The polygons are regular polygons. Find the area of the shaded region to the nearest tenth.
@dmezzullo i need help
please help
We need one more constraint for this to be possible.
what do you mean this is all the problem is
Unless I'm missing something (which could be the case), there needs to be an extra constraint...
this is all it gave me
please help
I reckon I can have a crack if you still need help?
i need help
Ok, so I think we can break down the shapes in to two squares and 2 triangles.|dw:1366259113710:dw|We know the length of the diagonal side, which is 4. With me so far?
still with you
Great, we also know that the sides of a square are all the same. Let's call them \(x\)|dw:1366259222870:dw|Looking at the triangle with the sides I have marked x, we can use Pythagoras to calculate x. |dw:1366259261591:dw|What is the value of x?
4
Nope use Pythagoras' theorem.\[a^2+b^2 = c^2\]In this case our value for c is 4. And our value for a and b is x, making our equation\[x^2+x^2 = 4^2\]Does that make sense?
yes
Ok, so what do you get for x?
2x^2=16 right
So far so good
then do you divide 2 to make x^2=8
yep making x...?
x=64
No, square root both sides. \[x^2=8 \rightarrow \sqrt{x^2}=\sqrt{8}\] Making \[x=\sqrt{8}\]
Oh, crap, missed the whole first part of the case "regular polygons." Thanks, mate, for catching that!
No probs
Does that make sense @sandbeach ?
yes
Great now we know the length of the square, we can think of the shaded region as simply three identical squares with length \[\sqrt{8}\]. Do you know how to find the area of a square?
lenth x width
Awesome, so what is the area of one of these squares of length \(\sqrt{8}\)?
i dont know the anser to sqrt8 x 4
I have to go now, but basically the size of a square of length \(\sqrt{8}\) is 8. We know that there are 3 squares making up our shaded region. \[3 \times 8 = 24\] That is your answer.
so that is the area to the shded region
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