Construct $∆$ LMN such that MN = 6.2 cm, M =50 $°$ and LN $-$ LM = 2.4 cm.

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Solution

Steps of construction:

1. Draw MN = 6.2 cm.
2. Draw ray MX, such that $∠ NMX = 50 °$ .
3. With M as centre and radius 2.4 cm, draw an arc intersecting ray MX in point D.
4. Join D and N.
5. Draw the perpendicular bisector of seg DN, intersecting seg MX(produced) at point L.
6. Join L and N.
7. Hence, $△$ LMN is the required triangle.

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∆

△ M N O

N O = 6.2 c m

∠ N = 50^{\circ}

M O - M N = 2.4 c m .

For constructing $Δ$ LMN exactly identical to $Δ$ ABC, choose from the options, the measurements required.

L M N

∠ M = 90^{\circ}, M N = 4 c m

L N = 6.2 c m

△ L M N

M N = 7 c m

∠ L M N = 75^{\circ}

L M + L N = 9 c m .

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