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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.

Sagot :

Eternal3fearidn
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
Apologiabiologyidn
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
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