Ducum ca Anianze
tuianze PQR where QR = 5cm LQ-60-
and PQ+ PR = 8 cm.
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Answer:
Draw the base QR = 6 cm.
At point Q draw a ray QX making an ∠ QXR = 60
o
.
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.
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