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    Can I assume <ABC is 45 degrees since AB is a straight line and AB is a straight line? I think that only applies to circle thereom.

    Also, work out the area of the original square piece of card.

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    (Original post by Moltenmo)
    Can I assume <ABC is 45 degrees since AB is a straight line and AB is a straight line? I think that only applies to circle thereom.

    Also, work out the area of the original square piece of card.

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    Yes.
    it is the exterior angle of a regular octagon and the sum of exterior angles of a polygon is 360
    Also you know AC and BC are the same length so it's an isosceles triangle and therefore the two base angles are equal


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    (Original post by gdunne42)
    Yes.
    it is the exterior angle of a regular octagon and the sum of exterior angles of a polygon is 360
    Also you know AC and BC are the same length so it's an isosceles triangle and therefore the two base angles are equal


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    Damn, I didn't even think about exterior angles for some stupid reason. Thank you.

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    1. Calculate the area of the octagon using sine rule.

    2. Calculate the area of the little right-angled triangle.
    1) Calculate the length AB using cosine rule.

    2) Calculate the length CB, given that angle ABC is 45 degrees.

    3) Given that AC=CB and the triangle is right-angled. Find the area.

    Add the area of the octagon and the triangles to get the total area.
    Good luck!
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    (Original post by Delia_*)
    The area of of original square should be approximately ...
    Please don't post dull slutions - it's against the rules of the Maths forum.
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    (Original post by Muttley79)
    Please don't post dull slutions - it's against the rules of the Maths forum.
    sorry, I never knew that there's such a rule. All I wanted was to help. Sorry again
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    There's different ways to work out the area of the square, but I personally would have used cosine rule to work out AB, then used pythagoras to work out AC and CB (because they'd be the same), and then added AB to 2AC and squared that. But that's just my way, other ways would work too.

    I think that would work, sorry if it doesn't, someone please correct me!
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    (Original post by IfYouCanDreamIt)
    There's different ways to work out the area of the square, but I personally would have used cosine rule to work out AB, then used pythagoras to work out AC and CB (because they'd be the same), and then added AB to 2AC and squared that. But that's just my way, other ways would work too.

    I think that would work, sorry if it doesn't, someone please correct me!
    How do you assume AC and CB are the same? It doesn't say or show that in the question?

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    (Original post by Moltenmo)
    How do you assume AC and CB are the same? It doesn't say or show that in the question?

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    Because of how you made them. You have removed a regular octagon from a square.
    And also because of the exterior angle business, both angles in each triangle are 45 degrees so you must have an isosceles triangle.

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    (Original post by Moltenmo)
    How do you assume AC and CB are the same? It doesn't say or show that in the question?

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    (Original post by gdunne42)
    Because of how you made them. You have removed a regular octagon from a square.
    And also because of the exterior angle business, both angles in each triangle are 45 degrees so you must have an isosceles triangle.

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    Yeah you summed it up well, I wasn't sure how to explain it. It might not explicitly say that they are the same but I think you'd be expected to be able to figure that out yourself if a question like this came up on an exam.
 
 
 
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