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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 $\frac{p^4}{A^2}$?

 

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

 

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?

 

4.)$A,B,C,D,$ and $E$ are points on a circle of radius $2$ in counterclockwise order. We know $AB=BC=DE=2$ and $CD=EA.$ Find $[ABCDE]$.

 Jun 5, 2019
 #1
avatar+103097 
+2

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

 

 

cool cool cool

 Jun 5, 2019
 #2
avatar+103097 
+3

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

 

 

cool cool cool

 Jun 5, 2019
 #3
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0

can u solve 3 plzzzzzzz

Guest Jun 5, 2019
 #4
avatar+103097 
+2

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

 

 

cool cool cool

 Jun 5, 2019
 #5
avatar+103097 
+2

4.)$A,B,C,D,$ and $E$ are points on a circle of radius $2$ in counterclockwise order. We know $AB=BC=DE=2$ and $CD=EA.$ Find $[ABCDE]$.

 

 

Call the center of the circle , O

If AB + BC + DE = 2

Then triangles  AOB, BOC and DOE   must be equilateral.....and their combined area  =

3 (1/2) (2)^2 * √3/2  =

3√3  units^2

 

Since angles AOB, BOC and DOE  = 60°

And CD = EA

Then angles COD and EOA  must   each equal  [ 360 - 3(60)]  / 2  =  180/ 2  = 90°

So....the total area of triangles COD and DOE  =   2 (1/2) * 2^2  =  4 units^2

 

So....the total area of  ABCDE  =   4+ 3√3  units^2

 

 

cool cool cool

 Jun 6, 2019

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