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    (Original post by Jas1947)
    Please attempt just question 8
    Psst you forgot the question
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    (Original post by Jas1947)
    Please attempt just question 8
    It's pretty straight forward but before I give you my working out, I'd like to see you try and achieve the answer.

    Hint:
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    Use the fact that the smaller triangle is enlarged by a factor of a on all sides. This means 4a=(x-4)+(4) and xa=x+3.
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    Okay so we can immediately eliminate 3 answers because they make no sense. For A to be correct X+3 = 5. That makes X=-2. That doesn't make sense for DB because the side would be -6cm. Similar issue using B or D, we get negative values for DB. So we can fairly assume that the answer is C or E. C would mean X = 6 and E would mean X = 1+2√7

    We also know that triangle ABC and triangle ADE are the same triangle but in different ratios. That means that DE/DA = BC/BA. Therefore:

    X+3/X = X/4

    Get rid of the fractions to get

    X² = 4X+12

    Turn that into X² - 4X -12 = 0

    Factorise into (X-6)(X+2), X = 6 or -2. X = -2 doesn't work as we established before (DB is negative again, as is BC and DA). Therefore X = 6 and DE = 9. So your answer is C which falls in line with the potential answers we established before.


    I googled UCLES2016 to see what would come up. Looks like papers related to Cambridge. I very much hope this isn't in any way related to entry requirements or getting into university.
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    (Original post by RDKGames)
    It's pretty straight forward but before I give you my working out, I'd like to see you try and achieve the answer.

    Hint:
    Spoiler:
    Show
    Use the fact that the smaller triangle is enlarged by a factor of a on all sides. This means 4a=x-4 and xa=x+3.
    Sorry, I think I just ruined your good intentions and posted full workings...
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    (Original post by Acsel)
    Sorry, I think I just ruined your good intentions and posted full workings...



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    (Original post by Acsel)
    Okay so we can immediately eliminate 3 answers because they make no sense. For A to be correct X+3 = 5. That makes X=-2. That doesn't make sense for DB because the side would be -6cm. Similar issue using B or D, we get negative values for DB. So we can fairly assume that the answer is C or E. C would mean X = 6 and E would mean X = 1+2√7

    We also know that triangle ABC and triangle ADE are the same triangle but in different ratios. That means that DE/DA = BC/BA. Therefore:

    X+3/X = X/4

    Get rid of the fractions to get

    X² = 4X+12

    Turn that into X² - 4X -12 = 0

    Factorise into (X-6)(X+2), X = 6 or -2. X = -2 doesn't work as we established before (DB is negative again, as is BC and DA). Therefore X = 6 and DE = 9. So your answer is C which falls in line with the potential answers we established before.


    I googled UCLES2016 to see what would come up. Looks like papers related to Cambridge. I very much hope this isn't in any way related to entry requirements or getting into university.
    I got something slightly different... Are your ratios correct? Or is my method flawed somehow? On one hand I want to say you're correct simply due to the nice answer but on the other I am right because it's cambridge. xD
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    Triangles are Homothetic find the scale factor.


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    (Original post by RDKGames)
    I got something slightly different...
    I want to say yours is probably correct, I haven't done this in years. But should your first line not read:

    4A=X and not 4A=X-4

    For AB to be a scale of AD don't you need to factor in that the entire length of AD = AB+BD? Or X-4+4 (Simplified to just X)?

    Surely if ABC is a smaller version of ADE then AB times something (the scalar) will be equal to DE, not BD. Or am I talking nonsense?


    EDIT: Obvious nonsense because it's super hard to explain what I'm trying to say. But basically I think you forgot to work out the length of AD and simply compared AB to BD.
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    (Original post by Acsel)
    I want to say yours is probably correct, I haven't done this in years. But should your first line not read:

    4A=X and not 4A=X-4

    For AB to be a scale of AD don't you need to factor in that the entire length of AD = AB+BD? Or X-4+4 (Simplified to just X)?

