Question

14. The perimeter of the right-angled triangle ABC is 80 cm. Its base is 6x and one of the side ribs x². Find x.

Answer 1

If one of the side ribs, which equals x² is hypotenuse, then the other rib length is: $b^2 = (x^2)^2 - (6x)^2 = x^4 - 36x^2.$

$b = \sqrt{x^4 - 36x^2}$

The perimeter of this triangle is:

$P = x^2 + 6x + \sqrt{x^4 - 36x^2} = 80,$

$\sqrt{x^4 - 36x^2} = 80 - x^2 - 6x,$

$x^4 - 36x^2 = 6,400 - 80x^2 - 480x - 80x^2 + x^4 + 6x^3 - 480x + 6x^3 + 36x^2,$

$12x^3 - 88x^2 - 960x + 6,400 = 0,$

$3x^3 - 22x^2 - 240x + 1,600 = 0,$

$(3x^3 - 30x^2) + (8x^2- 240x + 1,600) = 0,$

$3x^2(x - 10) + 8(x^2 - 30x + 200) = 0,$

$3x^2(x - 10) + 8(x - 10)(x - 20) = 0,$

$(x - 10)(3x^2 + 8x - 160) = 0,$

So, x1 = 10 is one of the roots of this equation.

D = 64 + 1920 = 1984.

$x2 = \frac{-8 + \sqrt{1984}}{6} = 6.09,$

$x3 = \frac{-8 - \sqrt{1984}}{6} = -8.76,$ which is not possible in our case.