Construct a trinagle with sides 5 …

 
To construct: To construct a triangle of sides 5 cm, 6 cm and 7 cm and then a triangle similar to it whose sides are of {tex}\frac{7}{5}{/tex} the corresponding sides of the first triangle.
Steps of construction:

1. Draw a triangle ABC of sides 5 cm, 6 cm and 7 cm.
2. From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
3. Locate 7 points B₁, B₂, B₃, B₄, B₅, B₆ and B₇ on BX such that BB₁ = B₁ B₂ = B₂ B₃ = B₃ B₄ = B₄ B₅ = B₅ B₆ = B₆ B₇.
4. Join B₅ C and draw a line through the point B₇, draw a line parallel to B₅ C intersecting BC at the point C'.
5. Draw a line through C' parallel to the line CA to intersect BA at A'.
   Then, A'BC' is the required triangle.
   Justification :
   {tex}\because CA||C'A'{/tex}  [By construction]
   {tex}\therefore {/tex} {tex}\triangle ABC \sim \triangle A'BC'{/tex} [AA similarity]
   {tex}\therefore \frac{{A'B}}{{AB}} = \frac{{A'C'}}{{AC}} = \frac{{BC'}}{{BC}}{/tex} [By Basic Proportionality Theorem]
   {tex}\because {B_7}C'||{B_5}C{/tex} [By construction]
   {tex}\therefore \triangle B{B_7}C' \sim \triangle B{B_5}C{/tex} [AA similarity]
   But {tex}\frac{{B{B_5}}}{{B{B_7}}} = \frac{5}{7}{/tex} [By construction]
   Therefore, {tex}\frac{{BC}}{{BC'}} = \frac{5}{7} \Rightarrow \frac{{BC'}}{{BC}} = \frac{7}{5}{/tex}
   {tex}\therefore \frac{{AB}}{{AB}} = \frac{{A'C'}}{{AC}} = \frac{{BC'}}{{BC}} = \frac{7}{5}{/tex}