Another test... PleaseHurry Please.

[Guest]

Well I have some problems that I don't get and I need to do well on the 2nd test (almost christmas break, so lots of tests!)

1:

2:

3:

Thanks!!!!!

 

[soroban]

Hello, AirForceOne!

1:

A regular (equilateral) triangle has three \(\displaystyle 60^o\) angles.

A regular quadrilateral (square) has four \(\displaystyle 90^o\) angles.

A regular pentagon has five \(\displaystyle 108^o\) angles (<u>not</u> a multiple of \(\displaystyle 30^o\))

A regular hexabon has six \(\displaystyle 120^o\) angles.

There are 18 vertex angles in the drawer.
. . Thirteen of them are multiples of \(\displaystyle 30^o\).

Therefore, the probability is \(\displaystyle \frac{13}{18}\).

2:

An exterior angle and an interior angle are supplementary; their sum is \(\displaystyle 180^o\).
. . So we have: \(\displaystyle \,x\,+\,\frac{1}{4}x\:=\:180\;\;\Rightarrow\;\;\frac{5}{4}x\,=\,180\;\;\Rightarrow\;\;x\,=\,144^o\)

Each interior angle is \(\displaystyle 144^o\). .How many sides does it have?


In a polygon of \(\displaystyle n\) sides, the sum of the interior angles is: \(\displaystyle \,180(n\,-\,2)\) degrees.

If the polygon is equiangular, each interior angle has: \(\displaystyle \,\frac{180(n\,-\,2)}{n}\) degrees.

Our polygon has an interior angle of \(\displaystyle 144^o\).

. . So we have: \(\displaystyle \,\frac{180(n\,-\,2)}{n}\:=\:144\;\;\Rightarrow\;\;180n\,-\,360\:=\:144n\;\;\Rightarrow\;\;36n\,=\,360\)

Therefore: \(\displaystyle \,n\,=\,10\;\) . . . It is a decagon.


Edit: Ha! .... miscounted on #1. .Thanks for catching it, jacket81.

 

[jacket81]

problem

I could be wrong, but didn't he/she make a mistake on the first problem. Isn't the probability (18-5)/18 = 13/18, not 11/18?

 

[soroban]

Hello, AF1!

In #3, the diagram indicates \(\displaystyle \Delta NTU\), but you said \(\displaystyle \Delta NTE\).

\(\displaystyle \Delta NTU\) is isosceles.
. . Quadrilaterals \(\displaystyle NOIT\) and \(\displaystyle TSEU\) are congruent.
. . Hence, \(\displaystyle NT\,=\,TU\)

 

[jacket81]

problem

ALso, on the 2nd problem, I dont undestand something.
I thought the exterior angle would have to be greater then interior angle.
But it said the exterior angle is 25% of interior angle!
Is that possible?
Maybe I'm not looking at it correctly!

 

[soroban]

Re: problem

Hello, jacket81!

I thought the exterior angle would have to be greater then interior angle.

You must have the concept confused with another . . .

         |             |
         *   polygon   *C
          \           /
           \         / θ
            *-------* - -
. . . . . . A . . . B

You can see that \(\displaystyle \theta\) is an exterior angle
. . and it is less than the interior angle \(\displaystyle ABC\).

 

[jacket81]

problem

okay, i was thinking the angle "all away around", in other words, i thought the exterior angle was (360-interior angle).