Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are `1 1/2` times the corresponding sides of the isosceles triangle.

Explanation

Setps of Construction :
1. Draw line segment BC = 8 cm.
2. Construct OQ the perpendicular bisector of line segment BC meeting BC at P.
3. Along PO cut off PA = 4 cm.
4. Join BA and CA. So, `DeltaABC` is the required isosceles triangle.
5. From B, draw any ray BX making an acute `angleCBX`.
6. Locate three points `B_(1),B_(2)and B_(3)` on BX such that `BB_(1)=B_(1)B_(2)=B_(2)B_(3).`
7. Join `B_(2)C` and from `B_(3)` draw a line `B_(3)N"||"B_(2)C` intersecting the extended line segment BC at N.
8. From point N, draw `NM"||"CA` meeting MA produced at M. Then, `DeltaMBN` is the required triangle.
Justification :
`because" "B_(3)N"||"B_(2)C" "("by construction")`
`therefore" "(BN)/(BC)=2/1`
`Now," "(BN)/(BC)=(BC+CN)/(BC)=1+(CN)/(BC)=1+1/2=1(1)/(2)`