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rtraylor3

Find the measure please, I got 3 but it's wrong?

  • one year ago
  • one year ago

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  1. rtraylor3
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    • one year ago
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  2. eseidl
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    find the measure of what?

    • one year ago
  3. eseidl
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    D?

    • one year ago
  4. rtraylor3
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    Sorry the area of triangle DEF

    • one year ago
  5. eseidl
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    ok, so A=12 is the area of the first triangle

    • one year ago
  6. rtraylor3
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    Yes. I went by the side of the first and second. The first was 4 and the second was 1 so I thought that the first triangle was 4 times greater than the second. 12/4=3

    • one year ago
  7. eseidl
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    each side of the triangle on the right is scaled down by a factor of 1/4.

    • one year ago
  8. rtraylor3
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    Correct...

    • one year ago
  9. eseidl
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    Area of a triangle is base*height/2 so,\[\frac{\left| AC \right|4*\sin A}{2}=12\]for the triangle on the left. The triangle on the right is:\[\frac{\left| DF \right|1*\sin D}{2}=?\]Sow the sides AC and DF are related. DF=(1/4)AC and the angles A and D (and therefore their sines) are equal. so, we can write the second equation above as:\[\frac{(1/4)\left| AC \right|1*\sin A}{2}=?\]comparing this with the first equation, we see that 2AC*sinA=12 and (1/8)AC*sinA=?. Now we see that the second equation is 1/16 of the first one. So the area of the second triangle must be 12/16=3/4

    • one year ago
  10. eseidl
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    So, in summary, if one side of a triangle is scaled by a factor of x, then the area of the second triangle is x^2*area of the first.

    • one year ago
  11. eseidl
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    where x=1/4 in this example.

    • one year ago
  12. rtraylor3
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    Thanks so much!! I got it now!

    • one year ago
  13. eseidl
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    no prob :)

    • one year ago
  14. rtraylor3
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    New fan= me(:

    • one year ago
  15. eseidl
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    and if you are scaling volumes, it is the same type of argument. Only now you use x^3 instead of x^2 because there is an extra dimension.

    • one year ago
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