1) AB=12/3=4cm; S=4*12=48cm^2
2) Sabcd=Sdbc; Sabcd=12+12=24cm^2
3) AH=BH; AD=6+2AH, 2AH+6=12, AH=3cm=BH
По теореме Пифагора AB=9cm
S=12+6/2+3=12cm^2
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Answers & Comments
S = AB×BC = 12×4 = 48
2. Sabsd = 2×Sabd = 2×12 = 24
3. BH = 6
S=0,5×BC×AD×BH = 216
1) AB=12/3=4cm; S=4*12=48cm^2
2) Sabcd=Sdbc; Sabcd=12+12=24cm^2
3) AH=BH; AD=6+2AH, 2AH+6=12, AH=3cm=BH
По теореме Пифагора AB=9cm
S=12+6/2+3=12cm^2