= 6 cm, PR-PQ=2 cm & ∠Q= 60°

Steps of construction:

1. Draw the base QR=6 cm.

At Point Q draw a ray QX making an  ∠XQR=60°

Here , PR -PQ= 2cm

PR>PQ

The side containing the base angle Q is less than third side.

 

2. Cut the line segment QS equal to PR-PQ=2 cm , from the ray QX extended on  opposite side of base QR.

3. Join SR and draw its perpendicular bisector ray AB which intersect SR at M.

4. Let P be the intersection point of SX and perpendicular bisector AB. Then join PR.

Thus ∆PQR is the required Triangle.

Answer 2

Answer:

see in the math's work book u will fine the answer I hope