Problem
of the Week
Problem
A and Solution
Disc
Golf Distance

Problem

Disc golf is a sport where players throw a flying disc towards a
target. A disc golf course is made up of several holes, each with their
own tee box and basket. Players start by standing in the tee box and
throwing the disc towards the basket.

Chippewa Public School is building a new nine-hole disc golf course
on their property. They drew a plan for the course on grid paper,
showing the tee box , basket , and hole number for each hole. The distance between grid
lines is \(20\) m, and the centre of
each tee box and basket is placed either on a grid line, or halfway
between two grid lines.

The plan is a grid
with 10 vertical grid lines and 8 horizontal grid lines. The nine holes
have varying lengths and are each placed along a horizontal grid line or
a vertical grid line. The locations of the tee box and basket for each
hole are given below.

Hole number

Tree box location

Basket location

1

First line

Fifth line

2

Sixth line

Between eighth and ninth lines

3

Between first and second lines

Between seventh and eighth lines

4

Between seventh and eighth lines

Third line

5

Between eighth and ninth lines

Second line

6

First line

Sixth line

7

Between third and fourth lines

Eighth line

8

Eighth line

Between second and third lines

9

Fifth line

Eighth line

The length of a hole is the distance from the centre of the tee box
to the centre of the basket. The length of a course is the sum of the
lengths of all the holes, not including the distances between the
holes.

Calculate the length of the disc golf course the school is planning
to build.

Solution

Since the distance between the grid lines represents \(20\) m, then the distance halfway from one
grid line to the next would be \(10\) m. Keeping this in mind, we can
calculate the length of each hole by counting the number of grid squares
between its tee box and basket.

Hole Number

Number of Grid Squares Between Tee Box
and Basket

Length of Hole (m)

\(1\)

four

\(4 \times 20
= 80\)

\(2\)

two and a half

\(2 \times 20
+ 10 = 50\)

\(3\)

five (a half, plus \(4\), plus a half)

\(5 \times 20
= 100\)

\(4\)

four and a half

\(4 \times 20
+10 = 90\)

\(5\)

six and a half

\(6 \times 20
+ 10 = 130\)

\(6\)

five

\(5 \times 20
= 100\)

\(7\)

four and a half

\(4 \times
20 + 10 = 90\)

\(8\)

five and a half

\(5 \times 20
+ 10 = 110\)

\(9\)

three

\(3 \times 20
= 60\)

So the length of the course is: \[80 + 50
+ 100 + 90 + 130 + 100 + 90 + 110 + 60 = 810~\text{m}.\]

Alternatively, we can calculate the total number of grid squares
between the tee boxes and the baskets, and then use this to calculate
the length of the course.

First we add the complete squares: \(4+2+4+4+6+5+4+5+3=37.\)

Then we add the half squares, using the fact that \(\frac{1}{2} + \frac{1}{2} = 1\). \[\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{2}=1+1+1+\tfrac{1}{2}=3+\tfrac{1}{2}\]
Then we add these together to obtain \(37+3+\frac{1}{2}=40+\frac{1}{2}\).

Finally we can calculate the length of the course in meters. Since
\(40 \times 20 = 800\), the length of
the course is \(800+10=810\) m.