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Answer :
Area of \(\triangle ABC = \frac{1}{2} × b×h = \frac{1}{2}×22×3 = 33^2 \)
Area of \(\triangle ADC = \frac{1}{2} × b×h = \frac{1}{2}×22×3 = 33^2 \)
Area of the quadrilateral ABCD
= Area of \(\triangle \)ABC + Area of \(\triangle \)ADC
\( = 33 cm^2 + 33 cm^2 = 66 cm^2 \)
Hence, the required area = \(66 cm^2. \)