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# Math Project on Trigonometry

The purpose of this presentation is to show how do we find the area of our school quadrangle

by

Tweet## HAO WEI CHONG

on 20 August 2011#### Transcript of Math Project on Trigonometry

The Quadrangle Problems

( Scenario 3 ) The Purpose -We are here to present to you

the methods of how we determined

the area of our school Quadrangle Questions: In order to find the area of our school quadrangle, we need to determined the lenght of each side of the quadrangle first. 1. We used measuring tap to measure the lenght of the side of the quadrangle. AB and ED A B C D E 2. AB=40.71m, and DE=38.56m We calculate the area using the Trigonomrtry formula.

Area of Triangle= 1/2 x Length between angle x Sin Angle btw length Areas: F 1. So, how do we find the lenght of each side of the Quandrangle? Finding the Length : To prevent inaccurate measurement, we decided to

measure the two side which is the easiest to measure. 3. Then we measure both AB and DE by using

ruler on MAP. AB= 3.1cm, DE= 2.8cm. 4. We find the length to see 1cm represent how many meters. Working: Line AB:

3.1cm --> 40.71m

1cm --> 13.13m

Line ED:

2.8cm --> 38.56m

1cm --> 13.77m

Then we find the Average

13.13m + 13.77m = 13.45m

Therefore, 1cm represent 13.45m So now we can find the other length.

We have to measure the length on map

in cm first.

BC=2.2cm, CD= 1.1cm, EF= 1.1cm, AF=1.3cm

BD=2.5cm, BE= 3.8cm We also divide the shape of the quadrangle into

different triangles. The lines are BD, AE, BE Find the Length in meters

BC= 2.2cmx13.45m

= 29.59m

CD= 1.1cm x 13.45m

= 14.795m

EF= 1.1cm x 13.45m

= 14.795m

AF= 1.3cm x 13.45m

= 17.485m

BD= 2.5cm x 13.45m

= 33.625m

BE= 3.8cm x 13.45m

= 51.11m A B C D E F 29.59m 14.795m 14.795m 33.625m 51.11m 17.485m We also need to determined the angles of the triangles as well.

2. So how do we find the angle of the triangle? The Angles: 1. We used protractor to measure

the angles on MAP. Angle ABE=29

Angle BCD= 86, Angle EFA=78 A B C D E F After finding the length and angles, we can calculate the Area

of the quadrangle.

3. How do we calculate the area of the quadrangle ? Method 1 : The Trigonometry Method Method 2 : The Map Scale method 1cm : 13.45m

1cm square : 179.56m square We will use the length we found initially Areas : Area of Tri BCD = 1/2 x 2.2 x 1.1

= 1.21cm square

>1.21cm square x 179.6m square

=217.3m square

Area of Tri FBD=1/2 x 2.8 x 2.6

= 3.64cm square

>3.64 x 179.6 =653.6m square

Area of Tri ABE= 1/2 x 3 x 1.8

= 2.7cm square

>2.7 x 179.6 = 484.8m square

Area of Tri AFE= 1/2 x 0.8 x 1.9

= 0.76cm square

>0.76 x 179.6 = 136.8m square

Total Area =217.3+653.6+484.8+136.8

= 1497m square

Therefore, the area of quadrangle is 1497m square. 40.71m 38.56m Credits Math Project on Trigonometry Group Members

-Hao Wei (Assistence Leader)

-Kelly (Leader)

-Dong Do

-Ivan

-Nicole From Class,

3E4 Software used:

- Prezi The Quadrangle Problem

( Scenario 3 ) A B C D E F 3.1cm 2.8cm 29 78 Area of Triangle BCD= 1/2 x 29.59m x 14.795m x (Sin86 )

= 218.4m square

Area of Triangle BDE= 1/2 x 33.625m x 38.56m x ( Sin90 )

= 648.3m square

Area of Triangle ABE= 1/2 x 51.11m x 40.71m x ( Sin29 )

= 504.4m square

Area of Triangle AEF= 1/2 x 14.795m x 17.485m x ( Sin 78 )

= 126.52m square

Total Area of Quadrangle ABCDEF = 218.4+648.3+504.4+126.52

= 1497.6m square

Therefore, the area of our school quadrangle is 1497.6m square The Graph Method: Methods 1. Draw out the shape of our school quadrangle.

