Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}3\)times the corresponding sides of the given triangle.

Steps of Construction the figure by the given data:
Step 1. Construct a \(\triangle \) ABC, with BC = 4 cm, CA = 3 cm and ∠BCA = 90°.

Step 2. By making an acute angle with BC, draw a ray BX .

Step 3. Mark five points \(B_1, B_2, B_3, B_4 \)and \(B_5 \)on BX, such that \(BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5\) and join \(B_3C\).

Step 6. Through \(B_5\), draw \(B_5C’\) parallel to \(B_3C\) intersecting BC produced at C’.

Step 7. Through C’, draw C’A’ parallel to CA intersecting AB produced at A’.

Thus, \(\triangle \) A’BC’ is the required right triangle.



JUSTIFICATION:

In \(ΔA’BC’ and ΔABC\),

\(∠ABC=∠ A’BC’ \)(common)

\(∠ACB=∠ A’C’B\) (corresponding angles)

\(∴ ΔABC∼ΔA’BC’ \) (By AA similarity)

\(∴ \frac{AB}{A’B}=\frac{AC}{A’C’}=\frac{BC}{BC’}\)

But, \(\frac{BC}{BC’}=\frac{BB_3}{BB_5}=\frac{3}5 ∴ \frac{BC’}{BC}=\frac{5}3 \)

\(⇒ \frac{A’B}{AB}=\frac{A’C’}{AC}=\frac{BC’}{BC}=\frac{5}3 .\)

\(\therefore \) Sides of the new triangle formed are \(\frac{5}{3} \) times the corresponding sides of the first triangle.