Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5} \) of the corresponding sides of the first triangle.

Steps of construction by the given data are:
Step 1. Draw a \(\triangle \) ABC with sides as AB=5 cm, BC= 7 cm, AC= 6cm.

Step 2. At point B, draw an acute angle CBX below BC.


Step 4. Join \(B_5\) and C.

Step 5.Draw \(B_7C'\) parallel to \(B_5C\), where C` is a point on extended line BC.

Step 6. Draw A'C' parallel to AC, where A' is a point on extended line BA.

Hence,A’BC’ is the required triangle.



JUSTIFICATION:

In \(\triangle \) ABC and \(\triangle \) A’BC’,

AC || A’C’

\(\frac{AB}{A’B}=\frac{BC}{BC’}\)
(By the basic Proportionality Theorem)...(i)

In \(ΔBB_5C \)and \(ΔBB_7C’\),

\(B_5C || B_7C’\)

\(\frac{BC}{BC’}=\frac{BB_5}{BB_7}=\frac{5}7\) (By the basic Proportionality Theorem)...(ii)

Equating (i) and (ii) \(\frac{AB}{A’B}=\frac{BB_5}{BB_7} \Rightarrow \frac{AB}{A’B}=\frac{5}7 \)

\(\therefore A’B=\frac{7}5AB\)

\(\therefore \) Sides of new triangle is \(\frac{7}5 \)times the corresponding sides of the first triangle.