Answer:
21cm; 28cm; 14cm
Explanation:
There is no info in the problem/s text which one of the triangle's side is 21 cm. That is why we have to try all possible variants.
Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.
Let O is AK and BM cross point.
Have a look to triangle ABM. AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)
=> triangle ABM is isosceles => AB=AM (1)
1. Let AC=21 So AM=21/2=10.5 cm
So AB=10.5 cm as well. So BC= P-AB-AC=63-21-10.5=31.5 cm
Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill. AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)
2. Let AB=21 So AM=21 and AC=42 .So BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.
3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm
We know (1) that AB=AM so AC=2*AB. So AB+AC=AB+2*AB=3*AB
=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.
Let check if this triangle exists ( if the triangle's inequality fulfills).
BC+AB>AC 21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.
This variant is the only possible solution of the given problem.