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#1 2023-01-27 12:05:32

aquafina09
Member
Registered: 2022-08-14
Posts: 4

Help me please! Area of Polygons

Hello,

I've been having a lot of trouble with a lesson called Areas of Polygons. The lesson it gives me is really hard to
understand. Could somebody help show me how to figure these questions out?

For #1-7, calculate the area for each of the polygons described below. Round answers to the nearest hundredth and remember to include the unit of measure.

- For #3, 4, and 5, divide the polygon into triangles as shown in the lesson.
- You must show the work for any values not provided in the problem, including the height for the triangles in #1, 3, 4, and 5.

1. An equilateral triangle with a side of 1 inch

2. A square with a side of 2 feet 

3. A regular pentagon with a side of 3 centimeters

4. A regular hexagon with a side of 10 cm

5. A regular heptagon with a side of 7 inches.

6. A trapezoid where the height is 18 cm, base 1 = 16 cm and b2 = 8 cm.

7. A trapezoid where the height = 7 mm, base 1 = 26 mm and base 2 = 9 mm.

For #8 and #9, fill in the missing information for the following trapezoids. SHOW YOUR WORK to solve for the missing value.

8.
height = 19.8 cm
b1 = ________
b2 = 14.4 cm
area = 401.94 cm2


9.
height = ________
b1 = 20 cm
b2 = 21 cm
area = 205 cm2


10. If the area of a parallelogram is 690.84 m2 and the height is 20.2 m, what is the length of the base?

11. If the base of a rectangle is 28 cm and the area is 588 cm^2, what is the height of the rectangle?

12. What is the area of a parallelogram with height 26 cm, base 16 cm, and side length 28 cm? 

13. What is the area of this polygon?  Show all of your calculations.



ls_XF    =    53 mm    ls_XV    =    72 mm    ls_VR    =    16 mm
ls_FB    =    31 mm    ls_BT    =    31 mm    ls_EU    =    47 mm
ls_UL    =    31 mm    ls_TL    =    88 mm    ls_DE    =    16 mm
ls_RM    =    70 mm    ls_MC    =    21 mm    ls_DC    =    70 mm








14. What is the area of this rectangle?  Show all of your work.



15. What is the area of this polygon?  Show all of your work.

Last edited by aquafina09 (2023-01-27 12:09:22)

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#2 2023-01-27 22:21:34

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 46,731

Re: Help me please! Area of Polygons

Hi aquafina09,

Welcome to the forum!

I can help you with one or two, not many.

1) Area of an Equilateral Triangle is

where a is length of a side.

2) Square of side 2 feet is

You may use the link Polygons.


It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#3 2023-01-28 05:39:07

Bob
Administrator
Registered: 2010-06-20
Posts: 10,474

Re: Help me please! Area of Polygons

hi aquafina09

These look like Compu High questions. Whatever,  the method is the same for all of them.

If a polygon is regular that means all its sides are equal and all its angles.

So if you fnd the centre of the shape (call it point O) and draw lines radiating out to the vertices, A, B, C, D, E,  etc
then OAB, OBC, OCD, ODE etc are all identical isosceles triangles. So, if you can calculate the area of one of these and then times by how many there are, you get the total area of the whole polygon.

Follow these steps:

(1) Let n be the number of sides and s be the length of one side.

(2) Angle AOB = 360/n so you can calculate the angle at the centre.

(3) Let M be the midpoint of AB.

(4) As AOB is isosceles and the angle sum of any triangle is 180 you can calculate OAM.

OAM = (180 - AOB) / 2

(5) AM = s/2

(6) Use trig to calculate the height of the triangle AOB.

h =  s/2 tan(OAM)

(7) Area of any triangle is half base times height.  In our triangle

area AOB = half x s x [s/2 tan(OAM)]

(8) times by n

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#4 2023-02-02 13:56:58

aquafina09
Member
Registered: 2022-08-14
Posts: 4

Re: Help me please! Area of Polygons

so for #1
360/3 = 120
(180-120)/2
=60/2
=30

.5tan(30°)
=.288

1/2(1)(.288)
=.144 x 3
=0.432

Did I do this right?

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#5 2023-02-02 14:12:55

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 46,731

Re: Help me please! Area of Polygons

Hi aquafina09,

Applying the value 1 inch,

required area is approximately 0.433 sq in.


