Question
Construct an angle $ABC = 90^\circ$ . Locate a point $P$ which is $2.5 \ cm$ from $AB$ and $3.2 \ cm$ from $BC$.
✓
Answer
Steps of construction :
- Draw $\angle ABC = 90^\circ$
- From $AB$, cut $BD = 3.2 \ cm.$
- Through point $C,$ draw $CH\perp BC$. From $CH$, cut $CE = 3.2$. Join $DE$. Now $DE$ is a line parallel to $BC$ and at a distance of $3.2 \ cm$ from $BC.$
- From $BC$ cut $BM = 2.5 \ cm.$
- Through point $A$, draw $AK \perp AB$. From $AK$ cut $AN = 2.5 \ cm$. Join $NM$. Therefore $NM$ is parallel to $AB$ and at a distance of $2.5 \ cm$ from $AB$.
- $DE$ and $MN$ intersect each other at $P$. Thus $P$ is the required point which is $2.5 \ cm$ from $AB$ and $3.2 \ cm$ from $BC$.
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