Construct a triangle $ \mathrm{PQR} $ in which $ \mathrm{QR}=6 \mathrm{~cm}, \angle \mathrm{Q}=60^{\circ} $ and $ \mathrm{PR}-\mathrm{PQ}=2 \mathrm{~cm} $.
Given:
$QR=6\ cm, \angle Q=60^o$ and $PR-PQ=2\ cm$.
To do:
We have to construct a $\triangle PQR$.
Solution:
Steps of construction:
(i) Let us draw a line segment $QR$ of length $6\ cm$.
(ii) Now, construct an angle $RQX$ such that $\angle RQX=60^o$
(iii) Now, by taking a measure of $PR-PQ=2\ cm$ with the compasses, let us draw an arc from point $Q$ and mark it as Point $Y$. Since $PQ-PR$ is negative, the line $QY$ will be below the line segment $QR$.
(v) Now, by taking compasses let us draw another arc from point $Q$ on $QX$.
(vi) Now, let us join $YR$. Then by taking compasses let us draw a perpendicular bisector of the line $YR$ and mark the intersection point of the bisector with the ray $QX$ as $P$
(v) Now, let us join $PR$. Therefore, $PQR$ is the required triangle.
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