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anonymous

  • one year ago

Find the missing measures.

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  1. anonymous
    • one year ago
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    |dw:1437458115319:dw|

  2. mathstudent55
    • one year ago
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    What do you know about triangles IFE and IHG?

  3. anonymous
    • one year ago
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    I know that HI=\[3\sqrt{3}\]

  4. mathstudent55
    • one year ago
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    That is correct, but what about my question?

  5. anonymous
    • one year ago
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    IFG has 2 side missing and IHG has 1 missing side?

  6. mathstudent55
    • one year ago
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    No, that's not what I mean. Look at the triangles when you separate them. See the figure below. |dw:1437458809533:dw|

  7. nincompoop
    • one year ago
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    understand the concept of proportionality or similarity and congruence.

  8. nincompoop
    • one year ago
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    if you want, you can solve the smaller portion of right triangle's adjacent side (adjacent to the angle), and then work your way from there

  9. mathstudent55
    • one year ago
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    Now look. Triangle IFE and triangle IHG are both right triangles. That means that each one has a right angle. Right angles are congruent, so you already have one pair of congruent angles.

  10. nincompoop
    • one year ago
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    |dw:1437459025019:dw|

  11. mathstudent55
    • one year ago
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    Now also notice that angle I of triangle IFE and angle I of triangle IHG are congruent because they are the same angle. That is a second pair of congruent angles. That makes the triangles similar.

  12. mathstudent55
    • one year ago
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    @nincompoop If instead of writing, you did some more reading, you'd see he already found that out.

  13. mathstudent55
    • one year ago
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    Since we now know we have similar triangles, we use the fact that the corresponding sides have proportional lengths. We set up proportions and find the missing lengths.

  14. mathstudent55
    • one year ago
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    |dw:1437459177765:dw|

  15. mathstudent55
    • one year ago
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    We can use the two sides in black above to establish a ratio of lengths: \(\dfrac{large~triangle}{small~triangle} = \dfrac{5}{3} \)

  16. mathstudent55
    • one year ago
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    |dw:1437459273682:dw|

  17. mathstudent55
    • one year ago
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    \(\dfrac{5}{3} = \dfrac{EG + 6}{6} \) Do you understand this proportion?

  18. nincompoop
    • one year ago
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    basic principle: |dw:1437459224575:dw|

  19. nincompoop
    • one year ago
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    |dw:1437459427659:dw|

  20. nincompoop
    • one year ago
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    |dw:1437459476749:dw|

  21. mathstudent55
    • one year ago
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    \(\dfrac{5}{3} = \dfrac{EG + 6}{6}\) \(6 \times 5 = 3 \times (EG + 6) \) \(30 = 3EG + 18\) \(3 EG = 12\) \(EG = 4\)

  22. triciaal
    • one year ago
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    |dw:1437459565905:dw|

  23. mathstudent55
    • one year ago
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    |dw:1437459606710:dw|

  24. mathstudent55
    • one year ago
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    We just need to find FH.

  25. mathstudent55
    • one year ago
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    \(\dfrac{6}{3\sqrt 3} = \dfrac{4}{FH} \) \(6FH = 4 \times 3 \sqrt 3\) \(6 FH = 12 \sqrt 3\) \(FH = 2 \sqrt 3\)

  26. nincompoop
    • one year ago
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    use tangent ratio

  27. nincompoop
    • one year ago
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    page 41 http://www.scribd.com/doc/22374733/Mathematics-Teach-Yourself-Trigonometry

  28. mathstudent55
    • one year ago
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    We can find FH another way. We can solve for FI using the Pythagorean theorem with the large triangle, then subtract HI from it. \((FI)^2 + (FE)^2 = (EI)^2\) \((FI)^2 + 5^2 = 10^2\) \((FI)^2 + 25 = 100\) \((FI)^2 = 75\) \(FI = 5\sqrt3\) \(FH = FI - HI\) \(FH = 5 \sqrt 3 - 3\sqrt 3\) \(FH = 2 \sqrt 3\) As you can see, we get the same result for FH as we did above.

  29. mathstudent55
    • one year ago
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    @SimiYami I hope you can follow my explanations above. If you have any questions, just ask.

  30. nincompoop
    • one year ago
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    @triciaal |dw:1437460115571:dw|