The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm. Find the area of the triangle.

The perimeter of a right triangle $=40cm$
Hypotenuse measures $=17cm$.
Let x and y are the another sides.
$40=17+x+y$
$x+y=23$
or $y=23-x$
So, we have 3 sides as $x,23-x$ and $17.$
Apply Pythagoras theorem:
Hypotenuse${}^2$ $=$ Base${}^2+$ Perpendicular${}^2$
$17^2=x^2+(23-x{)}^2$
$289=x^2+529+x^2-46x$
$x^2-23x+120=0$
$(x-8)(x-15)=0$
$x=8$ or $x=15$
If $x=8$ then $y=23-8=15$
If $x=15$ then $y=23-15=8$
Now,
Area of triangle $=\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times 8\times 15$
$=60$
Therefore, Area of triangle is $60c{m}^2$