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Finding area of a triangle

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

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Reply 1
Avatar for johnny09
johnny09
area=(0.5)*(9*square_root(10)/10)*(α)square_root(10)

combine the solutions of questions (iii) and (iv)
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.
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 A=0.5∗h∗bA=0.5 * h * b

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.

=>A=0.5∗6∗3=9=> A = 0.5 * 6 * 3 = 9


is this the answer?
(edited 11 years ago)
Reply 4
Avatar for Sitrix
Sitrix
10
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!
Reply 5
Avatar for Magenta96
Magenta96
OP
16
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
(edited 11 years ago)
Reply 6
Avatar for BabyMaths
BabyMaths
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 = ?
Reply 7
Avatar for Magenta96
Magenta96
OP
16
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
Reply 8
Avatar for Magenta96
Magenta96
OP
16
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
Reply 9
Avatar for TenOfThem
TenOfThem
18
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
Reply 10
Avatar for Robbie242
Robbie242
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.
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
you can split it into two right angle triangles

Then use 1/2 (base*height) for each triangle and add them together
Reply 13
Avatar for Magenta96
Magenta96
OP
16
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:

OB=(root)(6−0)2+(1−0)2OB = (root)(6-0)^2+(1-0)^2
OB=(root)(6)2+(1)2OB = (root)(6)^2 + (1)^2
OB=(root)(36)+(1)OB = (root)(36)+(1)
OB=(root)37OB = (root)37
(edited 11 years ago)
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
(edited 11 years ago)
Reply 15
Avatar for Robbie242
Robbie242
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.
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
Reply 17
Avatar for BabyMaths
BabyMaths
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!]
Reply 18
Avatar for Robbie242
Robbie242
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:

OB=(root)(6−0)2+(1−0)2OB = (root)(6-0)^2+(1-0)^2
OB=(root)(6)2+(1)2OB = (root)(6)^2 + (1)^2
OB=(root)(36)+(1)OB = (root)(36)+(1)
OB=(root)37OB = (root)37
1/2 x OA x 1
(edited 11 years ago)
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
(edited 11 years ago)

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