Question

Construct a $△PQR$ in which $PQ=6.2$ cm, $\angle Q={60}^o$ and $RQ-PR=3$ cm.

Answer 1

Given $PQ=6.2\text{cm}$, $\angle Q={60}^{\circ }$ & $RQ-PR=3\text{cm}$

1) Draw $PQ$ of length $6.2\text{cm}$ using scale.

2) At point $Q$, draw $\angle Q={60}^{\circ }$ using protractor. Let the ray be $QX$.

3) Cut the line segment $QY=3\text{cm}$ (equal to $RQ-PR=3\text{cm}$) on line $QX$.

5) Join $PY$ and draw a perpendicular bisector $AB$ of $PY$ which cuts the line $QX$ at point $R$.

6) Join $P$ and $R$

Now, $△PQR$ is the required triangle.