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    No point replying now , she got her answer , until next time....
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    (Original post by Hydeman)
    Nice one! Did you you round the awkward number you get by square rooting 84.5? It later gives a clean 1 d.p. number when you use trigonometry and the final answer I got was exactly 97.5 cm2 although yours is so close that I don't think any examiner would care.

    Edit: Out of interest, what level is this? Is it GCSE, A Level, or one of the Scottish/Irish equivalents of those?
    This is a Westminster School 13+ exam paper
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    (Original post by User23)
    This is a Westminster School 13+ exam paper
    No worries, I've already been pointed in that direction. xD I was just curious.
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    (Original post by boojiebaba)
    Attachment 472685
    I know you've already figured it out, but here's my solution anyway
    Hey, I really like your solution.
    But, I don't get how did you do 13/2 = 6.5 and make that one of the sides of the right-angled triangle.
    I get how you did 13 cm as the side opposite the right angle is the hypotenuse and then you used pythagoras.
    But, you done 13/2 = 6.5? Can you please explain
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    (Original post by User23)
    This is a Westminster School 13+ exam paper
    So it's for people over 13? This is like a tough GCSE question!
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    (Original post by Hydeman)
    No worries, I've already been pointed in that direction. xD I was just curious.
    is it possible if u could help me with another question please?
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    (Original post by User23)
    is it possible if u could help me with another question please?
    Post more please !
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    (Original post by User23)
    is it possible if u could help me with another question please?
    Yeah sure.
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    (Original post by Hydeman)
    Yeah sure.
    thanks so much
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    (Original post by User23)
    thanks so much
    So what do you think the first couple of answers are? I'll tell explain if you get it wrong.
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    s = 1/2 * the sum of all the sides of the triangle
    Make this into an equation s = 1/2(.....)
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    (Original post by Hydeman)
    So what do you think the first couple of answers are? I'll tell explain if you get it wrong.
    this is my working out
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    (Original post by User23)
    this is my working out
    The second part where you're rewriting s in terms of a, b and c isn't quite right. You should be getting a formula, not an integer as an answer. I'm not sure why the triangles are there since the question is algebraic but you'd need to play around with the equation (Heron's formula) given. Think about how you can collapse all those s symbols into one (hint: factorise -- but first square both sides).
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    Not the only one expecting this to be dear cavy' second bye bye speech?
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    (Original post by Hydeman)
    The second part where you're rewriting s in terms of a, b and c isn't quite right. You should be getting a formula, not an integer as an answer. I'm not sure why the triangles are there since the question is algebraic but you'd need to play around with the equation (Heron's formula) given. Think about how you can collapse all those s symbols into one (hint: factorise -- but first square both sides).
    yeah but i swear it tells u s is half the perimeter of a triangle so i found s to be 12 for the second triangle.:
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    (Original post by User23)
    yeah but i swear it tells u s is half the perimeter of a triangle so i found s to be 12 for the second triangle.:
    It does tell you that but it doesn't ask you to do anything with the triangles just yet. Use the formula they've given and try to rearrange it to make s the subject. The triangles are for the next question.

    Your answer should be algebraic, not a number.
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    (Original post by Hydeman)
    It does tell you that but it doesn't ask you to do anything with the triangles just yet. Use the formula they've given and try to rearrange it to make s the subject. The triangles are for the next question.

    Your answer should be algebraic, not a number.
    oh no ok ill try make s the subject, please don't laugh at my working out if it's wrong
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    (Original post by Hydeman)
    It does tell you that but it doesn't ask you to do anything with the triangles just yet. Use the formula they've given and try to rearrange it to make s the subject. The triangles are for the next question.

    Your answer should be algebraic, not a number.
    S = (Area)^ / (1-a)(1-b)(1-c)
    ?


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    (Original post by User23)
    oh no ok ill try make s the subject, please don't laugh at my working out if it's wrong
    Hmm. I may have misled you on this one. D: I'm thinking it's just s = (a + b + c)/2 as in the first question, even though that makes little sense... For the second equation, you've got that capital A as well which is more or less impossible to get rid of so I'd think it's just the simple answer from part I.
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    (Original post by Chittesh14)
    S = (Area)^ / (1-a)(1-b)(1-c)
    ?


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    No, I'm thinking that's wrong because it includes area, which isn't in the question. I think it's just the simple s = (a + b + c)/2 from earlier although I've no idea why it's repeated. The first question does say calculate but mentions a hypothetical triangle with sides a, b and c so I don't see how you could give a different answer for that...

    Is there a markscheme for this anywhere? It shouldn't really be taking this long if it's for 13-year-olds so I'm going to assume that the long way is wrong... (by the way I don't think (s - a) factorises to s(1 -a); it would factorise to s(1 - a/s), which expands back to (s - a)).

    Sorry guys.
 
 
 
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