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# Grade 9 Geometry/Algebra Problem watch

1. Hey everyone I'm back with another question (hopefully it's as good as my others). I hope you like it and can give me feedback below. Feel free to attempt in and please spoiler your answers. This is aimed at Grade 9 students but anyone is welcome. I'd probably give it 7 marks because it's slightly lengthy. This question is calculator. Anyways I hope you enjoy and have a great day!

2. (Original post by Y11_Maths)
Hey everyone I'm back with another question (hopefully it's as good as my others). I hope you like it and can give me feedback below. Feel free to attempt in and please spoiler your answers. This is aimed at Grade 9 students but anyone is welcome. I'd probably give it 7 marks because it's slightly lengthy. This question is calculator. Anyways I hope you enjoy and have a great day!

Not sure but is it these?
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i) 414 square miles
ii) 40.3
3. (Original post by 3pointonefour)
Not sure but is it these?
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i) 414 square miles
ii) 40.3
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Part I) is correct however ii) is incorrect
4. (Original post by Y11_Maths)
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Part I) is correct however ii) is incorrect
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Is part ii) 50.8?
5. (Original post by 3pointonefour)
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Is part ii) 50.8?
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No it isn’t. You’re quite far away from it. Would you like a hint?
6. (Original post by Y11_Maths)
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No it isn’t. You’re quite far away from it. Would you like a hint?
I'll get there in the end
7. (Original post by 3pointonefour)
I'll get there in the end
Perseverance is key. Side note: It also took me longer to find the angle because I couldn’t spot it haha. I was bashing myself in the head because I couldn’t find it for ages lol
8. (Original post by Y11_Maths)
Perseverance is key. Side note: It also took me longer to find the angle because I couldn’t spot it haha. I was bashing myself in the head because I couldn’t find it for ages lol
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Not sure where I went wrong:

CG = 22.8034
GD = 4
Angle CGD = 43
So used cosine rule to get CD = 20.0643
Then got a right-angled triangle with the perpendicular length being 15.5519, and CD.
Used sin to get theta = 50.8?
9. (Original post by 3pointonefour)
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Not sure where I went wrong:

CG = 22.8034
GD = 4
Angle CGD = 43
So used cosine rule to get CD = 20.0643
Then got a right-angled triangle with the perpendicular length being 15.5519, and CD.
Used sin to get theta = 50.8?
Spoiler:
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hmm let me have a think. I’ve got 129 degrees as I used a different method. I worked out angle GCD and subtracted it from 137. Bear with me
Edit: If you think about it, if both methods are legitimate, can ask the student to find both possible value of angle BCD? Which would make it better lol
10. (Original post by Y11_Maths)
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hmm let me have a think. I’ve got 129 degrees as I used a different method. I worked out angle GCD and subtracted it from 137. Bear with me
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Yeah I know where you went wrong. I got that as part of my working. It can't be 129 since the angle is clearly acute. Here's how it works:

If you look at the graph of sin(x), you see it is positive from 0 to 180
This means there's 2 possible solutions for angles in a triangle.
For example, both sin(30) and sin(150) = 0.5
The rule is sin(x) = sin(180-x).
It's because we both used sine, but we took different roots.

https://brilliant.org/wiki/sine-rule...ambiguous-case
11. (Original post by Y11_Maths)
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hmm let me have a think. I’ve got 129 degrees as I used a different method. I worked out angle GCD and subtracted it from 137. Bear with me

Edit: If you think about it, if both methods are legitimate, can ask the student to find both possible value of angle BCD? Which would make it better lol
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Try finding out the remaining angles of triangle CGD. I don't think it adds to 180 using your angle.
12. (Original post by 3pointonefour)
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Yeah I know where you went wrong. I got that as part of my working. It can't be 129 since the angle is clearly acute. Here's how it works:

If you look at the graph of sin(x), you see it is positive from 0 to 180
This means there's 2 possible solutions for angles in a triangle.
For example, both sin(30) and sin(150) = 0.5
The rule is sin(x) = sin(180-x).
It's because we both used sine, but we took different roots.

https://brilliant.org/wiki/sine-rule...ambiguous-case
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Aah thank you so much that’s a fantastic explanation! I should’ve know it couldnt be obtuse haha. How did you find it? Any feedback?
13. (Original post by Y11_Maths)
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Aah thank you so much that’s a fantastic explanation! I should’ve know it couldnt be obtuse haha. How did you find it? Any feedback?
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Well I think it's one of the best questions you've made! Like your other questions, it's too hard and doesnt give enough hints for GCSE (Its also maybe too long to be put with 20 other GCSE questions lol), but it's great to give as a challenge. Very nice .
I'd just say the 27.2799344 side just seems a bit dodgy and out of place. Can you maybe round it to still get the side of the rhombus as 40?
But otherwise, well done! Great test of problem solving and trigonometry!
14. (Original post by 3pointonefour)
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Well I think it's one of the best questions you've made! Like your other questions, it's too hard and doesnt give enough hints for GCSE (Its also maybe too long to be put with 20 other GCSE questions lol), but it's great to give as a challenge. Very nice .
I'd just say the 27.2799344 side just seems a bit dodgy and out of place. Can you maybe round it to still get the side of the rhombus as 40?
But otherwise, well done! Great test of problem solving and trigonometry!
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Thank you so much. Yes I dislike the height I used but I originally gave the side of 40 but turns out it’s better if I make them work it out themselves. Thank you for completing it and giving me feedback. Have an awesome day!
15. Can someone please explain this to me? I've been getting 8s and 9s but I've never seen a question as bad as this - thank you!
16. (Original post by IReckonItsDanyal)
Haha.
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Start by using the perpendicular height and a bit of trig to work out the length of the rhombus since they’re all the same side. Then use the cosine rule to work out the length of the diagonal and subtract 32 to get one length of the trapezium. It’s all about spotting triangles and using repeated sine, consine rules and trig.

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