Answer:
Steps of construction - (i) Draw a line segment \[BC=6\text{ }cm\]. (ii) Construct \[\angle XBC=60{}^\circ \]. (iii) With B as centre and radius equal to 5 cm, draw an arc intersecting XB at A. (iv) Join AC. Thus, \[\Delta \text{ }ABC\]is obtained. (v) Draw an acute angle \[\angle CBY\] below of B. (vi) Mark 4-equal parts on BY as \[{{B}_{1}},{{B}_{2}},{{B}_{3}}\] and \[{{B}_{4}}\]. (vii) Join \[{{B}_{4}}\]to C. (viii) From \[{{B}_{3}}\] draw a line parallel to \[{{B}_{4}}C\]intersecting BC at C. (ix) Draw another line parallel to CA from C?, intersecting AB at A?. (x) \[\Delta \,A'BC'\] is required triangle which is similar to \[\Delta \,ABC\] such that \[BC'=\frac{3}{4}BC\].
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