Solution:-

From the question it is given that,

ABCD is a parallelogram, AM ⊥ DC and AN ⊥ CB

AM = 6 cm

AN = 10 cm

The area of parallelogram ABCD is 45 cm²

Then, area of parallelogram ABCD = DC × AM = BC × AN

45 = DC × 6 = BC × 10

(i) DC = 45/6

Divide both numerator and denominator by 3 we get,

= 15/2

= 7.5 cm

Therefore, AB = DC = 7.5 cm

(ii) BC × 10 = 45

BC = 45/10

BC = 4.5 cm

(iii) Now, consider ∆ADM and ∆ABN

∠D = ∠B … [because opposite angles of a parallelogram]

∠M = ∠N … [both angles are equal to 90^o]

Therefore, ∆ADM ~ ∆ABN

Therefore, area of ∆ADM/area of ∆ABN = AD²/AB²

= BC²/AB²

= 4.5²/7.5²

= 20.25/56.25

= 2025/5625

= 81/225

= 9/25

Therefore, area of ∆ADM : area of ∆ANB is 9: 25