Construct an angle ABC = 90∘. Locate a point P which is 2.5 cm from AB and 3.2 cm from BC.
Steps of construction :
1. Draw ∠ABC=90∘
2. From AB, cut BD = 3.2 cm.
3. Through point C, draw CH ⊥ BC. From CH, cut CE = 3.2. Join DE. Now DE is a line parallel to BE and at a distance of 3.2 cm from BC.
4. From BC cut BM = 2.5 cm.
5. Through point A, draw AK ⊥ 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.
6. DE and MN intersect each other at P. Thus P is the required point which is 2.5 cm from AB and 3.2 from BC.