Ta có: ΔCAB vuông tại A
=>\(CA^2+AB^2=CB^2\)
=>\(CB^2=36^2+48^2=3600\)
=>\(CB=\sqrt{3600}=60\left(cm\right)\)
Xét ΔCAB có MN//AB
nên \(\dfrac{MN}{AB}=\dfrac{CN}{CB}=\dfrac{CM}{CA}\)
=>\(\dfrac{CN}{60}=\dfrac{CM}{48}=\dfrac{24}{36}=\dfrac{2}{3}\)
=>\(CN=60\cdot\dfrac{2}{3}=40\left(cm\right);CM=\dfrac{2}{3}\cdot48=32\left(cm\right)\)
Ta có: CM+MA=CA
=>MA+32=48
=>MA=16(cm)
Ta có: MN//AB
AB\(\perp\)AC
Do đó: MN\(\perp\)AC
=>ΔCMN vuông tại M
=>\(S_{CMN}=\dfrac{1}{2}\cdot24\cdot32=12\cdot32=384\left(cm^2\right)\)
Vì ΔABC vuông tại A
nên \(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC=\dfrac{1}{2}\cdot36\cdot48=864\left(cm^2\right)\)
Ta có: \(S_{CMN}+S_{AMNB}=S_{ABC}\)
=>\(S_{AMNB}+384=864\)
=>\(S_{AMNB}=480\left(cm^2\right)\)
