What is the perimeter of a regular heptagon inscribed...

Question

What is the perimeter of a regular heptagon inscribed in a circle of radius 7? Round your answer to 2 decimal places.

Answer 1

We can determine the perimeter of a regular heptagon which is inscribed in a circle in the following way.

 

1.a regular heptagon inscribed in a circle will form 7 isosceles triangles, each of which will have an apex angle of ≅ 51.43^o (360^o ÷ 7)

2. two of the legs of the triangles will have a length of 7, which is the radius of the circle

3. using the side angle side formula, we can calculate the length of the third leg, which we will label a

4. that formula is a² = b² + c² - (2bc x cos 51.43^o)

5. our equation becomes a² = 7² + 7² - (2 x 7 x7 x 0.623)

6. expanding, we get a² = 98 - 61.05 ≅ 36.95

7. solving, a ≅ 6.08

8. we then multiply that length by 7 to obtain the perimeter, which is ≅ 42.55

Answer 2

This problem has several stages.

 

Use the formula for sum of interior angles in a heptagon to start.

 

Sum = 180(n - 2), where n = 7 for heptagon.

 

Sum = 180(5) = 900.

 

 

The heptagon can be partitioned into seven isosceles triangles.

 

The measure of each interior angle of the heptagon equals the sum of the two base angles of one isosceles triangle.

 

The two base angles = 900/7 = 128.57 degrees.  One base angle is 128.57/2 = 64.28 degrees.

 

 

Half of one of the isosceles triangles is a right triangle of hypotenuse 7 and base equal to one-fourteenth the perimeter of the heptagon.  The angle between the base and the hypotenuse is 64.28 degrees.   Use the definition of cosine to find the length of the base of the right triangle.

 

7cos(64.28) = 3.04

 

Multiply by 14

 

Perimeter of heptagon = 14(3.04) = 42.56

 

 

Answer 3

draw 2 radii to form a triangle

use the law of cosines

c^2=a^2+b^2-2abcos(360/n) where n=7   (360/n is a central angle)

a and b are both radii so call a and b both r

c is one side of the regular heptagon

c^2=r^2+r^2-2*r*r*cos(360/7)

c^2=2r^2-2r^2cos(360/7)

c^2=r^2(2-2cos(360/7)

take the square root of both sides

c=r√(2-2cos(360/7))

r=7 and cos(51.4285º)=0.62349

c=7√(2-2(0.62349))

c=7√(2-1.24698)

c=7√0.75302

c=7*0.867767

c=6.074369 is the length of one side of the heptagon

multiply this value by 7 to get the perimeter and round off

7*6.074369=42.520583

rounded off you get 42.52 units for the perimeter of the heptagon  (answer)

Answer 4

There are several ways to do this one.  One way is to use the sine ratio.  I wish I could include a picture, but see if you can follow this description.  I suggest you try to create your own diagram.

 

We can view the heptagon as being made up 7 congruent isosceles triangles.  The legs of these triangles are radii of the circle, so we know they are 7 units long.  The vertex angle of each of these triangles is 360/7 = 51.43 degrees.  Ok so far?  We are going to use this info to find the length of the base of one of the triangles, then multiply that number by 7 to get the perimeter.

 

Let's focus on just one of those seven triangles.  If you divide it down the middle (from its vertex angle to the midpoint of its base) you create two right triangles.  The hypotenuse is 7 units long, and the angle opposite the leg that is half of the base of our isosceles triangle in one half of 51.43, or 25.71 degrees.

 

How's your diagram look?

 

Do you know that the sine of an angle in a right triangle = the opposite leg/the hypotenuse?  Let's set up a small equation using x to represent the leg of the right triangle opposite the 25.71 degree angle:

 

sin  =  opp / hyp

 

sin(25.71)  =  x / 7

 

0.4339  =  x / 7   (I got the sine of 25.71 using a scientific calculator)

 

7(.4339)  =  x

 

3.037 = x

 

Remember, this is just half of the base of our isosceles triangle.  Multiply 3.037 by 2 and we get the length of the full base, that is the length of one side of the heptagon.

 

3.037 x 2 = 6.074

 

Since the heptagon has 7 sides, we multiply 6.074 by 7 to get the total perimeter:

 

6.074 x 7 = 42.52 units (rounded to two decimal places.)

 

Fun!