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Geometry problem, fractions

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Picture: http://matematikk.net/res/eksamen/1T/kort/1T_V11.pdf
    Task 5, the one the with a triangle inside a square. I'ts not in English so i'll transalate. I managed to do task a

    The picture above shows a square ABCD. The sides in the square have length 1. E is the center of BC, and F is the center of CD.

    B Show that the area of triangle AEF is 3/8

    C Show that sin a=3/5



    2. Relevant equations
    area of a sqaure: s*s area of a triangle: 0.5*b*h or 0.5*a*b*sin angle

    3. The attempt at a solution
    At first I tried to find all the sides of the triangle, but i cant find a way to calculate EF. It also makes it hard when you have to mix alot with the fractions and answers to get the answer they want. That makes it extra hard to find a solution. Task c i really had no idea, sinus is only usable for right triangles as far as i know. (not the sinus sentence)
    I find these kinds of tasks very hard, so if you know of any place to learn geometry that would be nice also :)
     
  2. jcsd
  3. Apr 6, 2015 #2
    You know that the sides of the square have length 1. So, you should know what EC and FC are (since E and F are the midpoints of their sides). You also know the angle of C. So, use Pythogoras.

    To do c, use the law of cosines.

    To learn geometry, yo can use khanacademy.org. Also, many contributers upload great lectures on Youtube.
     
  4. Apr 6, 2015 #3

    SammyS

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    Capture_1.PNG

    It's pretty straight forward to find the area of triangles ADF, ABE, and ECF .

    Do you know the angle addition/subtraction identities, especially for the tangent ? i.e. tan(θ - φ)
     
    Last edited: Apr 6, 2015
  5. Apr 6, 2015 #4
    I see how it's possible to find EF, but i don't know how to find any angles in the triangle. You could probably use trigonometry, but this is a part 1 math problem on an exam. On part 1 you aren't allowed to use a calculator so you wouldnt be able to use asine to find angles.
     
  6. Apr 6, 2015 #5

    SammyS

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    What is the length of AE ?
     
  7. Apr 6, 2015 #6
    1.25 squared wich is 1.56
     
  8. Apr 6, 2015 #7
    Well, maybe you do not need any angles. Sure, they ask you to find the sine of an angle, but that doesn't mean you need to know the angle. Do you know any formula you can use that would include the sine of [itex]\alpha[/itex]?

    Do you know what [itex]sin^2 x + cos^2 x[/itex] is equal to?
     
  9. Apr 6, 2015 #8
    Unfortunately not, we haven't learned that yet. We started with trigonometry this year and we have only learned the very basics so far.
     
  10. Apr 6, 2015 #9
    That isn't really that advanced. I went to highschool before Kunnskapsløftet. However, I learned that before the law of sines. Of course, they have changed the curriculum since then. Are you sure it is not in your book? It is just a consequence of the definitions of sine and cosine, and Pythagoras theorem.
     
  11. Apr 6, 2015 #10
    Quite sure, in our book we only learn about finding unknown angles and sides in triangles.
     
  12. Apr 6, 2015 #11

    SammyS

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    It's the square root of 1.25 .

    That's the same as the length of AF .


    You then have this handy formula for the area of a triangle: 0.5*a*b*sin angle

    If you know the area of the triangle and the lengths, a & b, you should be able to get sin(α), without actually knowing the measure of angle α .
     
  13. Apr 6, 2015 #12
    Okay, well, I can tell you that [itex]cos^2 x + sin^2 x = 1[/itex] for all x. Using that with the law of cosines could solve the problem. However, it seems like we have to find a different method to solve this using stuff that is in your curriculum. Lets see...

    The image screams that we should use a symmety argument (since the entire figure is symmetric about a straight line from A to C).

    So, you can use symmetry. You know half the length of EF. Also, you know the length of AF. That way, you can find [itex]sin \frac{\alpha}{2}[/itex]. Of course, this depends on whether you know what sin2x is. Do you?
     
  14. Apr 7, 2015 #13

    ehild

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    EFC is a isosceles right triangle. And you do not need the length EF.

    All you need is the length of AE=AF, that you get from Pythagoras' Law. And you need the area of all yellow triangles. It is easy, as they are right triangles. By subtracting the yellow areas from the area of the square, you get the area of the blue triangle. To get sin(alpha), apply your last relevant equation (area of the triangle using two sides and the sin of the angle between them. )
    (This is the same as Sammy's hint Post #11 ) .

    squaretriangles.JPG
     
  15. Apr 7, 2015 #14
    I do not know what sin2x is, but i think the method with subtracting the areas seems like a good idea. The area formula is a good idea too, but this was a part 1 task on an exam, and since it is part 1 you aren't allowed to use a calculator. It is more or less impossible to calculate asine values in you head.
     
  16. Apr 7, 2015 #15

    SammyS

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    Try to follow post #11 and post #13.

    No calculator needed to find sin(α)
     
  17. Apr 7, 2015 #16

    ehild

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    Read the problem. You have to give sin(α). Nobody asks the angle.

    Can you answer what is the area of the blue triangle? And what is sin(α) if you know the area and the two sides AF and AE of the blue triangle?
     
  18. Apr 7, 2015 #17
    Triangle AEF is 3/8, so to find an expression for the angle i have to change the subject of the formula. So it will be sinalfa=area/(0.5*a*b) i think i got it now, thanks for the help :) Pretty challenging math problem when you don't have alot of time to do it though.
     
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