In trapezium ABCD, AB // DC. M is mid-point of AD and N is mid-point of BC.

(i) If AB = 8 cm and DC = 11 cm, find MN.

(ii) If AB = 5.7 cm and MN = 6.2 cm, find DC.

(i) Given: ABCD is a trapezium where AB // DC. M is mid-point of AD and N is mid-point of BC.

Construction: Draw diagonal BD which intersect MN at Q.

In triangle BDC, N is mid-point of BC and CD ∥ QN (as CD ∥ MN)

By the converse of mid-point theorem,

∴ Q is mid-point of BD.

⇒ QN = $\dfrac{1}{2}$ DC

Similarly, QM = $\dfrac{1}{2}$ AB

Adding above two equation,

⇒ QN + QM = $\dfrac{1}{2}$ (AB + DC)

⇒ MN = $\dfrac{1}{2}$ (AB + DC)

= $\dfrac{1}{2}$ (11 + 8)

= $\dfrac{1}{2}$ x 19

= 9.5 cm

Hence, the value of MN = 9.5 cm.

(ii) MN = $\dfrac{1}{2}$ (AB + DC)

⇒ 6.2 = $\dfrac{1}{2}$ (5.7 + DC)

⇒ 6.2 x 2 = 5.7 + DC

⇒ 12.4 = 5.7 + DC

⇒ DC = 12.4 - 5.7

⇒ DC = 6.7

Hence, the value of DC = 6.7 cm.