Math Is Fun Forum

Never mind, I found a calculator.

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

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.