Construct a triangle $A B C$ in which $A B + A C = 5.6 c m, B C = 4.5 c m$ and $∠ B = 45^{\circ}$

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Solution

1. Construct a line segment $B C$ of $4.5 c m$
2. Construct a $45^{o}$ angle at $B$
3. Take the length of $P B$ as $5.6 c m$ .
4. Join $P C$
5. Construct a bisector at $P C$ . The line will meet $B P$ at point $A$ . Join $A C$
We get the required triangle.

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