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Please help me I have 2 questions.

Please help me I have 2 questions.-example-1
User Tim Lytle
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1) You know that the area (A) of a regular polygon can be computed from the apothem (h) and the perimeter (P) by
.. A = (1/2)*h*P
You also know that a regular pentagon's perimeter (P) is 5 times the length of one side (s).
.. 37.2 m^2 = (1/2)*(3.2 m)*(5s)
.. 37.2 m^2 = (8 m)*s
.. (37.2 m^2)/(8 m) = s = 4.65 m

The length of one side is 4.65 m.


2) There are several ways you can figure this, all giving the same result.
a) The side length will be (12 cm)*√2, and the area is the square of that:
.. A = ((12 cm)*√2)^2 = 144 cm^2 * 2 = 288 cm^2

b) The area is half the product of the diagonal lengths.
.. A = (1/2)*(2*12 cm)^2 = (1/2)*576 cm^2 = 288 cm^2

c) The area is 4 times the area of the right triangle with sides of 12 cm.
.. A = 4*(1/2)*(12 cm)^2 = 2*144 cm^2 = 288 cm^2

d) The area is 4 times the area of one triangle with central angle 90°.
.. A = 4*(1/2)*(12 cm)^2*sin(90°) = 2*144 cm^2*1 = 288 cm^2

The area of the square is 288 cm^2.
User Jerad Rose
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