Calculate the PERIMETER of a regular pentagon inscribed in a circle with radius 5.4 cm. Not radius, damnit!?

Question

Step by step, please.

Answer 1

Since it's a regular polygon, it divides the circle into 5 72-degree (360/5) isoceles triangles.

Each triangle would have one 72-degree angle in the middle, and 2 54-degree angles ((180-72)/2).

The isoceles sides would be both a radius and the hypotenuse of a right triangle whose base is 1/2 the length of a side of the pentagon.

So each side would be 2(5.4cos54) = 6.35cm, and the total perimeter would be 5(6.35) = 31.75cm.

I think.

Answer 2

A circle with a radius of 5.4 is inside a regular pentagon.

If we created 5 isosceles triangle within the circle (with each base of the isosceles triangle being a side of the pentagon) we would end up with a triangle that has congruent sides with a length of 5.4
To find the degree of acute angle of the isoceles triangle= 360/5 = 72
To find the degree of the remaining angles 180-72 = 108
Since it is an isosceles triangle the bottom angles are equal so 108/2 = 54
If we divide the isosceles triangle in 1/2, we would get a right angle. Using Sin = opposite/hypotenuse we can determine the length of the opposite side.

So, because we divided the isosceles triangle in 1/2, the acute angle is now 36 (72/2).
So the angles of our triangle is 36, 54, 90.

Sin 36 = opposite/hypotenuse
Sin 36 = opposite/5.4
.5644 = opposite/5.4
3.1744 = opposite (the length of 1/2 side of the pentagon)

Since there are 5 sides to a pentagon, and each 1/2 side is 3.1744, then the perimeter is 3.1744 x 2 x 5= 31.74.

Answer 3

via fact the pentagon is inscribed, its 5 corners will touch the circle. (A circumscribed pentagon could have its factors touching the circle, meaning this is greater than the circle itself.) Take the pentagon and harm it into 5 equivalent pie slices on the middle of the circle. each slice is an isosceles triangle. the perspective on the middle of the circle is precisely 72° (it is, 360°/5). The chord c on the different end of the triangle is one edge of the pentagon. the different 2 factors of the triangle, the two touching the 72° perspective, are the radius of the circle--9 inches, or merely say r for now. The triangle can now be chop up in a million/2. Bisect the perspective and the chord and you will have a suitable triangle with an perspective of 36°, an opposite side c/2 (a million/2 the dimensions of the chord), and a hypotenuse r. c/(2r) = sin(36°) c = 2r sin(36°) via fact that each chord is barely one edge of the pentagon, and that they are all equivalent, the fringe is 5c. P = 5c = 10r sin(36°) P = 10(9 inches) sin(36°) P ? fifty two.9 inches

Answer 4

Draw it, and make a triangle by joining two consecutive vertices to the center. This makes a triangle with two sides of 5.4 cm, one side = a side of the pentagon (which you want) and an angle between the 5.4s of 360/5 = 72 degrees. So the other angles are (180-72)/2 or 54 each. You can use either law of sines or law of cosines to get the side of the pentagon then multiply that by 5 for perimeter.


Law of sines: sin 54/5.4 = sin 72/p (p= side of pentagon)

p sin 54 = 5.4 sin 72

= = 5.4 sin 72 / sin 54 = 6.34808...
times 5 is 31.7404...

The answer above is right also, without using sine law. Good answer!

Answer 5

P = 5R sqrt(10 - 2 sqrt[5])/2

P = 5(5.4)(sqrt(10 - 2sqrt(5))/2)
P = 27sqrt(10 - 2sqrt(5))/2
P = about 31.74cm