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**taylorn5683****Member**- Registered: 2017-02-01
- Posts: 10

Calculate the area for each of the polygons below. If you do not know an equation to use, divide the polygon into other shapes to determine the area.

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.

Fill in the missing information for the following trapezoids:

8. height = 19.8 cm

b1 = ________

b2 = 14.4 cm

area = 401.94 cm2

9. height = 23 mm

b1 = 23 mm

b2 = ________

area = 529 mm2

10. height = ________

b1 = 20 cm

b2 = 21 cm

area = 205 cm2

11. height = 28.9 m

b1 = 26.9 m

b2 = ________

area = 806.31 m^2

12. If the area of a parallelogram is 690.84 m^2 and the height is 20.2 m, what is the length of the base?

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

14. If the height of a rectangle is 26.1 m and the base is 17.3 m, what is the area of the rectangle?

15. If the height of a parallelogram is 34 cm and the base is 15 cm, what is the area of the parallelogram?

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

17. What is the area of a regular octagon with a side of 6 cm?

18. What is the area of this polygon?

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

19. What is the area of this rectangle?

20. What is the area of this polygon?

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**taylorn5683****Member**- Registered: 2017-02-01
- Posts: 10

8. height = 19.8 cm

b1 = ___26.2_____

b2 = 14.4 cm

area = 401.94 cm2

9. height = 23 mm

b1 = 23 mm

b2 = ______23__

area = 529 mm2

10. height = __10______

b1 = 20 cm

b2 = 21 cm

area = 205 cm2

11. height = 28.9 m

b1 = 26.9 m

b2 = __28.9______

area = 806.31 m^2

12. If the area of a parallelogram is 690.84 m^2 and the height is 20.2 m, what is the length of the base?

34.2 m

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

21 cm

14. If the height of a rectangle is 26.1 m and the base is 17.3 m, what is the area of the rectangle?

451.53m

15. If the height of a parallelogram is 34 cm and the base is 15 cm, what is the area of the parallelogram?

510

19. What is the area of this rectangle?

78

I need help showing work I did most of it on a calulator.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,442

Hi Taylor

Since you joined the forum you have started 6 threads Each consists of 20 homework questions I asked you to read the forum policy on help but you are still doing it several posts later. This is likely to result in you being banned. Please reply with a proper request for help making it clear where your difficulty lies.

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 Bundy

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**laurenwest144****Member**- Registered: 2017-11-15
- Posts: 8

Hi, I am working on the same assignment and I am having trouble working on finding the missing base of a trapezoid. Could you help me? If so that would be greatly appreciated.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 26,620

Hi laurenwest144,

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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

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**laurenwest144****Member**- Registered: 2017-11-15
- Posts: 8

I am still working on this assignment and I just can not figure out how to use trig functions to complete these problems. If someone could help teach me how I would really appreciate it. I need to find the area of all these shapes.

1. An equilateral triangle with a side of 1 inch

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.

15. What is the area of this polygon?

http://www.compuhigh.net/testeditor/upload/pics/geometry/lesson21/area3.jpg

The area of this would be 206 sq units because the area of the rectangle is 152 and the triangle area is 54 and when you add them together to get 206 which will be the answer:206 sq units. With this one you split the polygon into a rectangle and a triangle. I just compared the sizes of the line when I split the shapes up. I found that on the rectangle you do hb b/2 and that is 9 12/ 2= 54 so the area is 54. With the rectangle you know that the short side is equal to 8 because the top one is too and the other side would be 19. You do 19 times 8 and you get 152. 152+54=206 sq units.

For the last one I just need to know how to find the the length of the lines without comparing. I need to use trig.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,442

hi laurenwest144

An equilateral triangle has angles of 60. If you split it in half with a line from the vertex to the midpoint of the base, you will create a right angled triangle 30-60-90.

So you can use trig. to work out the height of the triangle and then half base x height to get the area. eg with a hypotenuse of 1 the height will be 1 x cos(30)

A regular hexagon can be split into six equilateral triangles so you can do the above and then x by 6.

For a regular pentagon draw 5 lines from the centre to the points on the perimeter. This will divide it into five isosceles triangles. The angle at the centre is 360, so each triangle will have a top angle of 360/5. Split each triangle in half like you did for the equilateral triangle. **edit: Then work out the third angle of the right angled triangle and use tan to calculate the height.** Once you have one triangle you can times by 5.

Q15. That image is not loading for me. I expect CompuHigh have it password protected as it will be subject to copyright. When you post back with your answers to the others for me to check, try describing the diagram in words eg. There's a rectangle, L by W, and a ..........

