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# 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.