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    Name:  image.jpg
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Size:  506.3 KBI attempted to redo a past paper question that I had gotten wrong a while back on Surds in Pythag but am not sure if the answer is correct since the AQA mark scheme has now been taken off the website, could anybody tell me if this is correct or if not, where I've gone wrong?

    Any help would be massively appreciated!
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    (Original post by Niltiac229)
    Name:  image.jpg
Views: 165
Size:  506.3 KBI attempted to redo a past paper question that I had gotten wrong a while back on Surds in Pythag but am not sure if the answer is correct since the AQA mark scheme has now been taken off the website, could anybody tell me if this is correct or if not, where I've gone wrong?

    Any help would be massively appreciated!
    Check your length AD.
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    1) you're forgetting the entire base which is 4rt3 so you're area is 2rt3*rt3 which is 6… this doesn't add up lol
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    (Original post by zayn008)
    1) you're forgetting the entire base which is 4rt3 so you're area is 2rt3*rt3 which is 6… this doesn't add up lol
    You're right, I've just realised and now I don't have a clue what I've done and how to solve it fmlName:  image.jpg
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    (Original post by Niltiac229)
    You're right, I've just realised and now I don't have a clue what I've done and how to solve it fmlName:  image.jpg
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    The area is 6 which is rt36, you can get 3rt4, or 2rt9 but that wouldn't make sense because they're perfect squares.. They've made an error in this question which is why it's been removed
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    (Original post by zayn008)
    The area is 6 which is rt36, you can get 3rt4, or 2rt9 but that wouldn't make sense because they're perfect squares.. They've made an error in this question which is why it's been removed
    No they were removed because they were past papers from 2012 and they always take old past papers from a while back off every year so the question has to be correct since it was an actual GCSE paper :/
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    (Original post by Niltiac229)
    You're right, I've just realised and now I don't have a clue what I've done and how to solve it fmlName:  image.jpg
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    As I said in my first post, your length AD is not correct -

    (3\sqrt2)^2 + b^2 = (3\sqrt3)^2 is correct but you haven't squared the terms correctly so b is not correct. You have to square the whole term, not just the square roots.
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    As above, remember that (3 \sqrt{2})^2 = 3\sqrt{2} \times 3\sqrt{2} = 9 \times 2 = 18
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    (Original post by SeanFM)
    As I said in my first post, your length AD is not correct -

    (3\sqrt2)^2 + b^2 = (3\sqrt3)^2 is correct but you haven't squared the terms correctly so b is not correct. You have to square the whole term, not just the square roots.
    Oh okay, how do I square the Surds then because I thought squaring the rt2 would get you 2 to which 3 was multiplied by for the first one, the same method also used for the second?
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    (Original post by Zacken)
    As above, remember that (3 \sqrt{2})^2 = 3\sqrt{2} \times 3\sqrt{2} = 9 \times 2 = 18
    Oh, I understand- so the whole of the thing is squares and not just the sq rts, thank you very much
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    (Original post by Niltiac229)
    Oh okay, how do I square the Surds then because I thought squaring the rt2 would get you 2 to which 3 was multiplied by for the first one, the same method also used for the second?
    You don't just apply the square function to the square root, you apply it to the whole thing.

    Eg (3\sqrt2)^2 = 3^2 \times (\sqrt2)^2 = ....
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    (Original post by SeanFM)
    You don't just apply the square function to the square root, you apply it to the whole thing.

    Eg (3\sqrt2)^2 = 3^2 \times (\sqrt2)^2 = ....
    Thanks very much- I used this method however have come to another problem- I'm unsure which is the right way of multiplying the Surds later on. I wrote out two ways but could you tell me which (if either!) are correct? Name:  image.jpg
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    (Original post by Niltiac229)
    Thanks very much- I used this method however have come to another problem- I'm unsure which is the right way of multiplying the Surds later on. I wrote out two ways but could you tell me which (if either!) are correct? Name:  image.jpg
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    3 \times 3\sqrt{2} = 9\sqrt{2} is the correct one. Leave the number inside the square root untouched.
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    (Original post by Niltiac229)
    Thanks very much- I used this method however have come to another problem- I'm unsure which is the right way of multiplying the Surds later on. I wrote out two ways but could you tell me which (if either!) are correct? Name:  image.jpg
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    You have very neat handwriting not a lot of maths students do

    it is the second one. You only need to multiply 1 part of the surd by 3, not both the number and the square root.
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    (Original post by Niltiac229)
    Thanks very much- I used this method however have come to another problem- I'm unsure which is the right way of multiplying the Surds later on. I wrote out two ways but could you tell me which (if either!) are correct? Name:  image.jpg
Views: 74
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    Hey, that's still not quite right, this is what I've got Name:  image.jpg
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    I then couldn't help but wonder if I was right and I stumbled across this! >> http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF

    ALSO: you're forgetting the other root2 on the right, so it's not 3rt2 its 4rt2, then you times that by 3 and you get 12rt2 and multiply it by 0.5 (only the number out of the root) and you get 6rt2, that's the only mistake you've made in your second answers, other than that it's great!

    Remember:
    1) it's half * BASE * height, base is the entire bottom of the triangle.

    2) When you square a surd you square the number on the outside as normal but you just release the number from the root, then multiply them. So (5rt2)^2 is (5)^2 * (rt2)^2 which is 25*2 so it's 50
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    (Original post by SeanFM)
    You have very neat handwriting not a lot of maths students do

    it is the second one. You only need to multiply 1 part of the surd by 3, not both the number and the square root.
    Why thank you haha and I did suspect it would be the second- it was a second thought after doing the first- thanks so much for the help, hopefully a similar question will come up in my exam tomorrow and I'll be able to do it!!
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    (Original post by zayn008)
    Hey, that's still not quite right, this is what I've got Name:  image.jpg
Views: 65
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    I then couldn't help but wonder if I was right and I stumbled across this! >> http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF

    ALSO: you're forgetting the other root2 on the right, so it's not 3rt2 its 4rt2, then you times that by 3 and you get 12rt2 and multiply it by 0.5 (only the number out of the root) and you get 6rt2, that's the only mistake you've made in your second answers, other than that it's great!

    Remember:
    1) it's half * BASE * height, base is the entire bottom of the triangle.

    2) When you square a surd you square the number on the outside as normal but you just release the number from the root, then multiply them. So (5rt2)^2 is (5)^2 * (rt2)^2 which is 25*2 so it's 50
    Oh I understand now, thank you so much- also I don't know how you found the mark scheme either, I spent ages looking haha!
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    (Original post by Niltiac229)
    Why thank you haha and I did suspect it would be the second- it was a second thought after doing the first- thanks so much for the help, hopefully a similar question will come up in my exam tomorrow and I'll be able to do it!!
    No worries yes, you can do it.
    Best of luck with your exam
 
 
 
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