Question

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Answer 1

Steps of Construction:
Step I: BC = 8 cm is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersects BC at O.
Step III: At a distance of 4 cm, a point A is marked on the perpendicular bisector of BC.
Step IV: AB and AC are joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step VI: 3 points $B_1,B_{2}and{B}_3$ are marked on BX.
Step VII: $B_2$ is joined with C to form $B_{2}C$.
Step VIII: $B_3{C}^{\prime }$ is drawn parallel to $B_{2}C$ and C'A' is drawn parallel to CA.
Thus, A'BC' is the required triangle formed.
Justification:
$\Delta A{B}^{\prime }{C}^{\prime }$ ~ $\Delta ABC$ by AA similarity condition.
$\therefore \frac{AB}{A{B}^{\prime }}=\frac{BC}{B^{\prime }{C}^{\prime }}=\frac{AC}{A{C}^{\prime }}$
Also,
$\frac{AB}{A{B}^{\prime }}=\frac{A{A}_2}{A{A}_3}=\frac{2}{3}$
$\Rightarrow A{B}^{\prime }=\frac{3}{2}AB,B^{\prime }{C}^{\prime }=\frac{3}{2}$
$BCandA{C}^{\prime }=\frac{3}{2}AC$