Once again, it's mee

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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. :s-smilie:

so for part b do i not get integers??

Reply 41

username1732133

For part c you need to notice something about the triangle with sides 6,8 and 10.

Reply 42

troubadour.

Original post by User23

so for part b do i not get integers??

No, it would be algebraic. The questions quite clear about that. I was wrong about it having to do with the area equation. It's just s = (a + b + c)/2 as far as I can see, unbelievable as it is.

Reply 43

username1732133

Original post by Hydeman

No, it would be algebraic. The questions quite clear about that. I was wrong about it having to do with the area equation. It's just s = (a + b + c)/2 as far as I can see, unbelievable as it is.

The answers to part b are numbers. One of them is an integer.

Reply 44

User23

Original post by Hydeman

No, it would be algebraic. The questions quite clear about that. I was wrong about it having to do with the area equation. It's just s = (a + b + c)/2 as far as I can see, unbelievable as it is.

are you sure it's algebraic?? because for the first triangle i got 7.4833 and the second triangle is pythagorean triples and as you can see for part c it asks for a simpler method which is b x h/2

Reply 45

Chittesh14

Original post by Hydeman

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. :colondollar:

I done this in my head lol. I realised I was wrong after and just stopped. I'll do this when I'm free.

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Reply 46

troubadour.

Original post by BuryMathsTutor

The answers to part b are numbers. One of them is an integer.

Damn I read it wrong. Ignore me, OP. :getmecoat:

It's integers. I was still thinking about the formula question.

Original post by User23

so for part b do i not get integers??

Reply 47

troubadour.

Original post by User23

are you sure it's algebraic?? because for the first triangle i got 7.4833 and the second triangle is pythagorean triples and as you can see for part c it asks for a simpler method which is b x h/2

Yeah, sorry about that. I was still thinking about the formula question. It's not algebraic, it's numbers because you have to use Heron's formula on the two triangles shown.

Reply 48

User23

Original post by Hydeman

Yeah, sorry about that. I was still thinking about the formula question. It's not algebraic, it's numbers because you have to use Heron's formula on the two triangles shown.

oh thank god for that haha, i nearly had a heart attack

Reply 49

username1828143

Original post by Chittesh14

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? :s-smilie:

Can you please explain

basically I divided 13 by 2 to get the length from X to y (or y to z), because the trapeziums are identical, so X to y and y to z should be equal.
the fact that I got 6.5 after doing pythagoras a second time to find the length of the left hand side of the trapezium is simply a coincidence.