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If the normal of an equilateral triangle is the square root of three-quarters what is the sum of the sides of the triangle and how is this proved?
TempusFugitive
The answer is 3.
Proof:
According to Pythagoras, for any right-angled triangle (one that has one angle equal to 90 degrees), the square of the longest side equals the sum the squares of the two shorter sides.
So, if the sides of the equilateral triangle are of length s, a normal from any apex will divide the opposite side into equal lengths that are each equal to a half of s (= s x 0.5)
The normal is part of a right-angled triangle which has its longest side of length s, the normal as its next-shorter side and its shortest side is of length (s x 0.5)
We have been told that the square of the normal = 0.75 (A)
We can calculate the square of the length of a half side as:
(s x 0.5) x (s x 0.5) = s^2 x 0.25 (s squared times a quarter) (B)
So the sum of these squares (A) + (B) = s^2 x 0.25 + 0.75
According to Pythagoras, (A) + (B) = the square of the longest side s = s^2
So s^2 = [(s^2 x 0.25) + 0.75]
Using algebra, we can reduce this to: [s^2 - (s^2 x 0.25)] = 0.75
So (s^2 x 0.75) = 0.75 so s^2 = 1 so s =1
So the answer to the question " what is the sum of its three sides" is 1 + 1 + 1 = 3 ---- ; Above Correct Answer and Proof by Martinel. : Joined: Thu Feb 07, 2008 12:00 am
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