2. Divide the quadrangle into different

triangles.

3. Count the number of small square (0.2x0.2)cm in each of the triangles.

4. Calculate the area for one small square.

3.1cm : 40.71m

0.2cm : 2.63m

0.04cm^2 : 6.92m^2 5. Calculate the area for each of the triangles.

Triangle of BCD= 6.92m^2 x 32

= 220.7m^2

Triangle of BDE= 6.92m^2 x 93

= 641.2m^2

Triangle of AEB= 6.92m^2 x 70

= 482.9m^2

Triangle of AEF= 6.92m^2 x21

= 144.8m^2

Total Area= 220.7+641.2+482.9+144.8

= 1494.72

= 1495m^2 A B C D E F A B C D E F Thank You ! =) Problems Encounter 1. The map of the quadrangle that we recieve is not accurate as due to the reconstruction. so it is harder to measure.

2. We used the measuring tape but due to human error when we measure. (the measuring tape can only measure 5m) so, when we continue measuring the tape may shift. Discovery and Reporting 1. We discover how to use the methods

of trigonometry to find the area of the

quadrangle

2. Use of other method to find the area.

3. Overall the answers obtained using different

method is around the same. Reflection Hao Wei : I learnt that we can not only use other method like map scale

methods ang graph methods, we can use the trigonometry method.

eto use the wonders of prezi.

Dong Do: I learnt trigonometry is very useful in solving various problems. I also learn the use if prezi programme.

Kelly:I learnt how to use the wonders of prezi and do a powerpoint.

Ivan: Altough the measuring of quadrangle is hard. I enjoy it.

Nicole: i learnt the importance of trigonometry in solving problems.

Full transcript( Scenario 3 ) The Purpose -We are here to present to you

the methods of how we determined

the area of our school Quadrangle Questions: In order to find the area of our school quadrangle, we need to determined the lenght of each side of the quadrangle first. 1. We used measuring tap to measure the lenght of the side of the quadrangle. AB and ED A B C D E 2. AB=40.71m, and DE=38.56m We calculate the area using the Trigonomrtry formula.

Area of Triangle= 1/2 x Length between angle x Sin Angle btw length Areas: F 1. So, how do we find the lenght of each side of the Quandrangle? Finding the Length : To prevent inaccurate measurement, we decided to

measure the two side which is the easiest to measure. 3. Then we measure both AB and DE by using

ruler on MAP. AB= 3.1cm, DE= 2.8cm. 4. We find the length to see 1cm represent how many meters. Working: Line AB:

3.1cm --> 40.71m

1cm --> 13.13m

Line ED:

2.8cm --> 38.56m

1cm --> 13.77m

Then we find the Average

13.13m + 13.77m = 13.45m

Therefore, 1cm represent 13.45m So now we can find the other length.

We have to measure the length on map

in cm first.

BC=2.2cm, CD= 1.1cm, EF= 1.1cm, AF=1.3cm

BD=2.5cm, BE= 3.8cm We also divide the shape of the quadrangle into

different triangles. The lines are BD, AE, BE Find the Length in meters

BC= 2.2cmx13.45m

= 29.59m

CD= 1.1cm x 13.45m

= 14.795m

EF= 1.1cm x 13.45m

= 14.795m

AF= 1.3cm x 13.45m

= 17.485m

BD= 2.5cm x 13.45m

= 33.625m

BE= 3.8cm x 13.45m

= 51.11m A B C D E F 29.59m 14.795m 14.795m 33.625m 51.11m 17.485m We also need to determined the angles of the triangles as well.

