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3.)$ABCDEF$ is a regular hexagon with area $1$. The intersection of $\triangle ACE$ and $\triangle BDF$ is a smaller hexagon. What is the area of the smaller hexagon?

For simplicity..I have constructed a hexagon with a side of 2

See...the image

A  = (1, √3)       and   C =   (1, - √3)

So...the distance from A to C  =   2√3

And, by symmetry,  GL is (1/3)  of this  = ( 2/3)√3

And this is one side of the smaller hexagon  GHIJKL

So....the ratio  of   the area of the smaller hexagon to the larger hexagon  =  

(side of smaller hexagon)^2

______________________    =

(side of larger hexagon)^2

    [ (2/3)√3]^2           (4/9) * 3            4 * 3              3                1

___________  =       ______    =     ______  =    ____  =       ____

         2^2                      4                   4 * 9             9                 3

So..in our case...the area of the smaller hexagon  = (1/3) /  1  =    1/3 units^2

It is leftover from when the asker copy and pasted the question. Other sites use $'s to enclose latex, but this site does not. The dollar sign shouldn't be there.

What is the symbol $ ? 

Why is 15$ and not 15.What is the difference? 

Thank you.

Yes they are on the problems I have trouble with...... 

No

area doghouse = (area triangle) + (area square)

area doghouse = (\(\frac{1}{2}\times\)base \(\times\)height ) + (side)^2

area doghouse = (\(\frac{1}{2}\times3\times5\)) + \((5)^2\)

area doghouse = \((\frac{15}{2})\) + \((25)\)

area doghouse = \(7.5+25\)

area doghouse = \(32.5\) square feet

Corect answer the 1st.If you don't understand something tell me or you can send me in message.

I hope I help you.

Not the "nearest" integer......but the "next"  integer

So    

1.35  rounds to   2

So...we need 2 bags

All these questions look like they are from some type of online test...

54/40 = 1.35

12.5+7.5 = 20

Is this right 

can u solve 3 plzzzzzzz

Yep....and then round up to the next integer [ since we can't buy "partial bags" ]

So....what's the correct answer  ???

2.)Isosceles triangle $OPQ$ has legs $OP = OQ$, base $PQ = 2$, and $\angle POQ = 45^\circ$. Find the distance from $O$ to $\overline{PQ}$.

Bisect base  PQ

Call the point of bisection M

Then angle OMQ  = 90

Angle MOQ = 22.5

MQ = 1

And angle OQM = 67.5

So....by the Law of Sines

OM/sin 67.5   =  1/sin (22.5)

OM =  sin 67.5 / sin 22.5 =   cos 22.5 / sin22.5  = cot 22.5  = ( 1 + cos 45) / sin 45 =

1 + √2/2              1                           2

  ______  =       ___     +   1   =      ___  + 1    =   √2  + 1

    √2/2              √2/2                       √2

Oh my gosh I see my mistake I am so sorry Cphill stuff dosent sink in that face I have problems with these facts but I can assure you I know multiplucation 

So 

9*6= 54

so divide 54 by 40???

5 + 5  = 10

5 * 5  = 25

1.)A regular hexagon has side length $6$. If the perimeter and area of the hexagon are $p$ and $A$, respectively, what is the value of $\frac{p^4}{A^2}$?

The perimeter  =  6*6  =  36

The area is composed of 6 equilateral triangles and is given by :

6 * (1/2) * (6)^2 * √3/2   =  

3√3 / 2 * 36 =

54√3

So

p^4              36^4              (36)(36)(36)(36)          (2)(2) (36) (36)     (4)(36)(36)   

___   =       _______  =    ______________  =    _____________=  ________ =

A^2             54^2 * 3          (54) (54) (3)                (3)(3)(3)                    27

(4)(4)(36)

________   =     16 * 12   =     192

      3

Is the square 5*5=10  

then

10/2=5

+

15/ 1/2= 7.5 

so 7.5+5=12.5 is my answer 

I belive this is right..... 

Thanks let me see....

Remember, Nickolas....the area of a square  =  (side)^2

Calculate this.....(you got the triangle part correct....)

Add the two answers together to get the final answer....

The triangle right as you say: \(\frac{3\times5}{2}=7.5\)

How you find the area of square? Do you remember the type?