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

AcademicMathematicsNCERTClass 7

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!
raja
Updated on 10-Oct-2022 13:35:38

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