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anonymous

  • 4 years ago

Need Help with Geometry and Ratio!!!!

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  1. anonymous
    • 4 years ago
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    In the diagram, AX = 10, XB=5, AY=4 and YC=8. What is the ratio of the area of AXY to the area of ABC? Express your answer as a common fraction.

  2. anonymous
    • 4 years ago
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    |dw:1327715430695:dw|

  3. Directrix
    • 4 years ago
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    Are the triangles given to be similar? If so, what is the correspondence of vertices? Would you check to determine if the diagram is correctly labeled? Thanks.

  4. anonymous
    • 4 years ago
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    The diagram (seems) not correct.

  5. Directrix
    • 4 years ago
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    I agree. I think that the SAS Similarity Theorem is needed for solution but the given data doesn't satisfy the theorem's hypothesis. Triangles are similar if two sides in one triangle are in the same proportion to the corresponding sides in the other, and the included angles are equal.

  6. anonymous
    • 4 years ago
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    this is all the given information

  7. anonymous
    • 4 years ago
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    I could draw the diagram again to clarify.

  8. anonymous
    • 4 years ago
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    |dw:1327717724546:dw|

  9. Directrix
    • 4 years ago
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    Are both the black segments of length h? It seems that they would differ in length. And, are the segments labeled h perpendicular to segment AB?

  10. anonymous
    • 4 years ago
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    How come the black segments be equal in length? or am I misunderstanding ?!

  11. anonymous
    • 4 years ago
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    yeah sorry, they don't have the same length and i believe they are perpendicular. That was the hint from my teacher.

  12. anonymous
    • 4 years ago
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    |dw:1327720461436:dw|

  13. anonymous
    • 4 years ago
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    are you sure that this are the only given?

  14. anonymous
    • 4 years ago
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    yes

  15. anonymous
    • 4 years ago
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    I think it has something to do with ratio

  16. anonymous
    • 4 years ago
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    i have here only assumptions. .

  17. anonymous
    • 4 years ago
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    ok

  18. anonymous
    • 4 years ago
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    the area of ACB-Ayx= area of CyxB the area of ACB-CyxB= area of Ayx

  19. anonymous
    • 4 years ago
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    Yes, but how do I find the area?

  20. anonymous
    • 4 years ago
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    this is what I've got 4/12 = p/h 4h/12=p |dw:1327722363321:dw|

  21. anonymous
    • 4 years ago
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    but i am not sure how to find ABC.

  22. anonymous
    • 4 years ago
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    your problem is ABC? find area of ACB: sinA=h/(8+4) therefor h=12sinA area = 1/2bh = 1/2(10+5)(12sinA) ; substitute the value of h in the formula = 1/2(15)(12sinA) therefor: the area of ACB is 90sinA

  23. anonymous
    • 4 years ago
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    Can you clarify sinA?

  24. anonymous
    • 4 years ago
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    no it,s not (8+4) for h, it's 10+5 , and (10+5) is 8+4, but anyway the answer is still 90 sinA

  25. anonymous
    • 4 years ago
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    sin A will be find if there's a given measure for the angle. .

  26. anonymous
    • 4 years ago
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    ok

  27. anonymous
    • 4 years ago
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    thanks

  28. anonymous
    • 4 years ago
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    you're welcome. .

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