2. So how do we find the angle of the triangle? The Angles: 1. We used protractor to measure

the angles on MAP. Angle ABE=29

Angle BCD= 86, Angle EFA=78 A B C D E F After finding the length and angles, we can calculate the Area

of the quadrangle.

3. How do we calculate the area of the quadrangle ? Method 1 : The Trigonometry Method Method 2 : The Map Scale method 1cm : 13.45m

1cm square : 179.56m square We will use the length we found initially Areas : Area of Tri BCD = 1/2 x 2.2 x 1.1

= 1.21cm square

>1.21cm square x 179.6m square

=217.3m square

Area of Tri FBD=1/2 x 2.8 x 2.6

= 3.64cm square

>3.64 x 179.6 =653.6m square

Area of Tri ABE= 1/2 x 3 x 1.8

= 2.7cm square

>2.7 x 179.6 = 484.8m square

Area of Tri AFE= 1/2 x 0.8 x 1.9

= 0.76cm square

>0.76 x 179.6 = 136.8m square

Total Area =217.3+653.6+484.8+136.8

= 1497m square

Therefore, the area of quadrangle is 1497m square. 40.71m 38.56m Credits Math Project on Trigonometry Group Members

-Hao Wei (Assistence Leader)

-Kelly (Leader)

-Dong Do

-Ivan

-Nicole From Class,

3E4 Software used:

- Prezi The Quadrangle Problem

( Scenario 3 ) A B C D E F 3.1cm 2.8cm 29 78 Area of Triangle BCD= 1/2 x 29.59m x 14.795m x (Sin86 )

= 218.4m square

Area of Triangle BDE= 1/2 x 33.625m x 38.56m x ( Sin90 )

= 648.3m square

Area of Triangle ABE= 1/2 x 51.11m x 40.71m x ( Sin29 )

= 504.4m square

Area of Triangle AEF= 1/2 x 14.795m x 17.485m x ( Sin 78 )

= 126.52m square

Total Area of Quadrangle ABCDEF = 218.4+648.3+504.4+126.52

= 1497.6m square

Therefore, the area of our school quadrangle is 1497.6m square The Graph Method: Methods 1. Draw out the shape of our school quadrangle.

2. Divide the quadrangle into different

triangles.

3. Count the number of small square (0.2x0.2)cm in each of the triangles.

4. Calculate the area for one small square.

3.1cm : 40.71m

0.2cm : 2.63m

0.04cm^2 : 6.92m^2 5. Calculate the area for each of the triangles.

Triangle of BCD= 6.92m^2 x 32

= 220.7m^2

Triangle of BDE= 6.92m^2 x 93

= 641.2m^2

Triangle of AEB= 6.92m^2 x 70

= 482.9m^2

Triangle of AEF= 6.92m^2 x21

= 144.8m^2

Total Area= 220.7+641.2+482.9+144.8

= 1494.72

= 1495m^2 A B C D E F A B C D E F Thank You ! =) Problems Encounter 1. The map of the quadrangle that we recieve is not accurate as due to the reconstruction. so it is harder to measure.

2. We used the measuring tape but due to human error when we measure. (the measuring tape can only measure 5m) so, when we continue measuring the tape may shift. Discovery and Reporting 1. We discover how to use the methods

of trigonometry to find the area of the

quadrangle

2. Use of other method to find the area.

3. Overall the answers obtained using different

method is around the same. Reflection Hao Wei : I learnt that we can not only use other method like map scale

methods ang graph methods, we can use the trigonometry method.

eto use the wonders of prezi.

Dong Do: I learnt trigonometry is very useful in solving various problems. I also learn the use if prezi programme.

Kelly:I learnt how to use the wonders of prezi and do a powerpoint.

Ivan: Altough the measuring of quadrangle is hard. I enjoy it.

Nicole: i learnt the importance of trigonometry in solving problems.