The radius of a circle is 100 inches. Find the perimeter and area of an inscribed regular pentagon.?

Answer 1

Picture the pentagon as five congruent triangles, constructed by connecting each vertex to the center. Now, look at just one of those triangles. The angle at the center of the figure is 360 / 5 = 72. You already know that two of the sides of the triangle have a length of 100 in. The remaining, unknown side is one fifth of the pentagon's perimeter. Take the triangle and bisect it into two right triangles. Now, half of the base of the original triangle will have a length of 100*sin(72/2) = 100*sin(36). So the full length of the base of the triangle is 200*sin(36), and the perimeter of the pentagon is 1000*sin(36).

For area, you also need to find the length of the line that bisected the triangle into right triangles, because that is the height of the triangle. That length is 100*cos(36). The area of the triangle is then one-half base times height, which is 0.5*200*sin(36)*100*cos(36) = 10000*sin(36)*cos(36). The total area of the pentagon is therefore 50000*sin(36)*cos(36).

Answer 2

Since the pentagon is inscribed, its five corners will touch the circle.
(A circumscribed pentagon would have its sides touching the circle,
meaning it's bigger than the circle itself.)
Take the pentagon and break it into five equal pie slices at the center of the circle.
Each slice is an isosceles triangle.
The angle at the center of the circle is exactly 72° (that is, 360°/5).
The chord c at the other end of the triangle is one side of the pentagon.
The other two sides of the triangle, both touching the 72° angle,
are the radius of the circle--100 inches, or just say r for now.
The triangle can now be split in half.
Bisect the angle and the chord and you'll have a right triangle with an angle of 36°,
an opposite side c/2 (half the length of the chord), and a hypotenuse r.
c/(2r) = sin(36°)
c = 2r sin(36°)
Since each chord is just one side of the pentagon, and they're all equal, the perimeter is 5c.
P = 5c = 10r sin(36°)
P = 10(100 inches) sin(36°)
P ≈ 587.8 inches

Answer 3

ANSWER:
Perimeter is 588 inches (rounded to the nearest inch)
Area is 23776 square inches (rounded to the nearest square inch)

By drawing five lines from the centre of the circle to the vertexes of the pentagon, you get 5 congruent triangles. Each of these triangles is isosceles since the radius of the circle make up two of the sides of the triangle.

The angle formed by the two equal sides of the isosceles triangle is 2pi/5 rad. The third side of this triangle can be found using the Cosine Rule.

If that third side is L inches,
L^2 = 100^2 + 100^2 - 2 x 100 x 100 x cos (2pi/5)
L = 117.6 inches

The perimeter of the regular pentagon is 5 times of 117.6 inches = 588 inches (rounded to the nearest inch)

The area of one isosceles triangle = 1/2 x 100 x 100 x sin (2pi/5)
= 4755.3 square inches

The area of the pentagon = 5 x 4755.3
= 23776 square inches (rounded to the nearest square inch)

I hope this helps.

Answer 4

the pentagon can be divided into 5 triangle with two equal sides. the angles will be 72 54 54. now side opposite 54 = 100 inches so side opposite 72 is 72/54 * 100 inches = 4/3*100 inches.
drop a perpendicular from the centre to the base. this is the height of the triangle. now we have a right angle triangle of angles 90 54 36. side opposite 90 is 100 inches so side opposite 54 is 90/54 * 100 inches = 3/5*100inches=60 inches.
area of triangle is 1/2 b*h = 1/2 * (4/3) * 100 *60 = 4000 sq inches. so area of 5 triangles 20000 sq inches.

Answer 5

From the centre of the circle divide the circle into five equal parts (360°/5 = 72°). Draw these line form the center of the circle to the circumference of the circle. Where the radii intersect with the circumference of the circle, these points will be the verteses of thePenagon. Join these consective points with straight lines, now you have your pentagon.
The pentagon now consists of five triangles. Now consider one triangle.
The centre internal angle is 72°. The remaining internal angles are [(180° - 72°)/2] = 54°.
The length of the third side of the triangle is:
a/sinA = b/ sinB
a = sinA*b/sinB
a = sin 72° *100/sin 54°
a = 0∙951 056 516 * (100)(0∙809 016 994)
a = 117∙557 0505 in. (Length of one side of Pentagon).

Parameter lenght of Pentagon:
Parameter lenght = Length of side x 5.
Parameter lenght = 117∙557 0505 in. x 5
Parameter lenght = 587∙785 2523 in.

Now calculate the perpendicular hight of the triange.
h = √[(AB)² - (Bd)²]
h = √[(100²) - (117∙557 0505 / 2)²
h = √[(100²) - (58∙778 525 25)²
h = √(6545∙084 97)
h = 80∙901 699 42 in

Area of ▲ = ½ b * h
Area of ▲ = ½ (117∙557 0505) * (80∙901 699 42)
Area of ▲ = 4755∙282 581

Area of Pentagon = Area of ▲ x 5
Area of Pentagon = 4755∙282 581 x 5
Area of Pentagon = 23 776∙412 92 in²

Answer 6

A regular pentagon can be divided into five congruent isoceles triangles, each having two sides of 100 inches and an included angle of 72 degrees.

The cosine law gives you the length of the side you weren't given.

c^2 = a^2 + b^2 - 2 a b cos(C*)
c = 117.557 inches

The perimeter of the pentagon is

P = 5c = 587.785 inches

The area of the pentagon is five times the area of one of the isoceles triangles.

The area of one of the isoceles triangles is twice the area of either of the right triangles formed from the bisection of side c.

The hypotenuse of either of those right triangles is 100 inches, of course. The base length is

base = c/2 = 58.7785 inches

The height of the right triangle is found from the Pythagorean theorem.

height = { (hypotenuse)^2 - (base)^2 }^0.5
height = 80.9017 inches

The area of one of those right triangles is half the base times the height.

Art = (base) (height) / 2
Art = 2377.64 square inches

The area of one of the isoceles triangles is twice the area of that right triangle.

Ait = 2 Art = 4755.28 square inches

The area of the pentagon is five times the area of the isoceles triangle.

Ap = 5 Ait = 23776.4 square inches