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# Construct the right angled $âˆ†PQR$, where $m\angle Q = 90^{\circ},\ QR = 8cm$ and $PR = 10\ cm$.

**Given: **$m\angle Q = 90^{\circ},\ QR = 8cm$ and $PR = 10\ cm$.

**To do: **To construct the right-angled $\triangle PQR$

**Steps of construction:**

- Draw a line segment $QR$ of length $8\ cm$.
- At the point $Q$, let us draw a perpendicular $QX$ such that $QX\perp QR$.
- Assuming $R$ as a center, let us draw an arc of radius $10\ cm$ which should intersect $QX$ at point $P$.
- Now, let us join $P$ and $R$.

$\triangle PQR$ is the required triangle.

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