Question

Construct a triangle $PQR$ in which $QR=6cm,\angle Q={60}^o$ and $PR-PQ=2cm$.

Answer 1

1) Draw the base $QR=6cm$.

At point Q draw a ray $QX$ making an $\angle XQR={60}^{\circ }$.

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 $PR-PQ=2cm$, from the ray $QX$ extended on opposite side of base $QR$.

3) Join $SR$ and draw its perpendicular bisector ray $LM$ which intersect $SR$ at $M$.

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

Thus, $△PQR$ is the required triangle.