Question

If the perimeter of a right triangle is 42 units and the length of the hypotenuse is 20 units, find the length of the other two sides of the triangle.
Can you please show work

Answer 1

So
a^2+b^2=c^2
c=hypotenuse
a^2+b^2=20^2=400

if the permiter is 42 then
p=hypotonuse+side+side
42=20+side+side
subtract 20 from both sides
22=side+side

a+b=22
subtract b from both sides
a=22-b
subsitute for a in equation below
a+2+b^2=c^2
(22-b)^2+b^2=400
b^2-44x+484+b^2=400
2b^2-44x+484=400
divide both sides by 2
b^2-22x+242=200
subtract 200 from both sides
b^2-22x+42=0
factor
(b+/-number)(b+/-number)
to find the numbers, find which 2 number multiply to 42 and add to get -22
the numbers must be neative since same signs multiplied=positive and negative+negative=negative
(b-number)(b-number)
facrors of 42=2,3,6,7,14,21
2,21
3,14
6,7
it's hard to find the number, but the numbers are -19.8882 and -2.11181
(b-19.8882)(b-2.11181)=0
if xy=0 then assume that x and y=0

b-19.8882=0
add 19.8882 to both sides
b=19.8882

b-2.11181=0
add 2.11181 to both sides
b=2.11181



the side legnths are 19.8882 units and 2.11181 units