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!

1 Answer
Oct 1, 2017

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,

(Hypotenuse)^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