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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