Report, edit, etc...Posted by Screwed on 2005-05-02 at 06:04:11
Well, I had this question for a mini exam and after the test I got it wrong. I want to know how to work it out though. Any bit of help is appreciated and please take me step by step through the process. Thank you.
Edit: I'll just type what I think I know. (Correct me if I'm wrong)
Shape ABCD should be a trapezium (Assuming it is), and X is the midpoint of A and B as stated. Y is one quarter of the way from D to C because CY = 3DY is on the same line, in the ratio of 1:3, therefore it must be 1/4 of the way from D and 3/4 from C on the line CD. Then the problem kicks in in which I don't know how to work out the area of XYA using the two area measurements given.
ReEdit: Some Changes in what I know
QUOTE
The sides BC and AD of a quadrilateral ABCD are parallel. X is the midpoint of AB and Y is a point on the side CD such that CY=3DY. Find the area of the triangle XYA if the area of the triangle ABC is 120cm squared and the area of the triangle ABD is 160cm squared.
Edit: I'll just type what I think I know. (Correct me if I'm wrong)
Shape ABCD should be a trapezium (Assuming it is), and X is the midpoint of A and B as stated. Y is one quarter of the way from D to C because CY = 3DY is on the same line, in the ratio of 1:3, therefore it must be 1/4 of the way from D and 3/4 from C on the line CD. Then the problem kicks in in which I don't know how to work out the area of XYA using the two area measurements given.
ReEdit: Some Changes in what I know
Report, edit, etc...Posted by BeeR_KeG on 2005-05-02 at 17:12:54
You sure you copied the whole problem as it is?
There has to be a given angle, shape or line lenght to be able to do this.
Since you can't really assume anything because almost nothing is given it is imposible to tell whether the quadrilateral is a trapeziod. All I got was that:
AD and CD are of same length because AC and BC are parallel. For all I know the quadrilateral can be a trapezoid or parallelogram.
Now the problem for people studying geometry(I assume that you are), is that you cannot find the lengths or area of that triangle because you have to assume that they can be any angle from 0°<angle<180°. Since you are not sure if it is 90° you cannot use the formula;
1/2(Length x height)
Therefore you would have to use a different Trigonometric fomula involving Trigonometric Identities which is above your level. I tried it using Trigonometry and still no solution.
There has to be a given angle, shape or line lenght to be able to do this.
Since you can't really assume anything because almost nothing is given it is imposible to tell whether the quadrilateral is a trapeziod. All I got was that:
AD and CD are of same length because AC and BC are parallel. For all I know the quadrilateral can be a trapezoid or parallelogram.
Now the problem for people studying geometry(I assume that you are), is that you cannot find the lengths or area of that triangle because you have to assume that they can be any angle from 0°<angle<180°. Since you are not sure if it is 90° you cannot use the formula;
1/2(Length x height)
Therefore you would have to use a different Trigonometric fomula involving Trigonometric Identities which is above your level. I tried it using Trigonometry and still no solution.
Report, edit, etc...Posted by CheeZe on 2005-05-02 at 18:15:06
Tell your teacher it's impossible to solve unless angles are given. Or...
Area of AXY = Square Root of (((AX+XY+AY)/2)(((AX+XY+AY)/2)-AX)(((AX+XY+AY)/2)-XY)(((AX+XY+AY)/2)-AY))
Yes, the above equation is correct
Area of AXY = Square Root of (((AX+XY+AY)/2)(((AX+XY+AY)/2)-AX)(((AX+XY+AY)/2)-XY)(((AX+XY+AY)/2)-AY))
Yes, the above equation is correct
Report, edit, etc...Posted by indecisiveman on 2005-05-02 at 19:17:35
Aren't the angles 90 degrees because it is a quadrilateral? Or was that something else...
Report, edit, etc...Posted by Mini Moose 2707 on 2005-05-03 at 14:55:59
A quadrilateral has four sides. It says nothing about length of sides or angles.
Report, edit, etc...Posted by indecisiveman on 2005-05-03 at 18:31:09
Yeah your right I just realized that.
P.S. When can we finish our so acclaimed "debate" on God Cheeze/Moose? I am eager to prove you wrong again
P.S. When can we finish our so acclaimed "debate" on God Cheeze/Moose? I am eager to prove you wrong again
Report, edit, etc...Posted by (U)Bolt_Head on 2005-05-05 at 12:36:55
QUOTE(BeeR_KeG @ May 2 2005, 04:12 PM)
AD and CD are of same length because AC and BC are parallel. For all I know the quadrilateral can be a trapezoid or parallelogram.
[right][snapback]199717[/snapback][/right]
[right][snapback]199717[/snapback][/right]
Its not a parrallelogram, if it was then the two given measurements would be the same.
Since ABD > ABC
Then AD > AC
Because we know the leanth of B doesn't change.
From here on is guess work.
Area of AXY would be found using ratios of the two given measurements.
We know that ABD = 160 and that ABC = 120
From that in concluded that ABY would be equal to 150 (the differance in area changes by 10cm sqaured for every 1/4th jump you move along the line DC.
So if another point Z was half way between D and C then the area of ABZ would be half way between the area of ABD and ABC (140)
Since X is halfway between A and B I concluded that the area of AXY = ABY/2.
So my final answer is 75cm sqaured.
PS. When drawing it i drew the line AB as a vertical line 90degrees to AD and BC. The angle is unimportant but it helped me to visualize it that way.
Report, edit, etc...Posted by RexyRex on 2005-05-05 at 18:42:10
CheeZe is the math god.
I, on the other hand, have trouble with seventh grade math.
I, on the other hand, have trouble with seventh grade math.
Report, edit, etc...Posted by (U)Bolt_Head on 2005-05-05 at 18:48:14
*Wishes he was the math god*
Although i did teach myself Geometry. (but not spelling)
Although i did teach myself Geometry. (but not spelling)
Report, edit, etc...Posted by CheeZe on 2005-05-05 at 22:00:26
Lol Bolt. I think your answer is what people usually do when they can't figure out problems.
The problem with the answer is that the question it self is unsolvable.
Watch:
Pretend this is what the thing looks like:
A------------------B
D---------------C
But then, it could turn into...
A-------------B
D--------C
The area of the triangles could still be the same, however, the length of the quadrilateral would be different! So angles must be given. Unless you're allowed to substitue Variables into the formula, in which case, refer back to my equation.
The problem with the answer is that the question it self is unsolvable.
Watch:
Pretend this is what the thing looks like:
A------------------B
D---------------C
But then, it could turn into...
A-------------B
D--------C
The area of the triangles could still be the same, however, the length of the quadrilateral would be different! So angles must be given. Unless you're allowed to substitue Variables into the formula, in which case, refer back to my equation.
Report, edit, etc...Posted by Screwed on 2005-05-05 at 22:09:04
Well, thanks to Cheeze, Beer, Bolt and everyone who helped out.
I guess there isn't any answer to this, I tried getting to the teacher saying this can't be worked out, but she kept saying the answer is 70 cm^2. I duno how she got it, but neways - thanks for everyone who spent their time typing the long explanations to help me.
I guess there isn't any answer to this, I tried getting to the teacher saying this can't be worked out, but she kept saying the answer is 70 cm^2. I duno how she got it, but neways - thanks for everyone who spent their time typing the long explanations to help me.
Report, edit, etc...Posted by CheeZe on 2005-05-05 at 22:17:12
Ask her how she got it, step by step. Write them down, explain them here. I would like to know.
There is no way this is beyond me >_<
There is no way this is beyond me >_<