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# Construct $âˆ†DEF$ such that $DE = 5\ cm$, $DF = 3\ cm$ and $\angle EDF = 90^{\circ}$.

**To do: **To construct $∆DEF$ such that $DE = 5\ cm$, $DF = 3\ cm$ and $\angle EDF = 90^{\circ}$.

** Steps of construction :**

- Let us draw a line segment $DE=5\ cm$.
- At the point $D$, let us draw a ray $DX$ making an angle of $90^{\circ}$ with $DE$ such that $\angle EDX=90^{\circ}$.
- Let us assume $D$ as the center and draw an arc of radius $3\ cm$ that intersects $DX$ at the point $F$.
- Let us join $EF$ to get the required triangle.

$\triangle DEF$ is the required triangle.

Hence constructed!

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