SOLUTION: Elliott street is 24ft wide when it ends at main street in brattleboro, vermont, a 40 ft long diagonal crosswalk allows pedestrians to cross main street to or from either corner o

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Elliott street is 24ft wide when it ends at main street in brattleboro, vermont, a 40 ft long diagonal crosswalk allows pedestrians to cross main street to or from either corner o      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 714769: Elliott street is 24ft wide when it ends at main street in brattleboro, vermont, a 40 ft long diagonal crosswalk allows pedestrians to cross main street to or from either corner of elliott street. determine the width of main street
Found 2 solutions by ankor@dixie-net.com, KMST:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Elliott street is 24ft wide when it ends at main street in brattleboro, vermont, a 40 ft long diagonal crosswalk allows pedestrians to cross main street to or from either corner of elliott street. determine the width of main street
:
Let m = the width of Main street
:
draw this out marking the width of Elliot as 24 and the width of Main as m
The diagonal, the distance to opposite corners, is 40
:
Find the angle (A) enclosed by the diagonal (40) and the width of Elliot (24)
Use the cosine for this
Cos(A) = 24%2F40
A = 53.13 degrees
:
Find the width of Main street using the sine of this angle
Sin(53.13) = m%2F40
.8 = m%2F40
m = .8 * 40
m = 32 ft is the width of main street

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Another way to solve it: Use Pythagoras.
Pythagoras says x%5E2%2B24%5E2=40%5E2 --> x=sqrt%2840%5E2-24%5E2%29
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%2A5=40 and 8%2A3=24.
The missing side length should be 8%2A4=32 feet.

The harder ways to the calculation:
x=sqrt%2840%5E2-24%5E2%29 --> x=sqrt%281600-576%29 --> x=sqrt%281024%29 --> x=32
maybe with
x=sqrt%281024%29 --> x=2%5E10x=2%5E5 --> x=32
Or
x=sqrt%2840%5E2-24%5E2%29 --> x=sqrt%28%288%2A5%29%5E2-8%2A3%29%5E2%29 --> x=sqrt%288%5E2%2A5%5E2-8%5E2%2A3%5E2%29 --> x=sqrt%288%5E2%2A%285%5E2-3%5E2%29%29 --> x=sqrt%288%5E2%29%2Asqrt%285%5E2-3%5E2%29 --> x=8%2Asqrt%285%5E2-3%5E2%29 --> x=8%2Asqrt%2825-9%29 --> x=8%2Asqrt%2816%29 --> x=8%2A4 --> x=32