Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
3. Section 9.3
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Exercise 89 Page 514

Practice makes perfect
a

Let's draw a diagram illustrating the Pentagon.

To find the area enclosed by the Pentagon we can divide it into five congruent isosceles triangles, each with a vertex angle of 360^(∘)5=72^(∘). Let's add this information to the diagram.

By dividing the pentagon like this we can find its total area. We will do so by finding the area of one of the triangles and multiplying by 5. In an isosceles triangle the altitude from the vertex angle is also the median, which bisects the opposite side.

With the information in the diagram we can find the altitude by using the tangent ratio.
tan θ = Opposite/Adjacent
tan 36^(∘) = 460.5/h
Solve for h
tan 36^(∘) h = 460.5
h = 460.5/tan 36^(∘)
h = 633.82387...
h ≈ 633.82
When we have the triangle's altitude we can determine its area. Then, if we multiply this number by 5 we get the Pentagon's area. 5 (1/2(921)(633.82))≈ 1.46 * 10^6 The Pentagon covers about 1.46* 10^6 square feet.
b
Converting between miles (mi.) and feet (ft) will involve using a conversion factor. 1mi./5280ft Note that we want to convert from square feet to square miles, so we need to use the square of the above factor. (1mi./5280ft)^2 Multiplying 1.46* 10^6 square feet by this conversion factor will convert it to square miles.
1.46* 10^6 ft^2 * (1mi./5280ft)^2
1.46* 10^6 ft^2 * 1^2 mi.^2/5280^2ft^2
1.46* 10^6 ft^2 * 1 mi.^2/27 878 400ft^2
1.46* 10^6 ft^2* 1^2 mi.^2/27 878 400ft^2
1.46* 10^6 ft^2* 1^2 mi.^2/27 878 400ft^2
1.46* 10^6 * 1^2 mi.^2/27 878 400
1.46* 10^6 mi.^2/27 878 400
0.052370... mi.^2
0.05 mi.^2