A ladder of length 2x+1 feet is positioned against a wall such that bottom is x-1 feet away from a wall. The distance between the floor and top of the ladder is 2x. Assume that a right angle is formed by wall and the floor.

Pythagorean theorem

c^2 = a^2 + b^2

c = 2x+1

a = 2x

b = x-1

plug in and get:

(2x+1) ^2 = (2x) ^2 + (x-1) ^2

4x^2 + 4x + 1 = 4x^2 + x^2 - 2x + 1 [distributed]

4x^2 + 4x + 1 = 5x^2 - 2x + 1 [added like terms]

x^2 - 6x = 0 [subtracted 4x^2 and 4x and 1]

x (x-6) = 0 [factored out x]

x can equal 0 or 6

0 would not make any sense physically

therefore x = 6

plug this back into the length of the ladder 2x+1

2 (6) + 1 = 12 + 1 = 13

therefore the ladder is 13 feet long

So leangh of ladder=2x+1

bottom edgre=x-1

wall edge=2x

so therefor, since this is a right triangle, use pythagorean theorem

a^2+b^2=c^2

c=hypotonues=longest side

b and a=sides touching the right angle

so x-1 and 2x are a and b

2x+1=c

subsitute

(x-1) ^2 + (2x) ^2 = (2x+1) ^2

x^2-2x+1+4x^2=4x^2+4x+1

add like terms

5x^2-2x+1=4x^2+4x+1

subtract 1 from both sdies

5x^2-2x=4x^2+4x

subtract 4x from both sdies

5x^2-6x=4x^2

subtract 4x^2 from both sides

x^2-6x = 0

factor out the x using distributive property which is

ab+ac=a (b+c)

x^2-6x=x (x-6)

(x) (x-6) = 0

if xy=0 then assume x and/or y=0

x=0

we remember that one of the side legnths is 2x and if x=0 then the side legnth=0 which is not possible, so we discard

x-6=0

add 6 to both sides

x=6

subsitute and solve

legnth of ladder=2x+1

x=6 subsitute

2 (6) + 1=12+1=13

legnth of ladder = 13 feet

height=2x

2 (6) = 12

height=12 feet

base=x-1

6-1=5

legnth of ladder=13 feet

height=12 feet

base=5 feet