Question 714769
Another way to solve it: Use Pythagoras.
{{{drawing(400,350,-1,7,-2,5,
triangle(0,0,4,0,4,3),rectangle(4,0,3.8,0.2),
red(line(4,0,7,0)),red(line(4,3,7,3)),
red(line(0,5,0,-2)),
red(line(4,3,4,5)),red(line(4,0,4,-2)),
locate(1.2,4.5,Main),locate(2,4.5,Street),
locate(5.5,2,Elliott),locate(5.5,1.5,Street),
locate(2,0.3,x),locate(2,1.5,40),
locate(3.7,1.5,24)
)}}} Pythagoras says {{{x^2+24^2=40^2}}} --> {{{x=sqrt(40^2-24^2)}}}
That {{{x}}} can be calculated many different ways,
with more or less difficulty,
using pencil and paper, or a calculator.
 
However, mental math is easy if we think of similar (scaled up or down) right triangles.
The numbers 3, 4, and 5 form the most popular of Pythagorean triples,
sets of three positive integers that satisfy the Pythagorean equation,
and can be the lengths of the sides of a right triangle.
Triangles with sides measuring 3, 4, and 5
(and scaled-up versions) are used a lot in math problems.
The right triangle in this problem is similar to all 3-4-5 right triangles
(right triangles whose sides' lengths are in the ratio 3:4:5).
This triangle's known side lengths (in feet) are {{{8*5=40}}} and {{{8*3=24}}}.
The missing side length should be {{{8*4=32}}} feet.
 
The harder ways to the calculation:
{{{x=sqrt(40^2-24^2)}}} --> {{{x=sqrt(1600-576)}}} --> {{{x=sqrt(1024)}}} --> {{{x=32}}}
maybe with
{{{x=sqrt(1024)}}} --> {{{x=2^10}}}{{{x=2^5}}} --> {{{x=32}}}
Or
{{{x=sqrt(40^2-24^2)}}} --> {{{x=sqrt((8*5)^2-8*3)^2)}}} --> {{{x=sqrt(8^2*5^2-8^2*3^2)}}} --> {{{x=sqrt(8^2*(5^2-3^2))}}} --> {{{x=sqrt(8^2)*sqrt(5^2-3^2)}}} --> {{{x=8*sqrt(5^2-3^2)}}} --> {{{x=8*sqrt(25-9)}}} --> {{{x=8*sqrt(16)}}}  --> {{{x=8*4}}} --> {{{x=32}}}