Divide a line segment of length $8 c m$ internally in the ratio $3 : 4.$

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Solution

STEPS OF CONSTRUCTION:

1.Draw line segment of length $A B = 8 c m .$

2. Draw any Ray AX making acute $∠ B A X$ with $A B$

3. Draw a parallel to $B Y$ parallel to $A X$ by making $∠ A B Y = ∠ B A X$ .

4. Make 3 points $A_{1}, A_{2}, A_{3}$ on $A X$ & 4 points $B_{1}, B_{2}, B_{3}, B_{4}$ on $B Y$ such that

$A A_{1} = A_{1} A_{2} = A_{2} A_{3} = B B_{1} = B_{1} B_{2} = B_{2} B_{3} = B_{3} B_{4}$

5. Join $A_{3} B_{4} .$ Then it intersect $A B$ at a point $P .$

Thus P is the point of dividing AB internally in the ratio $3 : 4$ .

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