Question

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

Answer 1

Steps of Construction:
Step I: BC = 6 cm is drawn.
Step II: At point B, AB = 5 cm is drawn making an $\angle ABC={60}^{\circ }$ with BC.
Step III: AC is joined to form $\Delta ABC$.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 are marked at equal distances on BX.
Step VII: $B_3$ is joined with C' to form $B_3{C}^{\prime }$.
Step VIII: C'A' is drawn parallel CA.
Thus, A'BC' is the required triangle.
Justification:
$\angle A={60}^{\circ }$ (Common)
$\angle C=\angle C^{\prime }$
$\Delta A{B}^{\prime }{C}^{\prime }$ ~ $\Delta ABC$ by AA similarity condition.
$\therefore \frac{AB}{A{B}^{\prime }}=\frac{BC}{B^{\prime }{C}^{\prime }}=\frac{AC}{A{C}^{\prime }}$
also,
$\frac{AB}{A{B}^{\prime }}=\frac{AA3}{AA4}=\frac{4}{3}$
$\Rightarrow A{B}^{\prime }=\frac{3}{4}AB,B^{\prime }{C}^{\prime }=\frac{3}{4}BCandA{C}^{\prime }=\frac{3}{4}AC$