In figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4

Draw AL ⊥ DC, BM ⊥ DC then,

AL = BM = 4 cm and LM = 7 cm.

In Δ ADL, we have

$A D^{2}=A L^{2}+D L^{2}$

$\Rightarrow 25=16+\mathrm{DL}^{2}$

⇒ DL = 3 cm

Similarly,

$\mathrm{MC}=\sqrt{\mathrm{BC}^{2}-\mathrm{BM}^{2}}=\sqrt{25-16}=3 \mathrm{~cm}$

∴ x = CD = CM + ML + LD = (3 + 7 + 3) cm = 13 cm

$\operatorname{ar}(\operatorname{trap} \cdot \mathrm{ABCD})=1 / 2(\mathrm{AB}+\mathrm{CD}) \times \mathrm{AL}=1 / 2(7+13) \times 4 \mathrm{~cm}^{2}=40 \mathrm{~cm}^{2}$