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# 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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