a, SABCD=24.24=576(cm2)SABCD=24.24=576(cm2)

b, ΔNDC&ΔMCBΔNDC&ΔMCBCó:

MB=NC,ˆB=ˆC=90o,BC=DCMB=NC,B^=C^=90o,BC=DC

⇒ΔNDC=ΔMCB(c.g.c)⇒ˆN1=ˆM1⇒ΔNDC=ΔMCB(c.g.c)⇒N1^=M1^

ΔMBCΔMBCCÓ: ˆM1+ˆB+C1=180oM1^+B^+C1=180o, mà góc B=90 độ

⇒ˆM1+ˆC1=90o⇒M1^+C1^=90o, mà ˆN1=ˆM1N1^=M1^

⇒ˆN1+ˆC1=90o⇒N1^+C1^=90o

=> góc NIC=90 độ

MB= AB/2 = 24/2 =12 (cm)

SMBC=12.242=144(cm2)SMBC=12.242=144(cm2)

ΔCIN&ΔCBMΔCIN&ΔCBMCÓ:

ˆCchung,ˆB=ˆI=90oC^chung,B^=I^=90o

⇒ΔCIN∞ΔCBM(g.g)⇒ΔCIN∞ΔCBM(g.g), mà NCBC=12NCBC=12

⇒SCINSCBM=(12)2=14⇒SCINSCBM=(12)2=14

⇒SCIN=SCBM4=1444=36(cm2)⇒SCIN=SCBM4=1444=36(cm2)

Chú ý: ∞∞là kí hiệu đồng dạng