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Q.11 Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM \(\perp \)AC, DN \(\perp \)AC.

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 = \(66 cm^2. \)