Question 1108272
The easiest way to calculate the area of a right triangle is
{{{area=(1/2)*leg[1]*leg[2]}}} ,
because the legs are perpendicular to each other,
so one leg can be considered the base, and the other is the height.
The difficult part is calculating the length of those legs.
The sides of the triangle measure
{{{leg[1]=36cm}}} , {{{leg[2]=48cm}}} and {{{hypotenuse=60cm}}} .
In this case,
{{{area=(1/2)*(36cm)*(48cm)=highlight(864cm^2)}}}
How did I know the leg length?
 
THE GET RESULTS WITHOUT MUCH WORK WAY:
Looking at a list of Pythagorean triples,
or looking at multiples of the popular 3-4-5 triple,
we figure that 36-48-60 (the 3-4-5 triple times 12)
add up to 144, and there are differences of 12 between the numbers.
The sum of the numbers, and differences between numbers in 3-4-5
are 12 and 1 respectively, so after multiplying everything times 12,
the sum and differences are
{{{23*12=144}}} and {{{1*12=12}}}
 
THE EXPECTED WAY:
If there is a figure, it may suggest that
the "base" is one leg of the right triangle.
If we think the base is the shorter leg,
"the adjacent" can be taken to mean the other leg or the hypotenuse.
If we think the base is the longer leg, then "the adjacent" must be the hypotenuse.
Either way, what follows is not fun.
If we think "the adjacent" is the hypotenuse,
the lengths of the sides of the triangle, in cm, are:
{{{x}}}= hypotenuse,
{{{x-12}}}= the "base" leg,
{{{sqrt(x^2-(x-2)^2)}}}=the other leg, and
the perimeter (in cm) is
{{{x+x-12+sqrt(x^2-(x-12)^2)=144}}}
{{{2x-12+sqrt(x^2-(x^2-24x+144))=144}}}
{{{2x-12+sqrt(x^2-x^2+24x-144)=144}}}
{{{2x-12+sqrt(24x-144)=144}}}
{{{sqrt(24x-144)=144+12-2x}}}
{{{sqrt(24x-144)=156-2x}}}
If we get smart, our next step is
{{{2sqrt(6x-36)=156-2x}}} ,
an then we simplify, dividing both sides by 2,to get
{{{sqrt(6x-36)=78-x}}} .
Squaring both sides of the equal sign, we get
{{{6x-36=6084-156+x^2}}}
{{{x^2-156x-6x+6084+36=0}}}
{{{x^2-162x+6120=0}}}
AT that point, we may be able to factor it as
{{{(x-60)(x-102)=0}}} .
The solutions to that equation are {{{x=60}}} and {{{x=102}}} .
We discard {{{x=102}}} ,
because it is not a solution of {{{sqrt(6x-36)=78-x}}} .
It is just an extraneous solution we introduced when we squared both sides.
The conclusion is that the hypotenuse measures {{{60cm}}} ,
the base measures {{{60cm-12cm=48cm}}} ,
and the length of the other leg, in cm, is
{{{sqrt(x^2-(x-12)^2)=sqrt(60^2-(60-12)^2)=sqrt(3600*48^2)=sqrt(1296)=36}}}
 
If there is another way that is simple, let me know.