Question

The longer leg of a right triangle is 6cm longer than the shorter leg and also 1cm longer than the hyptenese. find the perimeter of the triangle.

Answer 1

x - a leg (x > 6);
x - 6 - a leg;
x + 1 - a hypotenuse;
Use the Pythagorean theorem

`(x+1)^2=x^2+(x-6)^2`

Use:
`(a\pm b)^2=a^2\pm2ab+b^2`


`x^2+2x+1=x^2+x^2-12x+36\ \ \ |-x^2-2x-1\\\\x^2-14x+35=0\ \ \ |-35\\\\x^2-14x=-35\\\\x^2-2\cdot x\cdot7=-35\ \ \ |+7^2\\\\x^2-2\cdot x\cdot7+7^2=-35+7^2\\\\(x-7)^2=14\iff x-7=\pm√(14)\ \ \ |+7\\\\x=7-√(14) < 6;\ x=7+√(14)`


`x=7+√(14)\\x-6=1+√(14)\\x+1=8+√(14)`

The perimeter:


`7+√(14)+1+√(14)+8+√(14)=16+3√(14)`