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Geometry is it solvable?

  1. Aug 20, 2004 #1
    Geometry....is it solvable?

    from the diagram, if AB parellel to DC, and AB = 3cm , DC = 9cm , if the triangle AOB has area 6cm^2 , then find the area of the triangle BOC.

    IS this question has enough information to sovle ? if u feel yes, please show me u solution, thanks you.

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  2. jcsd
  3. Aug 20, 2004 #2


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    its doable, have fun.

    you gotta assume a few things here..

    first the triangles are isosceles.

    then 'some' angles are equal and then you gotta see a trapezoid there.. and the area of a trapezoid is.. and so on. you have to kno what the relationship between height and base of a triangle is and what it equals to, as well as proportions of the thing.

    and finally once you find area of the trapezoid, you substract the triangles and divide by two to find your BOC = AOD.

    hope it helps.. if not.. then i totally just foobared 30 minutes on this thing and donno anything (its probable)
    Last edited: Oct 8, 2005
  4. Aug 20, 2004 #3


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    The triangles AOB and COD are similar triangles. Since DC is 3 times as long as AB, OC is 3 times as long as AO and OD is 3 times as long as OB. That shouold help.
  5. Aug 20, 2004 #4
    Well I'll try to give some help,

    Just solve for the 'height' of Triangle AOB. After that you know that the height of triangle BOC is 3 times bigger, since similiar triangles as indicated by Halls.

    You know that triangle BDC - triangle DOC = triangle BOC

    The height of triangle BDC is equal to the height of the figure which is four times bigger than the height of triangle AOB, since h + 3h = 4h.

    now you have enough information to solve the problem.

    hope this helps.
  6. Aug 22, 2004 #5


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    No it does NOT have enough information to solve UNLESS you assume something. The problem is that AB and be in many places relative to DC as long as it is parrallel and has the right area. Think of it this way: If you move AB to the right in the picture, the area doesn't change (right?). But even if it did, you could just move the line up or down (always remember not to guess anything from the appearance of a drawing, assume they are out of scale etc etc)
  7. Aug 22, 2004 #6
    uhh, even if you move it anywhere, but it's still within the problem's conditions, aren't the triangles still proportional any ways? So why can't you figure it out?
  8. Aug 22, 2004 #7
    Since you can move the parallel lines relative to each other left and right all you want without affecting any of the areas, move them until AC is vertical, making CAB and ACD right angles. From the area of triangle AOB and the length of AB you already know that the HEIGHT (shortest distance between O and AB) is 4 cm. When (and only when) you have the figure skewed as I said, that distance equals AO. If you look at it, you have a "slope" diagram: you go 3 to the left and 4 down. You want to go 9 more to the left to get to the lower line, so OC (in the skewed figure only) is proportionately longer, 12 cm. Now, you have both the base and the height of triangle BOC: the base (in the skewed figure) is OC, which is 12, and the height (in the skewed figure) is AB, which is 3. So the area of triangle BOC is 18.
  9. Aug 22, 2004 #8
    i had the same answer as Bartholmew...
  10. Aug 28, 2004 #9


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    Here's an alternate method:

    The vertical separation between AB and DC is 16 which makes the area of ABC = 24 so the area of BOC = 24 - 6 = 18.
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