Question

Draw a triangle $ABC$ with $BC=6cm,AB=5cm$ and $\angle ABC={60}^o$ then construct a triangle whose sides are $4/5$ of the sides of the $△ABC$. Justify.

Answer 1

Draw base BC with scale 6cm.

Draw $\angle B={60}^{\circ }$

Draw an arc with scale 5cm from B and mark it as A.

Join AC.

$△ABC$ is the required triangle.

Draw any ray BX making an acute angle with BC on the opposite side of vertex A.

Mark 5 points (the greater of 4 and 5 in $\frac{4}{5})$ $B_1,B_2,B_3,B_4,B_5$ on BX so that $B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5$

Join $B_{5}C$ and draw another line parallel to $B_{5}C$ from $B_4$ to C'

Take it to A and mark it A'.

By construction,

$\frac{B{C}^{\prime }}{BC}=\frac{B{B}_4}{B{B}_5}$

SO, $\frac{A^{\prime }B}{AB}=\frac{A^{\prime }{C}^{\prime }}{AC}=\frac{B{C}^{\prime }}{BC}=\frac{4}{5}$

Thus, construction is justified.