Question 967262: Can an isosceles right triangle have all sides with integer length? Why or why not?
I Googled this question and received the answer no with an explanation I did not understand. I want to know if you agree that the answer is no and if so why, and if not. I'm having a hard time with geometry concepts so please "dummy down the answer".
I greatly appreciate the help.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! not sure what you are asking.
an isosceles triangle can have sides that are integers.
no problem there.
can it have all 3 sides of integer length.
i don't see why not.
the only requirements is that two of the sides have the same length.
we'll now construct a triangle with two sides of equal length of 7.
we can make the third side any length we want as long as the third side is not greater than or equal to 14.
if it was, then that would violate one of the properties of a triangle that the sum of the lengths of any two sides of the triangle must be greater than the length of the third side.
so our only restriction, if two of the sides are 7, is that the third side has to be less than 14.
the other restriction is that the length of the third side is greater than the difference between the lengths of the other two sides.
since the difference between the two sides of 7 each is 0, this means that the third side has to have a length greater than 0.
that means the third side can be any number between 0 and 14, but not including 0 or 14.
there are 13 integers between 0 and 14 and not including 0 and 14.
those integers are 1,2,3,4,5,6,7,8,9,10,11,12,13.
the third side can be any one of those.
the base angles of the isosceles triangle will be different, depending on the length of the third side.
since it is an isosceles triangle, then two sides are equal and the two base angles are equal.
the third angle of the triangle is the vertex of the triangle which is the angle between the two equal sides.
BUT YOU SAID AN ISOSCELES "RIGHT" TRIANGLE.
THAT'S DIFFERENT.
CONTINUE READING TO SEE WHY THE THIRD SIDE CANNOT BE AN INTEGER WHEN THE ISO9SCELES TRIANGLE IS A RIGHT TRIANGLE.
if it's an isosceles right triangle then the hypotenuse of the right triangle is the base of the triangle.
now there is a restriction.
the vertex angle has to be 90 degrees for the right triangle to be an isosceles right triangle.
the side of the triangle that is not one of the equal sides of the triangle is the hypotenuse of this right triangle.
the formula to find the length of the hypotenuse of a right triangle is:
c^2 = a^2 + b^2
c is the hypotenuse
a and b are the legs of the right triangole.
since the triangle is an isosceles right triangle, then the legs are equal to each other.
this means that x = b, so you can replace b in the formula with a because b is the same length as a.
the formuls of c^2 = a^2 + b^2 becomes c^2 = a^2 + a^2 which becomes c^2 = 2 * a^2.
you want to solve for c which is the length of the hypotenuse.
take the square root of both sides of the equation of c^2 = 2 * a^2 and you will get:
sqrt(c^2) = sqrt(2 * a^2)
sqrt(c^2) is equal to c.
formula becomes c = sqrt(2 * a^2)
since sqrt(2 * a^2) is equivalent to sqrt(2) * sqrt(a^2), then the equation becomes:
c = sqrt(2) * sqrt(a^2).
since sqrt(a^2) is equal to a, the formula becomes:
c = sqrt(2) * a
the square root of 2 is an irrational number, so sqrt(2) * a can never be a rational number.
since an integer is a special form of a ration number where the denominator is equal to 1, this means that sqrt(2) * a can never be an integer.
the answer provided to you is correct.
the third side of an isosceles right triangle can never be an integer.
let's take our triangle and swee how it works in numbers.
the equal legs are 7 apiece.
it's a right triangle so the hypotenuse squared is equal to 2 * 7^2
the hypotenuse is equal to sqrt (2 * 7^2) which is equal to sqrt(2) * sqrt(7^2) which is equal to sqrt(2) * 7.
sqrt(2) is not a rational number, so sqrt(2) * 7 can never be a rational number.
no cigars.
third side can't be a rational number.
therefore the third side can't be an integer because an integer is a special form of a rational number.
a rational number is defined as a fraction where the numerator and the denominator are both integers.
3/4 is a rational number because 3 and 4 are integers.
5 is a rational number because 5 is equivalent to 5/1 and 5 and 1 are integers.
sqrt(2) is not a rational number and so the third side can't be an integer because the third side of the isosceles right triangle will always be equal to any number * sqrt(2).
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