# 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 = $$66 cm^2.$$