Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are `7/5` of the corresponding sides of the first triangle.

Explanation

Steps of Construction :
1. Draw a line segment BC = 5 cm.
2. Taking B and C as centres draw two arcs of radii 7 cm and 6 cm interecting each other at A.
3. Join BA and CA, `DeltaABC` is the required triangle.
4. From B draw any ray BX downwared making an acute `angleCBX.`
5. Locate seven points `B_(1),B_(2),B_(3),B_(4),B_(5),B_(6)and B_(7)` on BX, such that `BB_(1)=B_(1)B_(2)=B_(2)B_(3)=B_(3)B_(4)=B_(4)B_(5)=B_(5)B_(6)=B_(6)B_(7).`
6. Join `B_(5)C` and from `B_(7)` draw a line `B_(7)M"||"B_(5)C` interecting the extended line segment BC at M.
7. From point M draw `MN"||"CA` intersecting the extended line segment BA at N. Then, `DeltaNBM` is the required triangle whose sides are `7/5` of the corresponding sides of `DeltaABC.`

Justifiction :
By construction,
`B_(7)M"||"B_(5)C`
`therefore" "(BC)/(CM)=5/2`
`Now," "(BM)/(BC)=(BC+CM)/(BC)=1+2/5=7/5`
`therefore" "(BM)/(BC)=7/5`
`Also," "MN"||"CA`
`therefore" "DeltaABC~DeltaNBM`
`and" "(NB)/(AB)=(BM)/(BC)=(MN)/(CA)=7/5`