Answer :
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
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
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
