Given : Hypotenuse = (2y-1)(2y−1) cm
Side 1 (say height ) = xx cm and
Side 2 (say base) = (y + 5)(y+5) cm
Perimeter = sum of all three sides = 30 cm
2y-1+x+y+5 =302y−1+x+y+5=30
3y +x +4 = 303y+x+4=30
3y +x = 30-43y+x=30−4
3y +x = 263y+x=26 -------- let this be equation (1)
This gives x= 26-3yx=26−3y
According to Pythagoras theorem,
(Hypote(Hypotenuse)^2 = (Height)^2 + (Base)^2use)2=(Height)2+(Base)2
(2y-1)^2 = x^2 + (y+5)^2 (2y−1)2=x2+(y+5)2
Substitute value of xx from equation(1)
4y^2 -4y +1 = (26-3y)^2 + y^2 + 10y + 254y2−4y+1=(26−3y)2+y2+10y+25
4y^2 -4y +1 = (676 -156y + 9y^2) + y^2 + 10y + 254y2−4y+1=(676−156y+9y2)+y2+10y+25
4y^2 -4y +1 = 676 -146y + 10y^2 + 254y2−4y+1=676−146y+10y2+25
Grouping like terms and applying transposition method
4y^2 -10y^2-4y +146y = 676 + 25-14y2−10y2−4y+146y=676+25−1
-6y^2+142y = 700−6y2+142y=700
-6y^2+142y - 700 = 0−6y2+142y−700=0 --------divide by (-2),
3y^2 - 71y + 350 = 03y2−71y+350=0
Find two numbers such that their product is equal to product of coefficients of first and last term(350\times 3= 1050350×3=1050) and their sum is equal to the coefficient of middle term(-71).
3y^2 - 21y - 50y + 350=03y2−21y−50y+350=0
3y (y - 7) -50(y-7) =03y(y−7)−50(y−7)=0
(3y-50)(y-7)=0(3y−50)(y−7)=0
i.e. (3y-50)= 0 (3y−50)=0
OR
(y-7)=0(y−7)=0
Therefore y can have values:
y = 50/3 = 16.66y=503=16.66
or
y = 7y=7
And x x can have values with (x =26-3y)x=26−3y) as:
x = 26-(3\times 50/3) = -24x=26−(3×503)=−24
or
x = 26- 3\times 7= 26-21= 5x=26−3×7=26−21=5
Therefore , the possible values of x and y are
x = -24 and y = 50/3x=−24andy=503
or
x= 5 and y = 7x=5andy=7
