Find the area enclosed by the figure (ii) given below. All measurements are in centimeters.

From figure,

BJ = 2 + 2 + 2 + 2 = 8 cm.

Area of rectangle ABJK = l × b

= AB × BJ = 2 × 8

= 16 cm².

From figure,

JH = KH - KI = 6 - 2 = 4 cm.

Area of trapezium FGHI = $\dfrac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\dfrac{1}{2}$ × (FI + GH) × JH

= $\dfrac{1}{2}$ × (2 + 2) × 4

= $\dfrac{1}{2}$ × 4 × 4 = 8 cm².

Area of trapezium CDEF = $\dfrac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\dfrac{1}{2}$ × (CF + DE) × BD

= $\dfrac{1}{2}$ × (2 + 2) × 4

= $\dfrac{1}{2}$ × 4 × 4 = 8 cm².

Total area enclosed = Area of rectangle ABJK + Area of trapezium FGHI + Area of trapezium CDEF

= 16 + 8 + 8

= 32 cm².

Hence, area enclosed by figure = 32 cm².