Answer: Sides of ΔLMN are LM = 4.4 cm, MN = 4.8 cm, LN = 3.6 cm.
Analysis: ΔABC ~ ΔLMN, so \(\dfrac{AB}{LM}=\dfrac{BC}{MN}=\dfrac{CA}{NL}=\dfrac{5}{4}\). ΔLMN is the smaller triangle.
\(LM = 5.5\times\tfrac45 = 4.4\) cm, \(MN = 6\times\tfrac45 = 4.8\) cm, \(LN = 4.5\times\tfrac45 = 3.6\) cm. Also \(\angle B = \angle M\), so ΔLMN can be drawn with M on B, seg MN on seg BC and seg ML on seg BA.
Steps of construction:
- Draw seg BC = 6 cm. With centre B and radius 5.5 cm, and with centre C and radius 4.5 cm, draw arcs meeting at A. Join AB and AC. This is ΔABC.
- Draw ray BT making an acute angle with BC, on the side opposite to A. Mark 5 equal parts on it: \(B_1, B_2, B_3, B_4, B_5\).
- Join \(B_5C\). Through \(B_4\) draw a line parallel to \(B_5C\), meeting seg BC at N.
- Through N draw a line parallel to CA, meeting seg BA at L.
- ΔLBN (with M at B) is the required ΔLMN: BN = 4.8 cm, BL = 4.4 cm, LN = 3.6 cm.
(ΔLMN can also be drawn separately by SSS using 4.4 cm, 4.8 cm and 3.6 cm.)