Question

Cho hình chữ nhật ABCD có AB = 42 cm, AD = 27 cm. Biết AM =  MD, NB =  AN. Tính diện tích tam giác MNC

Answer 1

\(S_{ABCD}=42\cdot27=1134\left(cm^2\right)\)
\(S_{ANM}=\left[\left(42:2\right)\cdot\left(27:2\right)\right]:2=141,75\left(cm^2\right)\)
\(S_{NBC}=\left[\left(42:2\right)\cdot27\right]:2=283,5\left(cm^2\right)\)
\(S_{MDC}=\left[\left(27:2\right)\cdot42\right]:2=283,5\left(cm^2\right)\)
\(S_{MNC}=1134-141,75-283,5-283,5=425,25\left(cm^2\right)\)
Vậy \(S_{MNC}=425,25\left(cm^2\right)\)

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