Draw a line segment $AB$ of length 7 cm. Using ruler and compass, find a point on AB such that $\frac{AP}{AB} =\frac{3}{5}$
Given: Length of a line $AB=7cm$.
To do: To draw the line AB and by using ruler and compass, to find the point P on the line such that $\frac{AP}{AB} =\frac{3}{5}$.
Solution: Here in the above question given a line segment AB with a length of 7 cm.
First by using ruler draw $AB=7$ cm
Here given that there is a point P on the line AB.
And $\frac{AP}{AB} =\frac{3}{5}$
$\Rightarrow AP=\frac{3}{5} AB$
And we know that $AB=AP+PB$
$\therefore PB=AB-AP=AB-\frac{3}{5} AB=\frac{2}{5} AB$
Then $\frac{AP}{AB} =\frac{\frac{3}{5} AB}{\frac{2}{5} AB} =\frac{3}{2}$
Now follow the steps to find the point P on the given line segment.
1.Draw an acute angle $\angle BAX$ on point A.
2.By using compass mark 5 points equally on ray AX such that
$AA_{1} =A_{1} A_{2} =A_{2} A_{3} =A_{3} A_{4} =A_{4} A_{5}$
3. Draw a line joining $A_{5} \ to\ B.$
4. From $A_{3}$, draw a parallel to$\ A_{5} B$, it intersects AB on P.
Thus P is the required point on the line AB.
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