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  • Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.

Q 1.     Construct \Delta DEF such that DE = 5 cm, DF = 3 cm and m\angle EDF=90^{\circ}.

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\Delta DEF such that DE = 5 cm, DF = 3 cm and m\angle EDF=90^{\circ}:

1. Draw a line DE = 5cm 

2. Draw a line DX making 90 degrees with DE.

3. cut the length DF =3cm using the compass on the line DX

4. Join E and F

DEF is the required triangle

Posted by

Pankaj Sanodiya

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