Application of analytical geometry

Real Application Of Geometry
Paper

Purposed remedy meets one of the tasks subjects Analytical Geometry with the
lecturer Dr. Turmudi M.Ed., M.Sc.

Created by:
Adhina Mentari Ashri (1002390)
Rahmi Hadri Farza (1005124)
Ridha Zahratun (1002524)

Mathematics Education Department
Faculty of Mathematics and Science Education
University Indonesia of Education
2011

Chapter I
Introduction

1.1 Background
Most of college students do not enthusiastic to learn something that they do
not know the importance and the using of it and it happens in analytic geometry.
Students do not know that analytic geometry is really used in real life around them
and they keep on thinking that from the beginning until now, they learn about
something abstract. It makes the enthusiasm and interest of them to learn it is very
low and their motivation to learn just to require their process to get title of degree
similar to the bachelors.
To give the real description about the using of analytic geometry in our daily
life, we need the directly observation to prove is it true that analytic geometry
concept is really used to make objects around us, especially buildings? Therefore, we
do this analyze.

1.2 Problems
1.2.1 What is the form of the chimney in JICA building?
1.2.2 What is the form of the sport track?
1.2.3 How is the volume of the three dimentional of the segment of circle in
West of JICA?
1.3 Purposes
This report is to analyze the formula from the shape of a chimney in JICA, the
sport track, and the volume of three dimentional of the segments of circle in west of
JICA in University Indonesia of Education. This report not only to analyse the form
of those things, but also to make us learn more about the analytical geometry.

1

Chapter II
Problem Analyzes
2.1 First Problem
Chimney is one of shape in JICA building which is located on west building.
Can you determine the form of that chimney? Find the equation!
Problem Solving:
We got the data about that chimney. Then, to make easier our task, we can
draw the sketch as the following picture:

A

B
D
E

C
Picture 1

Picture 2

Before we get the form of chimney, we consider chimney as a tube form, which its
diameter 10 cm. Then we cut the tube with the certain slope. Hence, we get the
chimney form and we know that AC will be the same with DE , which is 10 cm.
We assume that the angle between vertical line and BC is 45 . From the
chimney, we can draw a right triangle as following picture.
We have to find BC by using trigonometry we know that “Cosine is a comparison
between the lengths of the nearest side to the angle and its hypotenuse.” So, based on
that definition we get:

2

A

B

AC
BC

Cos 45 

10 cm

Cos 45 

10
x

45

1
10
 x  10 2
2
2
x

C

So, we get BC  10 2 cm. From the result, it’s clearly that the form of the
chimney is an ellipse, because AC  BC .
Then, how about the equation? First, we have to draw the ellipse.

BC  10 2 cm, OB  5 2 cm
DE  10 cm, OE  5 cm
Since BC  DE , the major axis of the ellipse is y-axis, a  5 2 cm and b  5 cm.
Y

B

E

D
O

X

C
So, the equation of that ellipse is:
x2 y2

1
b2 a2
x2 y2

1
25 50

3

2.2 Second Problem

It’s the figure of running track in UPI’s stadium. Determine the equation of the
outside line of the track!
Problem Solving:
Based on the data, we know that:
The length of football field

: 100 meters.

The width of football field

: 50 meters.

The width of running track
y (m)

: 4 meters.

Equation I.

Equation III
Equation II.

58
54

29

4 4
0

x (m)

29

129

158

Equation IV.
Picture 16.2

4

The outside of the track consists of 4 parts, two-semicircles and two horizontal lines.
Then we have to determine the 4 equations. For equation I and II, actually they are
same with the equation of circle. The standard equation of circle is

x  a2   y  b2  r 2
with (a,b) is center of the circle and r is radius. We know that coordinate of the center
of the circle is (29, 29) and radius of it is 29 meters. Then we get the equation I:

x  292   y  292  841
The equation I is only applied for interval 0  x  29 and 0  y  58 .
For equation II, we know that the center of circle is (129, 29) and the radius is 29
meters. Hence, we get equation II:

x  1292   y  292  841
The equation II is only applied for interval 129  x  158 and 0  y  58 .
For equation III and IV, we know that they are horizontal lines, then the standard
equation for them is y  c , with c is constant. So, the equation III is y  0 , for
interval 29  x  129 . And the equation IV is y  58 , for 29  x  129 .

2.3 Third Problem

Picture 5.2

5

The picture above is the west building of JICA. Look at the picture! There is a form
like three dimensional of the segment of circle. Find the volume of that structure!

Problem Solving:
To get the volume, we have to know the area of the segment of circle. Based
on the source, actually, that structure is a circle with its center at the center of JICA
building. And we have the data:
Radius (r)

: 37 m

Height (h)

: 100 m

Length (l)

: 20 m

The distance from

: 20,9 m

the center to
the building side (d)
To make our task easier, there is a sketch of that building if we see it from the
top.
The west building
of JICA

C
The center of JICA
building
The segment of the
circle

Picture 6.2

6

To know the segment of circle area, we have to find the sector area first. After
we get the sector area, we can find that segment of circle area. Look at the following
picture!

A

x
x

D

C

B
Based on the source, we have AC and BC as the radius of the circle which its length
is 37 m. and we also know that CD is 20,9 m
Then, to calculate the sector area, we have to know the angle of it. So, let’s we
divide ACB into two angle with the same size. Assume that the picture is true.
By using trigonometric function, we can find the angle.
Cos x 



CD
BC

Picture 7.2

20,9
 0,5648
37
x  55,6
2 x  101,2

So, ACB is 101,2o

7

Now, we have known the angle. It’s easy to find the sector area. The sector
area is

A

101,2 22
  37  37
360 7

 0,2811 4302,57
 1209,453 m

2

So, the sector area is 1209,453 m2.
After that, we have to find the area of a triangle in that sector. And the area is

ABC 

1
 20  20,9
2

 209 m2
The area of the triangle is 209 m2.
To find the area of segment of the circle, we subtract the sector area with the area of
triangle. So, the area of segment is
As = The sector area – the area of triangle
= 1209,453 m2 – 209 m2
= 1000,453 m2
Finally, we find the volume of the segment.
V  As  t
 1000,453  100
 100045,3 m

3

So, the volume of the segment is

100045,3 m3

8

Chapter III
Closing Statement

Conclusion
By assume that the angle between the line vertical and the oblique line as the
diameter of the chimney is 45 o, we get that the form of the chimney in JICA’s
2
2
building is an ellipse which the equation is x  y  1 .
25 50
Next, for the running track of
UPI’s stadium, the outside’s
consists of 4 parts, two semicircles and two horizontal lines. Then, the equation for
the first semicircle (left side) is
semicircle (right side) is

x  292   y  292  841

x 1292   y  292  841

,

for

the

second

, for the first horizontal line

(upper side) the equation is y=0, and the last the equation for the second line (bottom
side) is y=58 with the interval for both horizontal lines is

29  x  129 .

And the last, the volume of the three dimentional of the segment of circle of
west building of JICA is 100045,3m3.

9

References

Morril, W.K. (1967). Analytic Geometry. (Second Edition). Pennsylvania:
International Textbook Company.
Karso dan Darhim. (1983). Geometri Analitik. Bandung: Epsilon.

10

Application of analytical geometry

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