Challege Problem

[vampirewitchreine]

Hi guys,
I'm working on a challenge problem from a previous section(Areas of Regular Polygons and Circles)

Below is the figure that I'm working with:


I colored it in different so that it was easier to tell what the figure's edges are. I need to find the area of the shaded region (The diamond shape is the unshaded region, the 2 concave hexagon shapes are the shaded area.....)

Should I solve the diamond shape as a Parallelogram to get the right answer? The height of it should be 5 right?

Correct me if I'm wrong, isn't half the diagonal congruent to one side?

If the above statement is correct:
To find the apothem I have
\(\displaystyle a^2 = 5^2 - 2.5^2\)
\(\displaystyle a^2 = 25 + 6.25 \)
\(\displaystyle \sqrt {a^2} = \sqrt {31.25}\)
\(\displaystyle a ~\approx 5.59 \)

Leaving the area of each hexagons (if you see the diamond shape as where the two overlap)
\(\displaystyle Area = 6(\frac {1}{2}bh)\)
\(\displaystyle Area = 6 (\frac {1}{2} 5 * 5.59)\)
\(\displaystyle Area = 6 (2.5*5.59)\)
\(\displaystyle Area = 6 (13.98) \)
\(\displaystyle Area = 83.85 \)

(This is all that I have so far.... feel free to correct me)

 

[vampirewitchreine]

Hazzzzelllllllllll......go stand in the corner...

That's what I get for not proof reading.....


Corrected:
\(\displaystyle a^2 = 5^2 - 2.5^2\)
\(\displaystyle a^2 = 25 - 6.25 \)
\(\displaystyle \sqrt {a^2} = \sqrt {18.75}\)
\(\displaystyle a ~\approx 4.33 \)

\(\displaystyle Area = 6(\frac {1}{2}bh)\)
\(\displaystyle Area = 6 (\frac {1}{2} * 5 * 4.33)\)
\(\displaystyle Area = 6 (2.5*4.33)\)
\(\displaystyle Area = 6 (10.83) \)
\(\displaystyle Area = 64.95 \)

 

[lookagain]

Corrected:
\(\displaystyle a^2 = 5^2 - 2.5^2\)
\(\displaystyle a^2 = 25 - 6.25 \)
\(\displaystyle \sqrt {a^2} = \sqrt {18.75}\)
\(\displaystyle a ~\approx 4.33 \)

\(\displaystyle Area = 6(\frac {1}{2}bh)\)
\(\displaystyle Area = 6 (\frac {1}{2} * 5 * 4.33)\)
\(\displaystyle Area = 6 (2.5*4.33)\)


\(\displaystyle Area = 6 (10.83) \)

\(\displaystyle This \ must \ be \ 6(10.825). \)
\(\displaystyle Then \ this \ would \ follow \ to \ the \ next \ (last) \ line.\)


\(\displaystyle Area = 64.95 \)

.

 

[vampirewitchreine]

I see what you're saying... I've had my calculator set fixed on the second decimal place, so I didn't even really notice the difference (I tried it using 10.83 and came out with 64.98)