Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

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

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)*

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 42,222

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.

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,648

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

Offline

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

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?

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 42,222

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.

Offline

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

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!

Offline

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

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)*

Offline

Pages: **1**