Steps of constructions:
1. Draw a line segment BC = 7 cm, and draw ∠YBC = 45°.
2. Draw ∠ZCB = 30° ; and let CZ and BY intersect at A. Therefore, ΔABC is ready.
[ ∠C = 30°, as ∠A = 105° ; ∠B = 45° and we know, By angle sum property of triangle,
∠A + ∠B + ∠C = 180°
⇒ 45° + 105° + ∠C = 180°
⇒ ∠C = 30° ]
3. Below BC, make an acute angle ∠CBX.
4. Along BX mark 4 (greater of 3 and 4 in 3/4) points B1, B2, B3,B4 at equal distances.
5. Join B4C
6. From point B3, make a line B3C' parallel to B4C , meeting BC at C' [Choosing B3 because we need a 3:4 ratio]
7. From point B draw a line C'A' parallel to CA.
8.A'BC' is the triangle, whose sides are three - fourth of ∆ABC
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