    Surely if ABC is a smaller version of ADE then AB times something (the scalar) will be equal to DE, not BD. Or am I talking nonsense?


    EDIT: Obvious nonsense because it's super hard to explain what I'm trying to say. But basically I think you forgot to work out the length of AD and simply compared AB to BD.
    Yeah I think you're right. Late at night and my brain doesn't work so well, I took the x-4 as the entire side for some reason >_>

    Your method is correct.
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    (Original post by Acsel)
    I want to say yours is probably correct, I haven't done this in years. But should your first line not read:

    4A=X and not 4A=X-4

    For AB to be a scale of AD don't you need to factor in that the entire length of AD = AB+BD? Or X-4+4 (Simplified to just X)?

    Surely if ABC is a smaller version of ADE then AB times something (the scalar) will be equal to DE, not BD. Or am I talking nonsense?


    EDIT: Obvious nonsense because it's super hard to explain what I'm trying to say. But basically I think you forgot to work out the length of AD and simply compared AB to BD.
    (Original post by Jas1947)
    Please attempt just question 8
    Uploading correct working out for OP. Cheeky Cambridge people must've put that option solely due to my mistake lol. Note to self: dont do triangles at night

    Name:  ImageUploadedByStudent Room1469571246.270045.jpg
Views: 127
Size:  127.9 KB



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    (Original post by RDKGames)
    Yeah I think you're right. Late at night and my brain doesn't work so well, I took the x-4 as the entire side for some reason >_>

    Your method is correct.
    Hahaha yeah it's way too late for this. I was staring at it for ages with no idea where to even start and no idea if I was doing anything correct. Easy mistake to make though, hence why they added an answer to catch people out. At least you had proper terminology. I'm here fumbling around trying to remember the word "ratio".
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    (Original post by Acsel)
    Okay so we can immediately eliminate 3 answers because they make no sense. For A to be correct X+3 = 5. That makes X=-2. That doesn't make sense for DB because the side would be -6cm. Similar issue using B or D, we get negative values for DB. So we can fairly assume that the answer is C or E. C would mean X = 6 and E would mean X = 1+2√7

    We also know that triangle ABC and triangle ADE are the same triangle but in different ratios. That means that DE/DA = BC/BA. Therefore:

    X+3/X = X/4

    Get rid of the fractions to get

    X² = 4X+12

    Turn that into X² - 4X -12 = 0

    Factorise into (X-6)(X+2), X = 6 or -2. X = -2 doesn't work as we established before (DB is negative again, as is BC and DA). Therefore X = 6 and DE = 9. So your answer is C which falls in line with the potential answers we established before.


    I googled UCLES2016 to see what would come up. Looks like papers related to Cambridge. I very much hope this isn't in any way related to entry requirements or getting into university.
    This is from a cambridge admissions assesment, should i be worried? I think with practice I can get a decent score....
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    (Original post by RDKGames)
    Uploading correct working out for OP. Cheeky Cambridge people must've put that option solely due to my mistake lol. Note to self: dont do triangles at night

    Name:  ImageUploadedByStudent Room1469571246.270045.jpg
Views: 127
Size:  127.9 KB



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    Thank you! It seems so obvious now i dont know why i couldnt see it before , thanks
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    (Original post by Jas1947)
    This is from a cambridge admissions assesment, should i be worried? I think with practice I can get a decent score....
    If it's an official assessment that determines if you get your place then don't cheat. You could be disqualified if anyone found out and it won't help you on your course. Do not under any circumstances ask for help with any sort of official graded assessments.

    If you're just practicing then you'll need to be able to get a decent enough score to get in. Cambridge is competitive from what I've seen. Not saying not being able to answer this question is the difference between getting in or not but at a baseline you should be able to answer all the simple questions with absolutely no trouble. If you're struggling on lots of questions like this then practice loads. Although I don't understand why you'd need the practice. This is the sort of stuff you should realistically know if you're intending to go to Cambridge (assuming you'd have done a Maths A Level and the test is related to getting into Cambridge to do something Maths oriented).
 
 
 
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