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

1

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.

2

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:

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:

A

C

B

D

E

Picture 1 Picture 2

3

So, we get 210BC cm. From the result, it’s clearly that the form of the

chimney is an ellipse, because BCAC .

Then, how about the equation? First, we have to draw the ellipse.

210BC cm, 25OB cm

10DE cm, 5OE cm

Since DEBC , the major axis of the ellipse is y-axis, 25a cm and 5b cm.

So, the equation of that ellipse is:

12

2

2

2

a

y

b

x

15025

22

yx

Y

XD

B

E

C

O

A

C

B

45

10 cm

x

102

2

1

xCos

1045

210 x

BC

ACCos 45

4

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 : 4 meters.

)(my

Picture 16.2

0 29 129 158)(mx

5854

4 4

Equation I. Equation III

Equation II.

Equation IV.

29

5

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

222rbyax

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:

The equation I is only applied for interval 290 x and 580 y .

For equation II, we know that the center of circle is (129, 29) and the radius is 29

meters. Hence, we get equation II:

8412912922 yx

The equation II is only applied for interval 158129 x and 580 y .

For equation III and IV, we know that they are horizontal lines, then the standard

equation for them is cy , with c is constant. So, the equation III is 0y , for

interval 12929 x . And the equation IV is 58y , for 12929 x .

2.3 Third Problem

841292922 yx

Picture 5.2

6

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.

C

The west building

of JICA

The segment of the

circle

The center of JICA

building

Picture 6.2

7

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!

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 BC

CDx

5648,037

9,20

2,1012

6,55

x

x So, ACB is 101,2

o

Picture 7.2

CD

B

A

x

x

8

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

area is

37377

22

360

2,101

A

57,43022811,0

453,1209 m2

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

9,20202

1ABC

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 m

2

= 1000,453 m2

Finally, we find the volume of the segment.

tAV s

100453,1000

3,100045 m3

So, the volume of the segment is

100045,3 m3

9

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 45o, we get that the form of the chimney in JICA’s

building is an ellipse which the equation is .

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 , for the second

semicircle (right side) is , 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 .

And the last, the volume of the three dimentional of the segment of circle of

west building of JICA is 100045,3m3.

15025

22

yx

841292922 yx

8412912922 yx

12929 x

10

References

Morril, W.K. (1967). Analytic Geometry. (Second Edition). Pennsylvania:

International Textbook Company.

Karso dan Darhim. (1983). Geometri Analitik. Bandung: Epsilon.

11

Attachment