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प्रश्न
Construct a ΔABC in which CA = 6 cm, AB = 5 cm and ∠BAC = 45°. Then construct a triangle whose sides are `3/5` of the corresponding sides of ΔABC.
उत्तर
Steps of construction:
1. Draw AB = 5 cm. With A as the centre, draw ∠BAC= 45°. Join BC, ∠ABC is thus formed.
2. Draw AX such that ∠BAX is an acute angle.
3. Cut % equals arcs AA1, A1A2, A2A2, A2A4 and A4A3.
4. Join A3 to B and draw a line through A3 parallel to A3B which meets AB at B
Here, AB' = `3/5` AB
5. Now draw a line through B' parallel to BC which joins AC at C'
Here, BC' = 3/5 BC and AC = `3/5` AC
Thus, AB'C is the required triangle.
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