Question

Each leg of an isosceles triangle is 9 inches longer than one-half the length of the base. The perimeter is 146 inches, what is the one leg of the isosceles triangle?

Answer 1

Best way to solve tis for me is to sketch the isosceles triangle then label some algebra expression each side. Try to "interpreter" and "translate" it into algebra expression. then lable it to each sides like this:



now add all sides together and set it equal to 146 (because question said the perimeter is 146 inches)
\[(x) + \left(9 + \dfrac{1}{2}x\right) + \left(9 + \dfrac{1}{2}x\right) = 146\]\[x + \dfrac{1}{2}x + \dfrac{1}{2}x + 9 + 9 = 146\]\[2x + 18 = 146\]\[\boxed{x = 64}\]
Question asks you to find the length of leg, which is \(9 + \dfrac{1}{2}x\) so plug in 64 and solve.
\[9 + \dfrac{1}{2}(64) = 9 + 32 = \boxed{41}\] Hope this helps.