The two sides of a right triangle differ by one. If the hypotenuse is 29 cm, then determine the lengths of the two sides.

Let the sides be a and b

a^2 + b^2 = c^2 = 29^2 = 841

a^2 + (a + 1)^2 = 841

a^2 + a^2 + 2 a + 1 = 841

2 a^2 + 2 a = 840

a (a + 1) = 420

21 * 20 = 420     trial and error

So the sides are 20 and 21

20^2 + 21^2 = 29^2

im here

step-by-step explanation:

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet