Download Water stop - InterMath

InterMath | Geometry | Quadrilaterals
Title
Water Stop
Problem Statement
The rectangular field pictured has unknown dimensions. Tom and Paul are both at point A.
Tom walks straight from A to C. To get a drink, Paul walks from A to B, then from B to C.
How far does Paul walk if he travels 40 yards farther than Tom? What are the dimensions of
the field?
Problem setup
In the rectangle above Tom and Paul are standing at point A. Tom decides that he does not need
any water so he goes straight from point A to point C. However, Paul is thirsty so he walks from
point A to point B to get some water then from point B to point C. In the end Paul has walked 40
yards further than Tom. What are the dimensions of the field then?
Plans to Solve/Investigate the Problem
First I will draw a rectangle in GSP and label it as the rectangle is labeled above. Then I will
measure how far Tom walks by drawing a line segment from point A to point C. I will then
measure how far Paul walked from point A to B and point B to C. After finding the ratio of the
difference in how far Tom walked to how far Paul walked, I will determine the original
dimensions of the field in Paul walked 40 yards further using algebra.
Investigation/Exploration of the Problem
Constructing a rectangle:
1. Construct a line segment in GSP
2. Select one of the points on the segment AB and segment AB and construct a
perpendicular line.
3. Select the other point and segment AB and construct a perpendicular line.
4. Draw a point C on one of the lines perpendicular to line segment AB.
5. Select point C and line segment BC and construct a perpendicular line
6. Place a point (D) where line segment AD and CD intersect.
A
B
D
C
7. Label point B as water stop
8. Draw a line segment from point A to point C to show the path Tom took. Label this line
segment x.
A
B (Water Stop)
X
D
C
If x is the distance that Tom walked then AB  BC is the distance that Paul walked. AB  BC =
y so the distance Paul walked is represented by y.
The problem tells you that Paul walks 40 yards farther than Tom.
Therefore, y  x  40 . We can rewrite it as y  x  40 .
ABCD is a rectangle; therefore, a relationship exists between the sides. By using the
measurement function I found the length of x and the length of y. Then I divided y by x to find
y
the ratio of AB  BC to AC . This ratio was 1.35. In equation form this means:  1.35
x
This can be rewritten so that y  1.35 x .
By placing the two equations equal to each we can solve for our variable x.
x  40  1.35 x
40  1.35 x  1x
40  .35 x
114.29  x
We can substitute the value of x into the original equation to find the value of y.
y  x  40
y  114.29  40
y  154.29
Therefore, Paul walks 154.29 yards.
The ratio of side AB to side BC is 1.90. I determined this by measuring both sides of the
rectangle and divided the length of AB by the length of BC.
AB
AB
 1.90
 1.90
So, BC
BC
AB  1.90* BC AB  1.90* BC
If you remember from earlier AB  BC = y
AB  y  BC
AB  154.29  BC
We can place the equations equal to each other and solve for length AB.
1.90 BC  154.29  BC
2.90 BC  154.29
154.29
2.90
BC  53.20
BC 
The width of the rectangle is 53.20 yards.
AB  154.29  BC
AB  154.29  53.20
AB  101.09
The length of the rectangular field is 101.09 yards.
The dimension of the field is 101.09 by 53.20 yards.
Extensions of the Problem
What if the field was a square instead of a rectangle? How far would Paul walk if he walked 40
yards farther than Tom? What are the dimensions of the field?
A
B
D
C
Water Stop
The above field is a square so AB=BC=CD=AD
Therefore, AB  BC  y
A
B
Water Stop
x
D
C
If y is the distance Paul traveled and x is the distance that Tom traveled then,
y  x  40
y  x  40
To determine the ratio of sides AB+BC to AC(x), I divided the length of AB+BC by the length
of AC in GSP. This gave me 1.41. This is otherwise known as 2 . To determine why this
occurs I looked at the properties of a right triangle including the Pythagorean Theorem
(a2+b2=c2). In this example, a=AB, b=BC, and c=AC.
a 2  b2  c 2
ab
 2a 2  c 2
You can take the square root of both sides and get, a 2  c (Since a baseball field measurement
cannot be negative I am concentrating on the positive value even though you could have a
positive or a negative after taking the square root)
Now we can set the equations equal to each other to solve for our variables.
y  x  40
yx 2
x  40  x 2
40  x 2  x
40  x( 2  1)
40
x
2 1
96.57  x
To solve for y,
y  x  40
y  96.57  40
y  136.57
Paul walks 136.57 yards.
The field is a square so the length and the width are equal to each other. Since Paul walked from
A to B then from B to C, he walked the length and width of the field. Therefore, the length and
width of the field added together is 136.57 yards. Since the length and width are equal you can
divide 136.57 by two to determine the length and width of the field. This equals 68.28 yards.
Therefore, the field is 68.28 yards by 68.28 yards.
Author & Contact
Tabitha Davis – Middle Grades cohort
[email protected]