(1)由图形得:C1=2(a+b+c+b)=2a+4b+2c;C2=2(a-c+b+3c)=2a+2b+4c,
C1-C2=2a+4b+2c-2a-2b-4c=2(b-c),
∵b>c,∴2(b-c)>0,
则C1>C2;
(2)由图形得:S1=a2+b2;S2=2ab,
∴S1-S2=a2+b2-2ab=(a-b)2>0,
∴S1>S2.
(1)由图形得:C1=2(a+b+c+b)=2a+4b+2c;C2=2(a-c+b+3c)=2a+2b+4c,
C1-C2=2a+4b+2c-2a-2b-4c=2(b-c),
∵b>c,∴2(b-c)>0,
则C1>C2;
(2)由图形得:S1=a2+b2;S2=2ab,
∴S1-S2=a2+b2-2ab=(a-b)2>0,
∴S1>S2.