Question

Without using set-squares or protractor construct:
construct a $$\triangle ABC,\,AB=5.5cm,\,BC=3.2cm,\,\&\,AC=4.8cm$$
Find the locus which is equidistant from sides $$BC\,\&\,AC$$ and also at a distance of $$2.5cm$$ from $$B$$
Mark the point of intersection of the loci with the letter $$P$$ and measure $$PC$$.

Answer 1

Steps of construction;-
1. draw a line segment $$BC=3.2cm$$
2. taking $$B$$ as centre and $$radius=5.5cm$$ construct an arc
3. taking $$C$$ as centre and $$radius=4.8cm$$ draw an arc to cut the previous arc at $$A$$
4. join $$AB\,\&\,AC$$
5. Construct the bisector of $$\angle BCA.$$
6. Taking $$B$$ as centre and $$2.5$$ cm radius, construct an arc which intersects the angle bisector of $$\angle BCA.$$ at $$P$$ and $$P’$$.
Here $$P$$ and $$P’$$ are the two loci which satisfy the given condition.
By measuring $$CP$$ and $$CP’$$
$$CP = 3.6$$ cm and $$CP’ = 1.1$$ cm.