a. Ta tính được AM = MA = 30cm; DN = 30cm; NC = 45 cm.
\(S_{ABCD}=\frac{\left(45+75\right).60}{2}=3600\left(cm^2\right).\)
\(S_{ABM}=\frac{1}{2}.AM.AB=\frac{1}{2}.30.45=675\left(cm^2\right).\)
\(S_{MND}=\frac{1}{2}.MD.DN=\frac{1}{2}.30.30=450\left(cm^2\right).\)
\(S_{BNC}=\frac{1}{2}.NC.AD=\frac{1}{2}.45.60=1350\left(cm^2\right).\)
Vậy \(S_{BMN}=S_{ABCD}-S_{ABM}-S_{MND}-S_{BCN}=3600-675-450-1350=1125\left(cm^2\right).\)
b. Gọi chiều cao của tam giác BMN là h, chiều cao tam giác BCN là k. Ta có:
\(S_{BMC}=S_{BPM}+S_{BPC}=\frac{1}{2}.BP.h+\frac{1}{2}BP.k=BP.\frac{h+k}{2}.\)
\(S_{MNC}=S_{MNP}+S_{NPC}=\frac{1}{2}.NP.h+\frac{1}{2}NP.k=NP.\frac{h+k}{2}.\)
Lại có \(S_{MNC}=\frac{1}{2}.NC.MD=\frac{1}{2}.45.30=675\left(cm^2\right).\)
\(\Rightarrow S_{BMC}=S_{ABCD}-S_{ABM}-S_{ADC}=3600-1800=1800\left(cm^2\right).\)
Vậy thì \(\frac{BP}{PN}=\frac{S_{BMC}}{S_{MNC}}=\frac{1800}{675}=\frac{8}{3}.\)
