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  1. #1
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    help mi wif tis questions

    i duno how to do~
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  2. #2
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    Quote Originally Posted by xiaoz
    i duno how to do~
    the first one is all substitution.

    $\displaystyle A\;=\;\frac{\left(2n-4\right)\times90}{n}$ so substitute 8 for n

    $\displaystyle A\;=\;\frac{\left(2(8)-4\right)\times90}{8}$ and solve

    number 2 asks you to solve for n, so lets do it...

    $\displaystyle A\;=\;\frac{\left(2n-4\right)\times90}{n}$ use the distributive property

    $\displaystyle A\;=\;\frac{2n\cdot90-4\cdot90}{n}$ simplify

    $\displaystyle A\;=\;\frac{180n-360}{n}$ multiply both sides by n

    $\displaystyle An\;=\;180n-360$subtract 180n from both sides

    $\displaystyle An-180n\;=\;-360$ subtract

    $\displaystyle \left(A-180\right)n\;=\;-360$ divide both sides by A-180

    $\displaystyle n\;=\;\frac{-360}{A-180}$ simplify

    $\displaystyle n\;=\;\frac{360}{-A+180}$ simplify

    $\displaystyle n\;=\;\frac{360}{180-A}$ and thats what n equals. Substitute the given number for A (157.5)

    $\displaystyle n\;=\;\frac{360}{180-157.5}$ and solve, this will give you the answer for a.ii
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  3. #3
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    how to do a(iii) ?
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  4. #4
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    Hello, xiaoz!

    i duno how to do~

    a) The measurement $\displaystyle A^o$ of an interior angle of a regular polygon of $\displaystyle n$ sides
    . . . is given by the formula: .$\displaystyle A\;=\;\frac{(2n-4)\times 90}{n}$ **

    (1) Find the measure of an interior angle of a regular polygon with 8 sides.

    How about plugging in $\displaystyle n = 8$ ?



    (2) Write $\displaystyle n$ in terms of $\displaystyle A.$ .Hence, find the number of sides
    of the regular polygon with interior angles measuring 157.5 each.
    This requires some knowledge of Algebra I . . . too bad!

    We have: .$\displaystyle \frac{90(2n - 4)}{n} \;= \;A$

    Multiply both sides by $\displaystyle n:\;\;90(2n - 4) \;= \;An\quad\Rightarrow\quad 180n - 360 \;= \;An$

    Then: .$\displaystyle 180n - An \;= \;360$

    Factor: .$\displaystyle (180 - A)n \;= \;360\quad\Rightarrow\quad n \:=\:\frac{360}{180 - A} $

    We are told that $\displaystyle A = 157.5$

    Plug it in and determine $\displaystyle n.$



    (3) Find the regular polygon such that when the number of sides is doubled
    the measure of an interior angle is doubled.

    This is the only one that's tricky . . .

    For $\displaystyle n$ sides, the angle is: .$\displaystyle \frac{90(2n-4)}{n}\:=\:A\quad\Rightarrow\quad\frac{180n - 360}{n}\:= \:A$ [1]

    For $\displaystyle 2n$ sides, the angle is: .$\displaystyle \frac{90(4n-4)}{2n}\:=\:2A\quad\Rightarrow\quad \frac{360n - 360}{4n}\:=\:A$ [2]

    Equate [1] and [2]: .$\displaystyle \frac{180n - 360}{n} \:= \:\frac{360n - 360}{4n}\quad\Rightarrow\quad 720n^2$ $\displaystyle - 1440n \:= \:360n^2 - 360n $

    . . which simplifies to: .$\displaystyle 360n^2 - 1080n \:=\:0$

    . . which factors: .$\displaystyle 360n(n - 3) \:= \:0$

    . . and has the positive root: .$\displaystyle x = 3$


    This checks out.

    With $\displaystyle n = 3$ we have an equilateral triangle: interior angle $\displaystyle 60^o.$

    Double the number sides and we have a hexagon: interior angle $\displaystyle 120^o.$

    . . . . . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **
    This is a really stupid way of writing the formula!

    The proper form is: .$\displaystyle A\:=\:\frac{180(n - 2)}{n}$

    It can be shown that the sum of interior angles of any n-gon is: $\displaystyle 180(n - 2)$

    If all $\displaystyle n$ angles are equal, each angle measures: .$\displaystyle \frac{180(n-2)}{n}$ degrees.


    Someone is going to say, "What's the difference? .They're equal!"

    Well, that is certainly true.
    This means that we will have dozens of formulas to memorize . . .

    $\displaystyle A\;=\;\frac{(n - 2) \times 180}{n} \;= \;\frac{(2n - 4) \times 90}{n}\;=$ $\displaystyle \frac{(3n - 6) \times 60}{n} \;=\;\frac{(4n - 8) \times 45}{n}$

    . . . $\displaystyle = \;\frac{(5n -10) \times 36}{n} \;= \;\frac{(6n - 12) \times 30}{n} \;=$ $\displaystyle \frac{(9n - 18) \times 20}{n} \;\hdots$

    Catch my drift?

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  5. #5
    Grand Panjandrum
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    I have split this thread

    Xiaoz,

    I have split this thread so that your new question is in a thread
    of its own.

    It is a good idea to create a new thread when you have a new
    question, otherwise your thread may become messy and difficult to
    follow.

    RonL
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  6. #6
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    ok..
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