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# C1 question Solomon I

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1. It's Part C I'm having an issue with - for your info, the coords of Q and R are (15,0) and (0,15/4) respectively

I can't seem to get the required area, and don't understand why in the markscheme the triangles area is as simple as '1/2 * 15 * 15/4'

I know the area of a triangle is 1/2*b*h but I can't seem to understand the logic in the answer.

Thanks!
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2. Because to find the area of the triangle on a graph x and y are base and height and so it is simply just 1/2*X*Y.
Question sounds a lot harder than it actually is tbf
3. (Original post by Xzerzes)
Because to find the area of the triangle on a graph x and y are base and height and so it is simply just 1/2*X*Y.
Question sounds a lot harder than it actually is tbf
I still don't get it.. because I thought you could only do this if the base and height are perpendicular
4. Zacken accidently posted in A levels just realised sorry could you please move it (& also help pls )
5. (Original post by iMacJack)
Zacken accidently posted in A levels just realised sorry could you please move it (& also help pls )
Draw a sketch and post it up if it's still not screaming out at you. Remember that the axes are perpendicular to one another so it's a right-angled triangle.
6. (Original post by iMacJack)
It's Part C I'm having an issue with - for your info, the coords of Q and R are (15,0) and (0,15/4) respectively

I can't seem to get the required area, and don't understand why in the markscheme the triangles area is as simple as '1/2 * 15 * 15/4'

I know the area of a triangle is 1/2*b*h but I can't seem to understand the logic in the answer.

Thanks!
Draw a graph with those points and you will see that it is a right angled triangle (i.e half a square).
7. X and Y are perpendicular though.
It is a right angle triangle with (0,0) (0,15/4) (15,0)
X2-X1= 15-0 = Base
Y2-Y1= 15/4-0 = Height
8. It's simple

(Original post by iMacJack)
It's Part C I'm having an issue with - for your info, the coords of Q and R are (15,0) and (0,15/4) respectively

I can't seem to get the required area, and don't understand why in the markscheme the triangles area is as simple as '1/2 * 15 * 15/4'

I know the area of a triangle is 1/2*b*h but I can't seem to understand the logic in the answer.

Thanks!
Really simple, (1/2) x (15) x (15/4) = 225/8 = 28 1/8
9. (Original post by Zacken)
Draw a sketch and post it up if it's still not screaming out at you. Remember that the axes are perpendicular to one another so it's a right-angled triangle.

Rough sketch but it really isn't screaming out at me

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10. (Original post by iMacJack)

Rough sketch but it really isn't screaming out at me

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OQR not PQR
11. (Original post by techfan42)
It's simple

Really simple, (1/2) x (15) x (15/4) = 225/8 = 28 1/8
Well, yes I am aware of that. It's the actual logic behind it I am not understanding - I appreciate the input however

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12. (Original post by techfan42)
OQR not PQR
For.God.Sake I am an idiot oh lord lmao I SWEAR IT SAID PQR

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13. (Original post by iMacJack)

Rough sketch but it really isn't screaming out at me

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Read the question, it says the triangle OQR. What is O? Its the origin.
14. (Original post by iMacJack)
Well, yes I am aware of that. It's the actual logic behind it I am not understanding - I appreciate the input however

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You did PQR, question says OQR where O is (0,0)
15. Well I am rather embarrassed at my lack of ability to read the question xD....

I honestly thought it was the triangle PQR, in which case I was like what wtf why have they just done 1/2 * b * h.. oh lord
16. (Original post by iMacJack)
For.God.Sake I am an idiot oh lord lmao I SWEAR IT SAID PQR

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hahahaha it's fine, I always misread the question
17. Thanks every one! That was a bit of a ridiculous error on my behalf.. well that serves me right I guess!!

Just to clarify - if it was PQR, you'd get the distances with the distance formula etc wouldnt you
18. (Original post by iMacJack)
Thanks every one! That was a bit of a ridiculous error on my behalf.. well that serves me right I guess!!

Just to clarify - if it was PQR, you'd get the distances with the distance formula etc wouldnt you
Yep.
19. (Original post by iMacJack)

Just to clarify - if it was PQR, you'd get the distances with the distance formula etc wouldnt you
Yes
20. (Original post by Marxist)
Yes
When you do this with the length formulas, how do you know which ones to multiply together? Cheers

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