Question 180849
the diagram shows the a right angled triangle. based on the length of the sides of the triangle form a quadratic equation in term of y.
AB=(y+3)cm
BC=3y cm
AC= (4y-3) cm
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The answer depends on which of the lengths is the hypotenuse;
I'm going to assume that y+3 is the hypotenuse.
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(y+3)^2 = (3y)^2 + (4y-3)^2
y^2 + 6y + 9 = 9y^2 + 16y^2-24y+9
24y^2 -30y = 0
3y(8y-10) = 0
y = 0  or y = 10/8 = 5/4
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If y = 5/4 cm
AB = (5/4) + 3 = 5/4 + 12/4 = 17/4 cm
BC = 3*(5/4) = 15/4
AC = 4(5/4)-3 = 5-3 = 2 cm
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If one of the other sides is really the hypotenuse,
use Pythagoras, as I have, and solve for "y".

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Cheers,
Stan H.