Steps of construction
1. Draw a line segment BC=6 cm
2. Taking B and C as centres, draw two arcs of radii 4 cm and 9 cm intersecting each other at A.
3. Join BA and CA. `DeltaABC` is the required triangle.
4. From B, draw any ray BX downwards making an acute angle.
5. Mark three points `B_(1),B_(2),B_(3)` on BX, such that `BB_(1)=B_(1)B_(2)=B_(2)B_(3)`.
6. Join `B_(2)C` and from `B_(3)` draw `B_(3)M||B_(2)C` intersecting the extended line segment BC at M.
7. From point M, draw MN||CA intersecting the extended line segment BA to N.
Then, `DeltaNBM` is the required triangle whose sides are equal to `(3)/(2)` of the corresponding sides of the `DeltaABC`.
Justification
Here, `B_(3)M||B_(2)C`
`:. (BC)/(CM)=(2)/(1)`
Now, `(BM)/(BC)=(BC+CM)/(BC)`
`=1+(CM)/(BC)=1+(1)/(2)=(3)/(2)`
Also, MN||CA
`:. DeltaABC ~DeltaNBM`
Therefore, `(NB)/(AB)=(NM)/(AC)=(BM)/(BC)=(3)/(2)`
The two triangles are not congruent because if two triangles are congruent, then they have same shape and same size. Here, all the three angles are same but three sides are not same i.e, one side is different.