It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#6 2023-02-02 15:10:41

aquafina09
Member
Registered: 2022-08-14
Posts: 4

Re: Help me please! Area of Polygons

ganesh wrote:

Hi aquafina09,

Applying the value 1 inch,

required area is approximately 0.433 sq in.

Huh okay so different formula, but the same, (or very close) answer. Thank you!

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#7 2023-02-03 05:36:47

aquafina09
Member
Registered: 2022-08-14
Posts: 4

Re: Help me please! Area of Polygons

Ok I have managed to complete the questions. Thank you for the help.

1. An equilateral triangle with a side of 1 inch
Area =0.43 in

360/3 = 120
(180-120)/2
=60/2
=30

0.5tan(30°)
=0.288

1/2(1)(.288)
=0.144 x 3
=0.43 in

2. A square with a side of 2 feet 
Area = 4 ft

360/4 = 90
(180-90)/2
90/2
= 45°

2/2 = 1
1tan45°=1x4
= 4 ft

3. A regular pentagon with a side of 3 centimeters
Area = 15.45 cm

360/5 = 72
(180-72)/2
108/2
= 54°

3/2
1.5
1.5tan54°
= 2.06
1/2(3)(2.06)
3.09 x 5
A = 15.45 cm

4. A regular hexagon with a side of 10 cm
Area = 433 cm

360/6 = 60
(180-60)/2
120/2
=60

10/2 = 5
5tan60° = 8.66
1/2(10)(8.66)
=43.3 x 6
A = 259.80 cm

5. A regular heptagon with a side of 7 inches.

Area = 177.87 in

360/7 = 51.42
(180-51.42)/2
128.58/2
=64.29

7/2 = 3.5
3.5tan64.29°
= 7.26

1/2(7)(7.26)
25.41 x 7
= 177.87 in

6. A trapezoid where the height is 18 cm, base 1 = 16 cm and b2 = 8 cm.

Area = 216 cm

1/2(16+8)(18)
=216

7. A trapezoid where the height = 7 mm, base 1 = 26 mm and base 2 = 9 mm.

Area = 122.5 mm

1/2(26+9)(7)
=122.5

8.
height = 19.8 cm
b1 = ________
b2 = 14.4 cm
area = 401.94 cm2

b1=26.2cm

1/2(19.8)=9.9
401.94/9.9=40.6
40.6-14.4=26.2
1/2(26.2+14.4)(19.8)=401.94


9.
height = ________
b1 = 20 cm
b2 = 21 cm
area = 205 cm2

height = 10

1/2(20+21) = 20.5
205/20.5 = 10
1/2(20+21)(10) = 205

10. If the area of a parallelogram is 690.84 m2 and the height is 20.2 m, what is the length of the base?

base=34.2m

690.84/20.2=34.2
1/2(34.2)(20.2)=345.42 m2
345.42(2) = 690.84

11. If the base of a rectangle is 28 cm and the area is 588 cm^2, what is the height of the rectangle?

height = 21 cm

588/28=21
1/2(28)(21)=294
294(2)=588

12. What is the area of a parallelogram with height 26 cm, base 16 cm, and side length 28 cm?

area= 416 cm^2

1/2(16)(26) = 208
208(2) = 416

13. What is the area of this polygon?  Show all of your calculations.



ls_XF    =    53 mm    ls_XV    =    72 mm    ls_VR    =    16 mm
ls_FB    =    31 mm    ls_BT    =    31 mm    ls_EU    =    47 mm
ls_UL    =    31 mm    ls_TL    =    88 mm    ls_DE    =    16 mm
ls_RM    =    70 mm    ls_MC    =    21 mm    ls_DC    =    70 mm

area=8014mm

since its a rectangle all the shorter sides equaled the same as the corresponding side so:
81x31 = 2728
21x70 = 1470
53x72 = 3816
2728+1470+3816=8014

14. What is the area of this rectangle?  Show all of your work.

area=60

h^2=13^2-12^2
h^2=25
h=5
12x5=60



15. What is the area of this polygon?  Show all of your work.

area=220

1/2(12)(10)=60
(20)(8)=160
160+60=220

Last edited by aquafina09 (2023-02-03 09:58:19)

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