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anonymous

  • 5 years ago

Prove that the area of an isosceles right triangle is one-fourth the square of the length of the hypotenuse.

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  1. anonymous
    • 5 years ago
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    Two sides equal in length

  2. anonymous
    • 5 years ago
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    If I make a straight line dividing the isosceles triangle it makes two right triangles inside it

  3. anonymous
    • 5 years ago
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    In order to prove this you need to set up an isosceles right triangle, find the area, find the length hypotenuse.

  4. anonymous
    • 5 years ago
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    Thank you

  5. anonymous
    • 5 years ago
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    how when there are no numbers provided?

  6. anonymous
    • 5 years ago
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    You put in your own numbers, dummy numbers, or you can use letters to represent all numbers.

  7. anonymous
    • 5 years ago
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    ok... thank you.

  8. anonymous
    • 5 years ago
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    IF I plug in 5 for my base and 3 for my height I get an area of 7.5 I can calculate the hypotenuse 3^2 + 5^2 = c^2 9 + 25 = c^2 34 = c^2 5.83 as my third side...

  9. anonymous
    • 5 years ago
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    what is my next step of how to prove the area of an isosceles triangle is one-fourth the square of the length of the hypotenuse?

  10. dumbcow
    • 5 years ago
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    is this it?

  11. anonymous
    • 5 years ago
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    yes thank you

  12. dumbcow
    • 5 years ago
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    An isosceles right triangle is a 45-45-90 triangle with sides of x and hypotenuse of sqrt(2)x x^2 +x^2 = (sqrt(2)x)^2 2x^2 = 2x^2 We know the area of triangle is 1/2 base*height, where base and height are both x A = 1/2*x^2 Hypotenuse is sqrt(2)x -> Hypotenuse^2 = (sqrt(2)x)^2 = 2x^2 1/4*2x^2 = 1/2*x^2 Therefore Area is 1/4 of Hypotenuse^2

  13. anonymous
    • 5 years ago
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    oh my goodness... you did good... I just have to understand it in english now lol thank you!!

  14. dumbcow
    • 5 years ago
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    your welcome sorry i only speak english so cant help you there :)

  15. anonymous
    • 5 years ago
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    lol

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