if the perimeter of a regular pentagon is 1000 meters. what will be its area?

Answer 1

68,819.096 m^2

I did it by calculating the area of a triangle and then multiplying by 5, but the wikipedia link below gives a good approximation doing only one calculation.

Isha K, you should have used tangent instead of sine

Answer 2

Divide pentagon onto 5 triangles where angles at centre = 72°. Side opposite angle at centre is 200m,other two sides in triangle are x and other two angles in triangle are 54° each
By sine rule:-
200/sin72° = x/sin 54°
x = 200sin54°/sin72° = 170m
Area of triangle = 1/2 x 170 x 170 x sin72° m²
= 13743 m²
So area of pentagon = 5 x 13743 m² = 68715m²

Answer 3

Pentagon is regular.
Thus, each side = s = 1000/5 = 200 m

Now, this regular pentagon can be divided into 5 congruent isosceles triangles with base = s = 200 m

The vertex angle of each triangle = 360/5 = 72 degrees

Let, the height to be taken for each triangle = h

From a diagram of a regular pentagon, we can derive that,
sin (72/2) = (200/2) / h
=> h = 100 / sin 36
h = 170.13 m

Total area of the pentagon = 5 (area of one triangle)
= 5 (0.5 * s * h)
= 5 (0.5 * 200 * 170.13)
=> Total area = 85065 m^2

Answer 4

If a pentagon has a perimeter of 1000 meters, it follows that the length of each side is 1000/5 = 200 meters.

The area of a pentagon is given by the formula

A = (5/4) (s^2) cot(pi/5), where s is the length of one side. Therefore,

A = (5/4)(200)^2 cot(pi/5)
A = (5/4) (40000) cot(pi/5)
A= 50000 cot(pi/5)

A = 50000 sqrt[25 + 10sqrt(5)]

Answer 5

I reckon the area of a pentagon is (5/4)*(tan 54°)*side².

Since the side is 200m, this gives the area as 68,819m².

Answer 6

Each side is 200m.

A = 5a^2/4 cot π/5
A = a^2/4 sqrt(25 + 10 sqrt(5) )
A = 200^2/4 sqrt(25 + 10 sqrt(5) )
A = 10,000 sqrt(25 + 10 sqrt(5) )

using 1.72048 as an approximation for sqrt(25 + 10 sqrt(5) )
gives an area of roughly 17,204.8m^2.

Answer 7

i punched wrong # in calculator.

Pebble is correct, 68,819.