Question

The hypotenuse of a right-angled triangle is (2y - 1)cm long. The other two sides are x cm and (y+5)cm in length. the perimeter of the triangle is 30cm, what are possible values of x and y?

please help!

Answer 1

`x = -24 and y = 50/3`
or
`x= 5 and y = 7`

Explanation:

Given : Hypotenuse = `(2y-1)` cm

Side 1 (say height ) = `x` cm and

Side 2 (say base) = `(y + 5)` cm

Perimeter = sum of all three sides = 30 cm

`2y-1+x+y+5 =30`

`3y +x +4 = 30`

`3y +x = 30-4`

`3y +x = 26` -------- let this be equation (1)

This gives `x= 26-3y`

According to Pythagoras theorem,

`(Hypote`n`use)^2 = (Height)^2 + (Base)^2`

`(2y-1)^2 = x^2 + (y+5)^2`

Substitute value of `x` from equation(1)

`4y^2 -4y +1 = (26-3y)^2 + y^2 + 10y + 25`

`4y^2 -4y +1 = (676 -156y + 9y^2) + y^2 + 10y + 25`

`4y^2 -4y +1 = 676 -146y + 10y^2 + 25`

Grouping like terms and applying transposition method

`4y^2 -10y^2-4y +146y = 676 + 25-1`

`-6y^2+142y = 700`

`-6y^2+142y - 700 = 0` --------divide by (-2),

`3y^2 - 71y + 350 = 0`

Find two numbers such that their product is equal to product of coefficients of first and last term(`350\times 3= 1050`) and their sum is equal to the coefficient of middle term(-71).

`3y^2 - 21y - 50y + 350=0`

`3y (y - 7) -50(y-7) =0`

`(3y-50)(y-7)=0`

i.e. `(3y-50)= 0`
OR
`(y-7)=0`

Therefore y can have values:
`y = 50/3 = 16.66`
or
`y = 7`
And `x` can have values with (`x =26-3y)` as:
`x = 26-(3\times 50/3) = -24`
or
`x = 26- 3\times 7= 26-21= 5`

Therefore , the possible values of x and y are

`x = -24 and y = 50/3`
or
`x= 5 and y = 7`