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# Finding area of a triangle watch

1. The question is 11v on this exam paper link:

http://www.mei.org.uk/files/papers/c109ju_kusp.pdf

What distances do I need to find for the base & height?
I thought it'd be 1/2(distance OA x distance OB) but that doesn't give an area of 9 like it should
2. area=(0.5)*(9*square_root(10)/10)*(α)square_root(10)

combine the solutions of questions (iii) and (iv)
3. (Original post by Magenta96)
The question is 11v on this exam paper link:

http://www.mei.org.uk/files/papers/c109ju_kusp.pdf

What distances do I need to find for the base & height?
I thought it'd be 1/2(distance OA x distance OB) but that doesn't give an area of 9 like it should
Remember the area of a triangle is a half multiplied by the base multiplied by the height perpendicular to the base.
4. (Original post by Magenta96)
The question is 11v on this exam paper link:

http://www.mei.org.uk/files/papers/c109ju_kusp.pdf

What distances do I need to find for the base & height?
I thought it'd be 1/2(distance OA x distance OB) but that doesn't give an area of 9 like it should
remember that the area is given by

So to find the hight you would need to draw a line from point B to the Y axis and call it X so the distance between XB is 6 ( which is my hight) using the formula and I know the base is from OA which is 3.

5. (Original post by Magenta96)
The question is 11v on this exam paper link:

http://www.mei.org.uk/files/papers/c109ju_kusp.pdf

What distances do I need to find for the base & height?
I thought it'd be 1/2(distance OA x distance OB) but that doesn't give an area of 9 like it should
It does work...

Point O (0,0)
Point A (0,3)
Point B (6,1)

Distance of OA is the Height - therefore is 3
Distance of OB is the Length - therefore 6

6x3 = 18. Thinking of the rule 0.5xhxb
So 18 x 0.5 = 9!
6. (Original post by Sitrix)
It does work...

Point O (0,0)
Point A (0,3)
Point B (6,1)

Distance of OA is the Height - therefore is 3
Distance of OB is the Length - therefore 6

6x3 = 18. Thinking of the rule 0.5xhxb
So 18 x 0.5 = 9!
I keep getting (root)37 as the distance of OB because O(0,0) and B(6,1)
OB = (root)(0-6)^2 + (0-1)^2
OB = (root)(-6)^2 + (-1)^2
OB = (root)36 + 1
OB = (root)37
7. (Original post by Magenta96)
I keep getting (root)37 as the distance of OB because O(0,0) and B(6,1)
OB = (root)(0-6)^2 + (0-1)^2
OB = (root)(-6)^2 + (1)^2
OB = (root)36 + 1
OB = (root)37
Do a sketch.

View the line from (0,0) to (0,3) as the base of your triangle.

The height is the distance from the y axis to the point (6,1).

base = ?

height = ?

Area = ?
8. (Original post by BabyMaths)
Do a sketch.

View the line from (0,0) to (0,3) as the base of your triangle.

The height is the distance from the y axis to the point (6,1).

base = ?

height = ?

Area = ?
Yes I've got one length of 3 but for the distance of OB, I'm making some kind of error in the distance formula as I keep getting (root)37 rather than (root)6 which would give me 6
9. (Original post by steve2005)
Diagram ( Can you see the dots ?)

Yes, I can tell where the base and height are now and have one distance of 3. I can't seem to get 6 for the distance of OB though, I'm making some error in the distance formula even though I've checked my working loads
10. (Original post by Magenta96)
Yes I've got one length of 3 but for the distance of OB, I'm making some kind of error in the distance formula as I keep getting (root)37 rather than (root)6 which would give me 6
Why are you finding OB surely you can see that the HEIGHT of the triangle in that diagram is 6
11. (Original post by Magenta96)
Yes, I can tell where the base and height are now and have one distance of 3. I can't seem to get 6 for the distance of OB though, I'm making some error in the distance formula even though I've checked my working loads
http://www.purplemath.com/modules/distform.htm Is the distance formula, I can't type it out cause I suck at latex.
12. (Original post by robbie242)
http://www.purplemath.com/modules/distform.htm is the distance formula, i can't type it out cause i suck at latex.
not needed
13. you can split it into two right angle triangles

Then use 1/2 (base*height) for each triangle and add them together
14. (Original post by TenOfThem)
Why are you finding OB surely you can see that the HEIGHT of the triangle in that diagram is 6
but the diagram that Steve2005 has given isn't on the exam paper so we're not supposed to see it's 6 without working it out first, what do you type in the distance formula for the distance of OB because even when substituting the coordinates for O and B in, I get (root)37, I subsituted:

15. (Original post by Magenta96)
but the diagram that Steve2005 has given isn't on the exam paper so we're not supposed to see it's 6 without working it out first
You are given that B = (1,6)

What working out do you think is needed
16. (Original post by TenOfThem)
not needed
Yup I know, since the y coordinate of the height is 6, giving you 6, just for reference in the future.
17. (Original post by Robbie242)
Yup I know, since the y coordinate of the height is 6, giving you 6, just for reference in the future.
But he had found OB correctly just unnecessarily
18. (Original post by TenOfThem)
You are given that B = (0,6)

What working out do you think is needed
Actually B is (6,1).

I don't think Steve's diagram is very helpful. [His diagrams are usually good!]
19. (Original post by Magenta96)
but the diagram that Steve2005 has given isn't on the exam paper so we're not supposed to see it's 6 without working it out first, what do you type in the distance formula for the distance of OB because even when substituting the coordinates for O and B in, I get (root)37, I subsituted:

1/2 x OA x 1
20. (Original post by BabyMaths)
Actually B is (6,1).

I don't think Steve's diagram is very helpful. [His diagrams are usually good!]
Poop

Never the less ... I am sure that the OP can sketch a diagram

Something I have suggested on other threads he has asked on this topic

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