Question

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 1

let, the base of the given isosceles triangle be "x"
Now,
If the base is x, then the length of the legs will be
`9+ (1)/(2)x`
Now,
Perimeter = sum of all sides

`146 = (9+ (1)/(2)x )+(9+ (1)/(2)x)+x`


`146 = 18+x+x`


`146 - 18 = 2x`


`128 = 2x`


`(128)/(2) =x`


`64 =x`

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

legs =
`9+ (1)/(2)x`

=
`9+ (1)/(2)(64)`

=
`9+ 32`

=
`41` inches.

So, the answer to your question is B.

Answer 2

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