The area of ABCD is.

In Δ ADC, \(\angle\)D =90°

By using pythagoras theorem:

AC² = AD² + DC²

AC² = 6² + 8² = 36 + 64 = 100

AC = 10 cm

Now, 

area of Δ ADC = \(\frac{1}{2}\) AD × DC = \(\frac{1}{2}\) × 6 × 8 = 24 cm²

area of Δ ABC = \(\frac{1}{2}\) AC × CB = \(\frac{1}{2}\) × 10 × 7 = 35 cm²

area of quadrilateral ABCD = area of Δ ADC + area of Δ ABC = 24 + 35 = 59 cm²