SOLUTION: The regular hexagon ABCDEF has sides of length 2. The point P is the midpoint of AB. Q is the midpoint of BC and so on. Find the area of the hexagon PQRSTU.

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Question 1207466: The regular hexagon ABCDEF has sides of length 2. The point P is the midpoint of AB. Q is the midpoint of BC and so on. Find the area of the hexagon PQRSTU.
Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.
The regular hexagon ABCDEF has sides of length 2. The point P is the midpoint of AB.
Q is the midpoint of BC and so on. Find the area of the hexagon PQRSTU
~~~~~~~~~~~~~~~~~~~ 

The area of the regular hexagon ABCDEF is 6 times the area of an equilateral triangle 
with the side length 2.


So, the area of ABCDEF is   =  square units.


To get the area of the regular hexagon PQRSTU, we should subtract from the area ABCDEF 
the area of triangle PBQ 6 times.


PBQ is an isosceles triangle with the lateral side length of 1 and the concluded angle of 120 degrees.
Therefore, its area is

     = .


Thus the area of PQRSTU is  

     -  =  =  = 7.794228634 square units.    ANSWER

Solved.



Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
the area of a hexagon is equal to 3 * sqrt(3) / 2 * a^2, where a is the length of one side of the regular hexagon.

the area of your original hexagon is equal to 3 * sqrt(3) / 2 * 2^2 = 10.39230485.

when you connect the midpoints of two adjacent triangles formed by your hexagon, then the side of that new hexagon is equal to 1.732050808.

the area of the new hexagon is equal to 3 * sqrt(3) / 2 * 1.732050808^2 = 7.794228634.

that should be your answer.

the only question you might have is how i derived the value of the side of the new hexagon formed by the midpoints of the original hexagon.

that takes a little explaining.

see my worksheet below.



in the diagram:

K is the center of the hexagon.

AKB is one triangle formed from the center of the hexagon to the sides of the hexagon.

BKC is an adjacent triangle formed from the center of the hexagon to the sides of the hexagon.

KP is a line from the center of the hexagon to the midpoint of one side of the hexagon.
KQ is a line from the center of the hexagon to the midpoint of an adjacent side of the hexagon.

the length of AB is 2 and the length of BC is 2.

you form one side of the new hexagon by connecting point P to Q.
that forms triangle PKQ which is one of the 6 triangles of the new hexagon.

angle AKB is split in two by the line KP which terminates at the midpoint of line AB.

AK is the hypotenuse of right triangle AKP.
angle AKP is 30 degrees.
sine(30) = 1 / AK
AK = 1 / sine(30) = 2.
that's how the length of AK was derived to be 2.

PK = 1.732050808.
that's found because tan(30) = 1 / PK
PK = 1 / tan(30) which is equal to 1.732050808.

when you connect point P to Q, that line intersects with line KB at point R.

that forms right triangle PRB.
the hypotenuse of that right triangle is PB whose length is equal to 1.

cosine (RPB) = PR / 1
solve for PR to get PR = 1 * cosine(RPB)
that becomes cosine(30) = .8660254038.

the length of PQ is twice that to be equal to 1.732050808.

that's one side of your new hexagon.

area of your new hexagon is equal to 3 * sqrt(3) / 2 * 1.732050808 = 7.794228634.

that should be your answer.







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