Myres245
Answered

Each leg of an isosceles triangle is 9 inches
longer than one-half the length of the base. If
the perimeter of the isosceles triangle is
146 inches, what is the length of one leg of the
isosceles triangle?
A 64 in.
B 41 in.
C 50 in.
D 82 in.



Answer :

ArchimedesEleven
B) 41"
If the legs of 41" are 9" less than half the base, the base is:
(41-9) 2 = 64
All sides of the triangle added together (perimeter):
41" + 41" + 64" = 146"
Hope this helps! ~ArchimedesEleven
rabinshrestha41
let, the base of the given isosceles triangle be "x"
Now,
If the base is x, then the length of the legs will be [tex]9+ \frac{1}{2}x [/tex]
Now, 
Perimeter = sum of all sides
[tex]146 = (9+ \frac{1}{2}x )+(9+ \frac{1}{2}x)+x[/tex]

[tex]146 = 18+x+x[/tex]

[tex]146 - 18 = 2x[/tex]

[tex]128 = 2x[/tex]

[tex] \frac{128}{2} =x[/tex]

[tex]64 =x[/tex]

Now, substitute the value of x in the equation for legs of the triangle,

legs = [tex]9+ \frac{1}{2}x[/tex]
     
       = [tex]9+ \frac{1}{2}(64)[/tex]
   
       = [tex]9+ 32[/tex]
 
       = [tex]41[/tex] inches.

So, the answer to your question is B.


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