Question 204722: Two isosceles triangles have the same base length. The legs of one of the triangles are twice as long as the legs of the other. Find the lengths of the sides of both triangles if their perimeters are 23 cm and 41 cm.
I haven't done much with this problem because whenever I try to solve I wind up with a weird number. I know that the perimeter of a triangle is P= a+b+c so it would be 41=a+b+c and 23=a+b+c but that is pretty obvious. Please help!
Found 2 solutions by Alan3354, Earlsdon:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Two isosceles triangles have the same base length. The legs of one of the triangles are twice as long as the legs of the other. Find the lengths of the sides of both triangles if their perimeters are 23 cm and 41 cm.
----------------
s = length of legs of smaller triangle
2s = that of taller triangle
b = base
23 = b + 2s
41 = b + 4s
------------- Subtract
-18 = -2s
s = 9
------
b = 5
Answer by Earlsdon(6294)  (Show Source):
You can put this solution on YOUR website! For the first isosceles triangle:
 where b is the base and, as in any isosceles triangle, a = c, so we can write:
 or 
For the second isosceles triangle:
 where b is the base and, as in any isosceles triangle, 2a = 2c, so we can write:
 or 
Now the bases, b, are equal, so we can set these two equations equal:
 Simplify and solve for a. Add 4a to both sides.
 Subtract 23 from both sides.
 Divide both sides by 2.
 and 
First triangle:
a = 9, b = 5, and c = 9
Second triangle:
2a = 18, b = 5, and 2c = 18
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