Question

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?

Answer 1

LET THAT TWO TRIANGLES BE SIMILAR

Answer 2

(a+b+c)/(a'+b'+c')=11/13

Answer 3

@hero

Answer 4

What do I do next? I get what you are doing

Answer 5

u mean difference of all three corresponding sides are 1m or only difference of one corresponding sides is 1m

Answer 6

pair of corresponding sides.

Answer 7

So, all
a'-a=1 ====>a'=a+1
b'-b=1====>b'=b+1
c'-c=1====>c'=c+1

Answer 8

yep

Answer 9

difference between one corresponding sides is 1m

Answer 10

U mean only a'-a=1
NOT all

Answer 11

yes

Answer 12

that is correct

Answer 13

(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}

Answer 14

Similarly,
a'(13a/a' -11) + b'(13b/b' -11) +c' (13c/c' -11)=0
(a'+b'+c') { 13a/(a+1) -11}=0

Answer 15

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

Answer 16

So, is it 5.5 and 6.5

Answer 17

NOTE: I assumed a'-a=1

Answer 18

Thanks

Answer 19

welcome