A Triangle Rectangle Crazy

Let the Cathetus be a and b . We have to find ab. According to question, a+b= 24 Also since this triangle is right angled, a^2 + b^2= 400. We know (a+b)^2= a^2 +b^2+2ab Putting them we will get : ab= 176/2 =88

If the hypothenuse is $20$ and the perimeter $44$ , then the legs (or Catheti) of this right triangle measure $x$ and $24-x$ each. By the Pythagorean Theorem we have:

$x^2 + (24-x)^2 = 20^2 \Rightarrow 2x^2 -48x + 176 = 0 \Rightarrow x = \frac{48 \pm \sqrt{48^2 - 4(2)(176)}}{4} = 12 \pm 2\sqrt{14}$

The Catheti equal $12+2\sqrt{14}$ and $12-2\sqrt{14}$ , which have the product $(12+2\sqrt{14})(12-2\sqrt{14}) = 144-4(14) = \boxed{88}.$

p=perpendicular, b=base p+b+20=44 (perimeter) p+b=24

p^(2) +b^(2)=20^(2)
(p+q)^(2)=400+2pb putting value p+q=24 576=400+2pb pb=88

axb = 176/2 = 88

first equation a^2+b^2=20^2 . . .second equation a+b+20=44. .two eaquation two unknown..a=4.517, b=19.483, axb=88