Question

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

Answer 1

Steps of construction -
(i) Draw a line segment BC = 6 cm.
(ii) Construct $\angle XBC={60}^o$.
(iii) With B as centre and radius equal to 5 cm, draw an are intersecting XB at A.
(iv) Join AC. Thus, $\Delta ABC$ 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) $\Delta A^{\prime }B{C}^{\prime }$ is required triangle which is similar to $\Delta ABC$ such that $B{C}^{\prime }=\frac{3}{4}BC.$