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anonymous

  • 5 years ago

length of a rectangle is 20 meter and its are is 320 squaremeters .find the area of the square drawn on the diagonal of a square whose perimeter equals that of the given rectangle.

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  1. anonymous
    • 5 years ago
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    need help..

  2. anonymous
    • 5 years ago
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    still here?

  3. anonymous
    • 5 years ago
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    yups

  4. anonymous
    • 5 years ago
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  5. anonymous
    • 5 years ago
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    You need to find z in this set up given the info. you have.

  6. anonymous
    • 5 years ago
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    You can find z by finding x, and you find x using the information you have about the rectangle.

  7. anonymous
    • 5 years ago
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    The perimeter of the big square is equal to the perimeter of the rectangle.

  8. anonymous
    • 5 years ago
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    If the rectangle has area 320m^2 and one side 20m, then the other side of the rectangle must be found from the formula for the area:\[A=lw \rightarrow 320=20w \rightarrow w = 16\]

  9. anonymous
    • 5 years ago
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    So the perimeter of the rectangle is 16x2 + 20x2 = 72. But we're told this is the perimeter of the big square, so the x-value in the diagram is \[x=\frac{72}{4}=18\]since each side of a square is equal.

  10. anonymous
    • 5 years ago
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    Now we have to use information about the side length (x) and the angles that have been made by inserting that smaller square alongside the diagonal.

  11. anonymous
    • 5 years ago
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    ok..

  12. anonymous
    • 5 years ago
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    The diagonal of the square bisects the angle it cuts through, so we have the following (see angles):

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  13. anonymous
    • 5 years ago
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    I've called one side of one triangle h and the other k, so that x=h+k.

  14. anonymous
    • 5 years ago
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    hmm..ya ok

  15. anonymous
    • 5 years ago
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    In the bottom left triangle, we have\[\sin(45^o)=\frac{z}{h}\]but\[\sin(45^o)=\frac{1}{\sqrt{2}}\]so\[\frac{z}{h}=\frac{1}{\sqrt{2}}\]Also, in the bottom right, you have\[\sin(45^o)=\frac{z}{k}\rightarrow \frac{1}{\sqrt{2}}=\frac{z}{k}\]

  16. anonymous
    • 5 years ago
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    Since sin(45)=z/h=z/k, we must have h=k, and so h+k=2h=18, which means h=9.

  17. anonymous
    • 5 years ago
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    So from any one of those two sine equations, you have\[\frac{z}{h}=\frac{1}{\sqrt{2}}\rightarrow z=\frac{h}{\sqrt{2}}=\frac{9}{\sqrt{2}}\]

  18. anonymous
    • 5 years ago
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    The area of the square is\[z^2=\frac{81}{2}\]

  19. anonymous
    • 5 years ago
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    but lokisan i dont know trignometry...i am in standard 9..

  20. anonymous
    • 5 years ago
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    Hmmmm...let me see if there's another way.

  21. anonymous
    • 5 years ago
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    yes please

  22. anonymous
    • 5 years ago
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    Are you taught Pythagoras' Theorem? I don't know how the Indian system works.

  23. anonymous
    • 5 years ago
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    yes

  24. anonymous
    • 5 years ago
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    What are you currently learning? That might help me understand what your teacher's thinking.

  25. anonymous
    • 5 years ago
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    mensuration

  26. anonymous
    • 5 years ago
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    Are you 'allowed' to know that the diagonals of a square bisect the right angles?

  27. anonymous
    • 5 years ago
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    no

  28. anonymous
    • 5 years ago
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    man...

  29. anonymous
    • 5 years ago
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    oh yes.. i understood

  30. anonymous
    • 5 years ago
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    I know you understood...I'm just trying to stick to what you're allowed to know.

  31. anonymous
    • 5 years ago
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    understanding has no restrictions ;) ,

  32. anonymous
    • 5 years ago
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    yeah...have you covered congruence tests for triangles?

  33. anonymous
    • 5 years ago
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    yeahh

  34. anonymous
    • 5 years ago
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    Okay, I'll try and put something together with that.

  35. anonymous
    • 5 years ago
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    ok

  36. anonymous
    • 5 years ago
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    I might have something, but I want to double check.

  37. anonymous
    • 5 years ago
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    hey i think i'hv got the answer ..it's coming to 648..is it correct.. ??

  38. anonymous
    • 5 years ago
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    You have to scrap what I wrote before too - I stuffed one of the sines up. And your 648 is a number I'm getting too.

  39. anonymous
    • 5 years ago
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    hmm

  40. anonymous
    • 5 years ago
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    You can do it using congruent triangles. Only catch is that you need to know that the diagonal of a square bisects the right-angle in to 45 degrees.

  41. anonymous
    • 5 years ago
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    okay.. thanks a lot .need to go and practice some more problems.will bug u if in need.thanx again

  42. anonymous
    • 5 years ago
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    Oh...okay...anyway, I got \[z= \frac{\sqrt{648}}{3}\]so that the area is \[z^2=\frac{648}{9}=72\]

  43. anonymous
    • 5 years ago
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    Congruent triangles and Pythagoras' Theorem were used.

  44. anonymous
    • 5 years ago
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    yes

  45. anonymous
    • 5 years ago
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    Cool...well, if you're happy, I'm happy :)

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