Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \(\angle \)ABC=60° ;. Then construct a triangle whose sides are \(\frac{3}4\) of the corresponding sides of the triangle ABC.

Step 1: Draw a line segment BC=6cm and at point B draw an angle ABC=60°;.
Step 2: Cut AB=5cm and join AC so we obtain \(\angle \) ABC.

Step 3: Draw a ray BX making an acute angle with BC on the side opposite to the A.

Step 4: Locate 4 points \(A_1,A_2,A_3 \)and \(A_4\) on the ray BX so that \(BA_1=A_1A_2=A_2A_3=A_3A_4\).

Step 5: Join \(A_4\) to C.

Step 6: At \(A_3\), draw \(A_3C’ || A_4C\), where C’ is a point on the line segment BC.

Step 7: At C’, draw C’A’ || CA, where A’ is a point on the line segment BA.

\(\therefore \triangle \) A’BC’ is the required triangle.



JUSTIFICATION:

In \(ΔA’BC’ and ΔABC\),

\(A’C’ ∥AC\)

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

In \(ΔBA_3C’ and ΔA_4C\),

\(A_3C’ ∥A_4C \)

\(\frac{BC’}{BC}=\frac{BA_3}{BA_4}=\frac{3}4 \) (By the Basic Proportionality Theorem)

\(∴\frac{BC’}{BC}=\frac{3}4 \) …........(ii)

From(i) and (ii), we get

\(\frac{A’B}{AB}=\frac{3}4 \Rightarrow A’B=\frac{3}4 AB\)

\(\therefore \) Sides of the new triangle formed are \(\frac{3}{4} \) times the corresponding sides of the first triangle.