# help mi wif tis questions

• Jun 23rd 2006, 06:54 AM
xiaoz
help mi wif tis questions
i duno how to do~ :(
• Jun 23rd 2006, 07:11 AM
Quick
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
• Jun 23rd 2006, 07:29 AM
xiaoz
how to do a(iii) ?
• Jun 23rd 2006, 08:16 AM
Soroban
Hello, xiaoz!

Quote:

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 $\displaystyle n = 8$ ?

Quote:

(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 $\displaystyle n.$

Quote:

(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?

• Jun 23rd 2006, 10:51 AM
CaptainBlack