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Perimeter of one triangle is 11/13 of the perimeter of a similar triangle. The difference of the lengths of two corresponding sides is 1m. What are the lengths of these sides?

Mathematics
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|dw:1358609419748:dw| LET THAT TWO TRIANGLES BE SIMILAR
(a+b+c)/(a'+b'+c')=11/13

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Other answers:

What do I do next? I get what you are doing
u mean difference of all three corresponding sides are 1m or only difference of one corresponding sides is 1m
pair of corresponding sides.
So, all a'-a=1 ====>a'=a+1 b'-b=1====>b'=b+1 c'-c=1====>c'=c+1
yep
difference between one corresponding sides is 1m
U mean only a'-a=1 NOT all
yes
that is correct
(a+b+c)/(a'+b'+c')=11/13 13a+13b+13c=11a'+11b'+11c' (13a-11a') + (13b-11b') + (13c-11c')=0 a (13-11a'/a) +b (13-11b'/b) +c(13-11c'/c)=0 Now, a'/a =b'/b =c'/c = (1+a)/a So, (a+b+c) {13-11(1+a)/a}
Similarly, a'(13a/a' -11) + b'(13b/b' -11) +c' (13c/c' -11)=0 (a'+b'+c') { 13a/(a+1) -11}=0
So, (a+b+c) {13-11(1+a)/a}/(a'+b'+c') { 13a/(a+1) -11}=0 {13 -11(1+a)/a}=0 13a-11-11a=0 2a=11 a=5.5
So, is it 5.5 and 6.5
NOTE: I assumed a'-a=1
Thanks
welcome

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