The expression that represents the volume of the prism, in cubic centimetres is [tex]28.476(2x+1) \: \rm cm^3[/tex] approx
If the prism is such that if we slice it horizontally at any height smaller or equal to its original height, the cross section is same as its base, then its volume is:
[tex]V = B \times h[/tex]
where h is the height of that prism and B is the area of the base of that prism.
For this case, we are provided a prism which has pentagon as its cross sections, and its height is [tex]2x + 1[/tex] units.
Let B be the area of those pentagons, then:
Volume of the prism = [tex]B(2x + 1) \: \rm cm^3[/tex]
It is provided that the apothem of the pentagon is 2.8 cm.
An n sided regular polygon has internal angles' sum as 180(n-2)°
Thus, each internal angle would be of 180(n-2)/n°
For n = 5, this comes as 108°
Since the line AO is bisecting the internal angle, the angle OAB is o of 108/2 = 54°
Using the tangent ratio, we get:
[tex]\tan(54^\circ) = \dfrac{|OD|}{|AD|}\\\\|AD| = \dfrac{2.8}{\tan(54^\circ)} \approx 2.034 \: \rm cm[/tex]
Thus, area of the triangle AOD = [tex]\dfrac{1}{2} \times |AD| \times |OD| \approx \dfrac{1}{2} \times 2.034\times 2.8 = 2.8476 \: \rm cm^2[/tex]
There are 10 such triangles, all congruent, thus:
Area of pentagon = [tex]B \approx 10 \times 2.8476= 28.476 \: \rm cm^2[/tex]
Thus, volume of the pentagonal prism is:
[tex]V = B \times h \approx 28.476(2x+1) cm^3[/tex]
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