Use trigonometric ratios to express the area of a regular polygon with 9 sides and lengths of 8.

Marco Hudson

Marco Hudson

Answered question

2022-08-11

Using trigonometric ratios to express area of regular polygons
I am very confused. My book just asked me to use trigonometric ratios to express the area of a regular polygon with 9 sides and lengths of 8.
So far I have learned how to use the sin cos and tan in right triangles and have no idea how this applies to all polygons. Can someone please explain this to me

Answer & Explanation

Olivia Petersen

Olivia Petersen

Beginner2022-08-12Added 16 answers

Step 1
Sketch a regular 9-gon. Join the centre of the 9-gon to the 9 vertices. This divides the 9-gon into 9 congruent triangles. To find the area of the 9-gon, we will find the area of one of these triangles, and multiply the result by 9.
Concentrate now on one of these triangles AOB, where O is the centre and A and B are two consecutive vertices. We will find the area of A O B.
Step 2
We have A B = 8, and A O B = 360 9 = 40 .
Drop a perpendicular from O to AB, meeting AB at M. Note that A O M = 20 .
Also, A M O M = tan ( 20 ). But A M = 4, so O M = 4 tan ( 20 ) .
Now we know the base AB of A O B, and the height OM. The area of A O M is therefore 1 2 8 4 tan ( 20 ) .
Finally, multiply by 9 and simplify a bit. We get 144 tan ( 20 ) , or if you prefer, 144 tan ( 70 )
musicbachv7

musicbachv7

Beginner2022-08-13Added 4 answers

Step 1
If the polygon has 9 sides, the angle of each of the 9 sectors is 2 π 9 .
If you cut this angle in half and draw an altitude to one of the sides, you get a right triangle with base 4 (half of 8) and an angle of 2 π 18 (the angle is also cut in half). So the area of that piece is base*height/2 or 1 2 ( 4 ) ( 4 ) cot ( 2 π 18 )
Step 2
The area of the whole polygon is 18 times that, so A = 8 18 cot ( 2 π 18 ) and I'm bad at mental math so I'll let you finish simplifying.

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