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

 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 which intersect XB at A. (iv) Join AC. Thus, $\Delta \text{ }ABC$ is obtained. (v) Draw D on BC such that $BD=\frac{3}{4}BC=\left( \frac{3}{4}\times 6 \right)cm=\frac{9}{2}cm=4.5\,cm$ (vi) Draw $DE\parallel CA$, cutting BA at E. Then, $\Delta \text{ }BDE$is the required triangle similar to $\Delta \text{ }ABC$such that each side of $\Delta \text{ }BDE$ is $\frac{3}{4}$ times the corresponding side of$\Delta \text{ }ABC$.