What is the length of a diagonal brace that could be used for a wall 9 feet high and 12 feet long

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

isocolese triangle

we know that the line connect a and b is the hyptonuse

so first find the hypotonuse length using the distance formula, call the length c

since it is an isocolese triangle, the lengths of the legs are the same

using pythagorean theorem,

in this case, a=b since isocolese triangle so we get

or

we can find c so we can find the distances

we know that if we draw 2 circles, each has center at a and b, and find where they intersect, we can the 2 possible locations for c

if a has center at (xa,ya) and b has center at (xb,yb), then we can solve where

and

intersect to find where they intersect

i'm not going to put an example because i have to go to class. i hope this explanation is enough