• # question_answer Draw a triangle ABC with $BC=6\text{ }cm,\text{ }AB=5\text{ }cm$ and $\angle ABC=60{}^\circ$. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the $\Delta \,ABC$.

 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 $BC'=\frac{3}{4}BC$.