# Hexagon Question

• MHB
• Ilikebugs

#### Ilikebugs

I know AM is 10... and that's it. I don't know.

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I think what I would do to find the area of the hexagon, is to take the area of the entire square, and subtract the areas of $\triangle AMN$ and $\triangle CPQ$.

First, can you give the area of $\triangle AMN$?

50, because (10)(10/2)?

Yes, we have:

$$\displaystyle \triangle AMN=50\text{ cm}^2$$

Okay, now in the last diagram, consider the diagonal $\overline{AC}$ What is its length? If we look at where this diagonal intersects $\overline{MN}$ and $\overline{PQ}$ we see the it is divided into 2 equal segments and a third segment which is the altitude of $\triangle AMN$ with $\overline{MN}$ serving as the base. Can you find this altitude?

AC is the square root of 800, but I don't understand what else youre saying.

A square whose sides measure $x$ in length will have a diagonal of length $\sqrt{2}x$, which means $\overline{AC}=20\sqrt{2}\text{ cm}$, and this agrees with what you found when you simply.

Okay, so we now know the length of $\overline{AC}$. So next, we can subtract from this the altitude of $\triangle AMN$ where $\overline{MN}$ is the base. This will be half the distance of the diagonal of a square having side lengths of $10\text{ cm}$. Or, we can simply observe this is 1/4 of the length of $\overline{AC}$. So, we are left with 3/4 of the diagonal $\overline{AC}$ which we must cut in half to get the altitude of $\triangle {CPQ}$ where $\overline{PQ}$ is the base. What do you find?

3/8 of 20√2?

Let's observe that what we have is that:

$$\displaystyle x=\overline{PQ}=\frac{3}{4}20\sqrt{2}\text{ cm}=15\sqrt{2}\text{ cm}$$

Now, using our knowledge of the length of the diagonal of a square, we may state:

$$\displaystyle x=\sqrt{2}\overline{CP}$$

Can you now find $\overline{CP}$ and thus the area of $\triangle CPQ$?

15, so 15 times 7.5 which is 112.5?

Yes, that's correct...so what is the area of the hexagon?

400-162.5= 237.5?

Let's let $A$ be the area of the hexagon. We then have:

$$\displaystyle A=\left(20^2-\frac{1}{2}10^2-\frac{1}{2}15^2\right)\text{ cm}^2=\frac{475}{2}\,\text{cm}^2=237.5\text{ cm}^2\quad\checkmark$$

Good work! (Star)