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Geometry Questions

  1. Mar 18, 2006 #1
    Please provide help with these questions please!

    #1)

    [​IMG]

    WXYZ is a trapezoid in which WZ is parallel to XY and WZ:XY = 3:4. WY and XZ intersect at R. If triangle RXY = 100, calculate the area of trapezoid WXYZ.

    #2)

    [​IMG]

    PQRS is a parallelogram. The diagonal PR is produced to M so that RM = PR. Prove that the quadrilateral SRQM = parallelogram PQRS.


    any help/clues/advice is appreciated!!!
     
  2. jcsd
  3. Mar 18, 2006 #2

    Curious3141

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    1) You have to show some work before getting help, that's the policy here.

    2) This should be pre-calc work.
     
  4. Mar 18, 2006 #3

    0rthodontist

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    Well, not necessarily--you could use Cavalieri's principle for #2.
    (excuse me for saying that, my rationale is it probably doesn't really constitute helping him)
     
  5. Mar 18, 2006 #4

    Curious3141

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    OK, just a couple of hints to get you going, then you'll have to show real working.

    For question 1 :

    View the figure initially as 4 triangles. One pair are similar. Prove that. Now you can find the area of one triangle in the similar pair using what you should know about the ratios of areas of similar figures. You don't know the areas of the other two small triangles, but you can find that out, by observing that each of these shares a height with a triangle of known area. Then using the ratio of the bases, and the formula for area of triangle, find the area of each of the other two. Now you can add everything up.

    For the second part, this is very easy. Just think of the figure as two large triangles. Subdivide each of these into two smaller triangles and observe a common height and equal base. Now draw a conclusion about the areas of each triangle and you can prove the result with ease.
     
  6. Mar 18, 2006 #5
    For 1)

    triangle WRZ and triangle XRY are similar by inspection. however, I can't gather enough information to make that deduction (am I missing something)? angle WRZ and angle XRY are equal (opposite). if I had the length of another side, I could compare it to the proportion given in the question. I also don't think I can assume angle RWZ and angle RXY are equal because no where in the question does it say WY and XZ are diagonals of the trapezoid nor the trapezoid is isosceles. if you could help me identify another property of the triangles I am potentially mislooking, I could use the ratios to find the area of the top triangle. based on your second hint, the two side triangles share a height with triangle RXY, correct?

    For 2)

    when you say "Just think of the figure as two large triangles.", are you referring to PQMS or just SRQM?

    thanks for your help so far. sorry for coming off as if all i wanted was a quick answer.
     
  7. Mar 19, 2006 #6

    0rthodontist

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    To show similarity, consider instead angles RWZ and RYX. Also it can be done more simply by realizing that you are looking for 7/8 *b * h where b is the length of the lowest side and h is the height. You can find b * h from the area of the triangle given.
     
  8. Mar 19, 2006 #7

    Curious3141

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    "Inspection" is good enough for a suspicion, but proof needs more rigor. :smile:


    OK, this part is good.

    And in fact angles RWZ and RXY are not (necessarily) equal.

    But angle RWZ is equal to angle RYX. Can you see why (hint : think of the parallel lines WZ and XY and line WY intersecting each of them).

    You now have two angles equal, can you prove similarity ? Identify which vertices of one triangle correspond to which vertices of the other. And I think you know how to do the rest.


    Yes, correct. I think you can complete the problem from here on.

    I'm thinking of PQM and PSM as the two large triangles.

    It's cool.
     
  9. Mar 19, 2006 #8
    Ok... im glad whoever is logging onto my account is getting help, but for future reference, please make your own.

    btw, HeroOfTheDay is my account, and i have never posted these messages.

    Whoever is logging into my account, please email me, or send me a private message.
     
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