Question 874044
Is it a 1-foot tall tree?
Is it an h-foot tall tree?
How tall is that tree?
If you do not have a number for the tree's height,
it gets a little more complicated.
 
If the height of the tree (in feet) is {{{h}}} ,
{{{x}}} is the length of the shadow in feet, then
{{{x+1}}} is the diagonal distance from the tip of the shadow to the top of the tree, in feet,
and it is like this (drawing not to scale, lengths in feet):
{{{drawing(300,250,-5,1,-1,4,
triangle(0,0,0,3,-4,0),
rectangle(0,0,-0.2,0.2),
rectangle(0,0,0.1,0.5),
green(triangle(-0.5,0.5,0.52,0.5,0,3)),
locate(0.05,1.7,h),locate(-2.1,-0.01,x),
locate(-2.1,1.5,x+1)
)}}} {{{(x+1)^2=x^2+h^2}}}<--->{{{x+1=sqrt(x^2+h^2)}}}
 
If you do not know the height of the tree, all you can get is formulas to calculate those distances.
In that case,
a) {{{x+1=sqrt(x^2+h^2)}}} is the diagonal distance from the tip of the shadow to the top of the tree, in feet,
b) {{{(x+1)^2=x^2+h^2}}}<--->{{{x^2+2x+1=x^2+h^2}}}<--->{{{2x+1=h^2}}}<--->{{{2x=h^2-1}}}<--->{{{x=(h^2-1)/2}}} is the length of the shadow.
 
If you knew the height of the tree,
you could substitute the value sooner rather than later,
and the algebra would get easier.
For example for a 3-foot tall tree, {{{h=3}}} ,
{{{(x+1)^2=x^2+3^2}}}<--->{{{(x+1)^2=x^2+9}}}<--->{{{x+1=sqrt(x^2+9)}}} ,
and to solve for {{{x}}} we write
{{{(x+1)^2=x^2+9}}}<--->{{{x^2+2x+1=x^2+9}}}<--->{{{2x+1=9}}}<--->{{{2x=9-1}}}<--->{{{2x=8}}}<--->{{{x=8/2}}}<--->{{{x=4}}} .
 
For a 5-foot tall tree, {{{h=5}}} ,
{{{(x+1)^2=x^2+5^2}}}<--->{{{(x+1)^2=x^2+25}}}<--->{{{highlight(x+1=sqrt(x^2+25))}}} is "an expression for the diagonal distance from the tip of the shadow to the top of the tree."
B) To solve for {{{x}}} (the length of the shadow in feet) we write
{{{(x+1)^2=x^2+25}}}<--->{{{x^2+2x+1=x^2+25}}}<--->{{{2x+1=25}}}<--->{{{2x=25-1}}}<--->{{{2x=24}}}<--->{{{x=24/2}}}<--->{{{highlight(x=12)}}} .
The length of the shadow is {{{highlight(x12)}}} feet.