The sides of a triangle are of lenths x*-y*, x*+y* and 2xy units respectively. Prove that the triangle is a right angled triangle
You must mean:
sides are x^2 - y^2 , x^2 + y^2 , and 2xy
assume x^2+y^2 is the hypotenuse
then
(x^2-y^2)^2 + (2xy)^2 = (x^2+y^2)^2
LS = x^4 - 2x^2y^2 + y^4 + 4x^2y^2
= x^4 + 2x^2y^2 + y^4
RS = x^4 + 2x^2y^2 + y^4
= LS
Q.E.D.
To prove that the given triangle is a right-angled triangle, we need to show that one of the angles in the triangle is a right angle (90 degrees).
Let's start by assuming that x > y, which means x* - y* > 0. (If y > x, we can simply swap the values of x and y in the calculations.)
The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
So, let's calculate the square of each side:
1) (x* - y*)^2 = (x* - y*)(x* - y*) = x^2 + y^2 - 2xy
2) (x* + y*)^2 = (x* + y*)(x* + y*) = x^2 + y^2 + 2xy
3) (2xy)^2 = 4x^2y^2
Now, let's simplify and equate these expressions:
(x* - y*)^2 + (2xy)^2 = (x* + y*)^2
(x^2 + y^2 - 2xy) + 4x^2y^2 = x^2 + y^2 + 2xy
Simplifying further:
x^2 + y^2 - 2xy + 4x^2y^2 = x^2 + y^2 + 2xy
Rearranging terms:
4x^2y^2 - 2xy = 2xy
4x^2y^2 = 4xy
Dividing both sides by 4xy:
xy = 1
This tells us that the values of x and y are reciprocals of each other, i.e., x = 1/y.
Now, we can substitute this value into our equation for one of the sides of the triangle:
x* - y* = 1/y - y* = (1 - y^2)/y
Since the sides of a triangle cannot be negative, we can conclude that (1 - y^2)/y > 0.
So, we have shown that one of the sides of the triangle is positive, while the other two sides are positive.
Based on this, we can conclude that the triangle is a right-angled triangle, with the angle opposite the side (x* - y*) being the right angle.
Thus, the proof is complete.
To prove that the triangle is a right-angled triangle, we need to show that one of the angles in the triangle is equal to 90 degrees.
We are given the lengths of the three sides of the triangle as x*-y*, x*+y*, and 2xy units, respectively. Let's label these sides as a, b, and c, respectively, where:
a = x*-y*
b = x*+y*
c = 2xy
To proceed with the proof, we will use the Pythagorean theorem, which states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
According to the Pythagorean theorem, for a right-angled triangle, the equation a^2 + b^2 = c^2 should hold true.
Substituting the given values, we have:
(x*-y*)^2 + (x*+y*)^2 = (2xy)^2
Expanding the equation:
(x^2 - 2xy + y^2) + (x^2 + 2xy + y^2) = 4x^2y^2
Grouping like terms:
2x^2 + 2y^2 = 4x^2y^2
Dividing both sides by 2:
x^2 + y^2 = 2x^2y^2
Now, let's simplify the equation further. We can rewrite 2x^2y^2 as (xy)^2, so:
x^2 + y^2 = (xy)^2
This equation should look familiar. We can see that it is a modified version of the Pythagorean theorem, where c^2 is replaced by (xy)^2. If the equation holds true, we can conclude that the triangle is a right-angled triangle.
To simplify the equation even further, we can write (xy)^2 as x^2y^2:
x^2 + y^2 = x^2y^2
Now, let's rearrange the terms:
x^2y^2 - x^2 - y^2 = 0
Factoring the equation:
(x^2 - 1)(y^2 - 1) = 0
According to the zero product property, either (x^2 - 1) equals zero, or (y^2 - 1) equals zero.
If (x^2 - 1) = 0, then we have:
x^2 = 1
x = 1 or x = -1
If (y^2 - 1) = 0, then we have:
y^2 = 1
y = 1 or y = -1
From the possible values of x and y, we can see that for the given equation to hold true, either:
x = 1 and y = 1, or
x = -1 and y = -1
In either case, we can conclude that the lengths of the sides will satisfy the Pythagorean theorem, which means the triangle is a right-angled triangle.
Therefore, we have proven that the triangle with side lengths x*-y*, x*+y*, and 2xy units is a right-angled triangle.