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about what percent of the Tennessee flag is redwrite as a %

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bergcar | Student, Grade 11 | Valedictorian

Posted February 20, 2012 at 4:30 AM via web

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about what percent of the Tennessee flag is red

write as a %

Tagged with math, write in percent

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wards31 | College Teacher | (Level 1) Adjunct Educator

Posted February 20, 2012 at 4:44 AM (Answer #1)

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I would say that about 85 -90% of the Tennessee flag is red.  Some of the renderings show the middle circle a little larger than others, so it depends on which photo or real flag they are basing their question.

Sources:

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beckden | High School Teacher | (Level 1) Educator

Posted February 20, 2012 at 12:40 PM (Answer #2)

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From the Tennessee Blue Book,

An oblong flag or banner in length one and two thirds times it's width.

...the width of the white stripe to be one fifth that of the blue
bar, and the total width of the bar and stripe together to be equal to one-eighth of the width of the flag.

In the center of the red field shall be a smaller circular field of blue, separated from the surrounding red field shall be a smaller circular field of blue, separated from the surrounding red field by a circular margin or stripe of white of uniform width and of the same width as the straight margin or stripe first mentioned. The breadth or diameter of the circular blue field, exclusive of the white margin, shall be equal to one-half of the width of the flag.

So now we have enough to find the percentage of the flag that is red.

First let the width be w and then the length is 5/3w so the total area is 5/3w^2

The area of the white stripe and blue bar is w*(1/8w)=1/8w^2

The width of the white stripe is complicated, and we need it to find the diameter of the central circle.  First the white stripe is 1/5 the blue stripe and the total width of the two lines is 1/8 the width.  So if b is the width of the blue bar, then the white stripe width s=1/5b and since s+b=1/8w then 1/5b+b=1/8w so 6/5b=1/8w or b=5/48w and then s=1/48w.

So the diameter of the circle in the center is `1/2w+1/48w = 25/48w` , and then it's area is

In this case the diameter is `25/48w` so the radius is `25/96w` . So the area is

`pi(625/9216)w^2`

Now our percentage red is `(5/3w^2 - (1/8w^2+pi(625/9216)))/(5/3w^2)`

And this gives

`%red = (5/3 - (1/8+pi(625/9216)))/(5/3) *100 = 79.7168268% `

Which seems a lot more acurate from the pictures.

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