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 Bundy

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**laurenwest144****Member**- Registered: 2017-11-15
- Posts: 8

I think I have the first problem but I am not sure. My teacher said to me earlier that I need to be using trig functions. These are the notes she gave me, as well as the answer I came up with.

1. An equilateral triangle with a side of 1 inch

The first thing you do is split the triangle into two 30-60-90 triangles. The base of those triangles is going to be the side over 2 and that means it is ½. The side opposite the 60 degree angle will be the square root of 3 times the side 1 over 2. So it is √3 ½. So I already know the base is one and now I have to find the height by doing the sqr of 3 times 1 over 2. I also know that to find the area of a triangle you have to do ½ of the base times the height. So you multiply the numerators. Which is Sqr 3 1 times 1= sqr 3 1

And then you multiply the denominators. 2 times 2=4

you end up with √3 ¼ and multiply them and you get the answer. The sqr of 3 is 1.7320 and the you multiply it by 1 over 2. When you simplify it the answer is 0.433

Recall that the area of a triangle is (1/2)*Base*Height. The base of this triangle is s. You can determine the height by breaking the triangle into two right triangles and using the trig function for tangent (you'll need to figure out what the appropriate angle measure is if it's not given).

Area=(1/2)*Base * Height

Find the height of the triangle using trigonometry. (The height would be opposite of the base angle. The side of the right triangle will be 1/2 of the side of the polygon.)

tan(angleº) = opposite/adjacent

tan(angleº) = h / (s/2)

h = [tan(angleº)] * (s/2)

You now have the base, s, and the height, [tan(angleº)] * (s/2). Substitute these values in the formula

Area = (1/2)*Base*Height

If s is 1 meter, the area would be 0.344 square meters. If s is 5 inches, the area would be 25*0.344 square inches, or 8.6 square inches.

You just need to use trig to find the height of the triangle, then use that to find the area of the triangle, then multiply it by the number of triangles in the polygon.

The trick when faced with any polygon is to be able to find the missing information, whether it's the missing height of a triangle or the missing base of a trapezoid. By using the theorems you've learned for area and the properties of right triangles you'll be able to do this for many different shapes.

Also here is the link for the picture. http://i561.photobucket.com/albums/ss51/Kevinvm35/polyarea.jpg

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,442

hi laurenwest144

Thanks for the image. If you extend the 10 line until it reaches the 20 line you will see this splits the shape into a rectangle and a triangle. By subtracting 8 from 20 you'll get the size of one side of the triangle. Even though it points downwards in the image, let's call that the height. You also need the base of the triangle. Easiest way to get this is by using Pythagoras' theorem. With two sides you can calculate the third (the base).

Going back to the others, for an equilateral triangle and a hexagon divided into equilateral triangles, all sides are equal so you can use sin cos or tan. I think cos is the easiest but there's not much difference in the calculation. I've realised that it'll have to be tan for the pentagon because the distance to the centre of the pentagon is not the same as the side so cos is not possible. Sorry about that. I have edited my earlier post so it doesn't put any other reader off.

Q1. I agree with your answer for this.

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 Bundy

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**laurenwest144****Member**- Registered: 2017-11-15
- Posts: 8

The problem is I know all the answers but I do not know how to get there. I do not have a calculator and I do not understand tan, cos, and sin. I have been trying to teach myself how to do this problem and I am really bad at math so it is not working. This is what I have for number 3 of the worksheet. 3. A regular pentagon with a side of 3 centimeters

With the pentagon you divide the shape into 5 equilateral triangles. You divide the triangles in two and then you take the side which is 3 cm and put it over 2 so you end up with 3/2 and the side that is opposite the 60 degree angle will be the square root of 3 times the 3/2. I know the base of the triangle is 3 centimeters because the side of the pentagon is 3cm. So now I just need to find the height. Once you multiply the numerators you get 3 squared and you multiply it by 0.344 and the multiply that answer by 5 because there are five triangles in the pentagon and you get the answer 15.48.

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**laurenwest144****Member**- Registered: 2017-11-15
- Posts: 8

Never mind, I found a calculator.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,442

hi laurenwest144

Yes, you'll definitely need a calculator for these. But there are plenty of on-line sites that will do the calculation as well.

When you make those triangles for a pentagon they are not equilateral. They are isosceles. 360/5 = 72 so not 60-60-60 but 72-54-54. When you split it in half the half triangle is 36-90-54 and you need to get the height by 1.5 x tan(54)

Bob

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 Bundy

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**laurenwest144****Member**- Registered: 2017-11-15
- Posts: 8

okay thank you so much!

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