Since we know 6 and 8 are the lengths of the biggest triangle. We can say that the 3rd side is 10 in length. We are given that ED = 5; Therefore, AE = 5.
That is correct, knowing that ABD is a right triangle and its hypotenuse is unknown. Note, that if the hypotenuse was 8 and one side was 6, then the other side would be √(8² – 6²) = 2√7 , even though the triangle is a right triangle. Remember, that in 3-4-5 right triangle, 5 is the hypotenuse.
… Therefore, AE = 5. Triangles AFE and DCE are right angled; Therefore they are (3, 4, 5) in length.
This conclusion is groundless. Knowing just one side of a right triangle is not enough to calculate the other two.
If we know that a hypotenuse is 5, then there are infinitely many such triangles. For example the sides can be: 5, 5√2 / 2, 5√2 / 2 5, 4, 3 5, 2√6, 1
Plug in a value for x ( 0< x < 5 ) to make your own example: 5, x, √(25 – x²)
Indeed, in this problem these triangles are 5-4-3 triangles, but its a lucky coincidence. Some more reasoning is needed to prove that.
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