triangle prob: find 2 possible lengths for base of iso. tria

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Math wiz ya rite 09

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The perimeter of an isosceles triangle is 286. Two of the sides are in a ratio 3 : 5. Find the two possible lengths for the base.
 
if the ratio 3:5 is the length of one of the two equal sides to the length of the base, then let the equal sides be 3x and the base 5x

3x + 3x + 5x = 286
11x = 286
x = 26

so, the two equal sides are 78 each and the base is 130.

now ... you figure out the other possibility.
 
skeeter said:
if the ratio 3:5 is the length of one of the two equal sides to the length of the base, then let the equal sides be 3x and the base 5x

3x + 3x + 5x = 286
11x = 286
x = 26

so, the two equal sides are 78 each and the base is 130.

now ... you figure out the other possibility.
Click to expand...
would it be...

6x + 6x +12 x =286

then 24x = 286

then x = 11.916?
 
no ... 6x:12x is not 3:5 (or 5:3 {hint}), is it?

read the original problem again.
 
no ... the other possibility is if the ratio of one of the two equal sides to the base is 5:3

5x + 5x + 3x = 286
 
Math wiz ya rite 09 said:
would it be 6x + 6x + 12 x = 286
then 24x = 286; then x = 11.916?
Click to expand...
Math wiz ya rite 09 said:
oh so 26 * 3 = 78 or 26 * 5 = 130
Click to expand...
Are you just guessing...? If not, it would be helpful if you included your reasoning, like the tutors have been doing for you.

Thank you.

Eliz.
 
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