Question #22db6

I would call the longer leg `x` so that the shorter becomes `x-7`.
Using Pythagoras we get:
`13^2=x^2+(x-7)^2`
so:
`169=x^2+x^2-14x+49`
`2x^2-14x-120=0`
Using the Quadratic Formula:
`x_(1,2)=(14+-sqrt(196+960))/4=(14+-34)/4`
so:
`x_1=12`
`x_2=-5`
we use the positive one, giving the length of the longer length as `12` and shorter: `12-7=5`

Since this is a right triangle , we can use`color(blue)" Pythagoras's theorem "`

If h represents the hypotenuse and a , b the other 2 sides then

`h^2 = a ^2 + b^2`

here , let the longer arm = x , so shorter one is then (x-7)

substitute values into formula :

hence `13^2 = x^2 + (x-7)^2 = x^2 + x^2-14x + 49`

so `2x^2 - 14x + 49 = 169`

This is a quadratic function so equate to zero to solve.

`2x^2 - 14x - 120 = 0`

factorising: `2(x^2 - 7x - 60 ) = 0`

To factor `x^2-7x-60` require factors of -60 which sum to -7
These are 5 and - 12

`rArr 2(x-12)(x+5) = 0 rArr x = -5 , x = 12`

now x > 0 hence x = 12 and so short leg = x-7 = 12 - 7 = 5

Consider the diagram

Use the pythagoras theorem

`color(brown)(x^2+y^2=h^2`

Where,

`h=`Hypotenuse
`x` `and` `y=` other two sides

`:.(x)^2+(x-7)^2=13^2`

Use the formula `color(brown)((a-b)^2=a^2-2ab+b^2`

`rarrx^2+x^2-14x+49=169`

`rarr2x^2-14x+49=169`

`rarr2x^2-14x+49-169=0`

`rarr2x^2-14x-120=0`

Factor it out

`rarr2(x-12)(x+5)=0`

Remove the `2`

`rarr(x-12)(x+5)=0`

If you solve for it you get,

`x=12,-5`

`x>0:.x=12`

`x=12,x-7=12-7=5`