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

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#### Solution

The rough sketch of the required ΔDEF is as follows.

The steps of construction are as follows.

1) Draw a line segment DE of length 5 cm.

2) At point D, draw a ray DX making an angle of 90° with DE.

3) Taking D as centre, draw an arc of 3 cm radius. It will intersect DX at point F.

4) Join F to E. ΔDEF is the required triangle.

Is there an error in this question or solution?

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