anonymous
  • anonymous
Prove that the area of an isosceles right triangle is one-fourth the square of the length of the hypotenuse.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Two sides equal in length
anonymous
  • anonymous
If I make a straight line dividing the isosceles triangle it makes two right triangles inside it
anonymous
  • anonymous
In order to prove this you need to set up an isosceles right triangle, find the area, find the length hypotenuse.

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anonymous
  • anonymous
Thank you
anonymous
  • anonymous
how when there are no numbers provided?
anonymous
  • anonymous
You put in your own numbers, dummy numbers, or you can use letters to represent all numbers.
anonymous
  • anonymous
ok... thank you.
anonymous
  • anonymous
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...
anonymous
  • anonymous
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?
dumbcow
  • dumbcow
is this it?
anonymous
  • anonymous
yes thank you
dumbcow
  • dumbcow
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
anonymous
  • anonymous
oh my goodness... you did good... I just have to understand it in english now lol thank you!!
dumbcow
  • dumbcow
your welcome sorry i only speak english so cant help you there :)
anonymous
  • anonymous
lol

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