Area of a pentagon with a apothem?? help please!?

Question

Find the area of a regular pentagon with sides 7 millimeters long and apothem 408mm long. (A=1/2aP)

can you please show how you got your answer the best you can so i can understand how to do this stuff***

Answer 1

There is an error in the question. A regular pentagon with 7 mm sides has an apothem approximately 4.8 mm long.

A = (1/2)ap

A = (1/2)(4.8 mm)((5)(7 mm))

A = (1/2)(4.8 mm)(35 mm)

A = (2.4 mm)(35 mm)

A = 84 mm squared

Answer 2

coz ur pentagon consists of 5 equal triangles so u just need to calculate the area of 1 triangle:
7 x 408 x 1/2 = 1428 (sp. mms)
then just mutiply by five times then u get the total area f the pentagon:
1428 x 5 = 7140 (sp.mms)
But there must be something wrong. How can a pentagon with 7 mm edge can have a apothem of 407 mms? It's just contradictory!!!

Answer 3

Since the pentagon is regular, its perimeter is 5*7=35mm; therefore, the are is (35*408)/2. However, I think that you might have written this problem down wrong because the apothem could not possibly be 408mm if the sides are only 7mm long.

Answer 4

Divide the pentagon radially into 5 isosceles triangles with height 13 cm. each and each triangle has a vertex attitude of 360/5 = seventy two° and base angles of (a hundred and eighty - seventy two)/2 = fifty 4°. portion of triangle = 13(13cot(fifty 4°)) = 122.8 cm² portion of pentagon = 5(122.8) = 614 cm²

Answer 5

a regular n-gon have the central angle as great as 360/n, therefore the area of a n-gon is
Area = n*sides*sides*sin(90-180/n)

the sides is 7mm and the n-gon is a 5-gon therefore

Area =5*7mm*7mm*sin(90-36)
Area =5*49mm2*sin(54)
Area =198.209mm2

I know a software which can calculate the area or volume of any regular geometry please refer to the source.