Construct the right angled $∆PQR$, 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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