To prove the equation [tex]\frac{1}{YN}^2 = \frac{1}{XY}^2 + \frac{1}{YZ}^2[/tex] for the right triangle XYZ with the right angle at Y and YN as the perpendicular to XZ, you can use the Pythagorean theorem.
According to the Pythagorean theorem, for a right triangle, the sum of the squares of the two shorter sides (legs) is equal to the square of the hypotenuse. In this case, XY and YZ are the legs, and XZ is the hypotenuse.
So, we have:
[tex]XY^2 + YZ^2 = XZ^2[/tex]
Now, since YN is a perpendicular drawn from the right angle vertex Y to the hypotenuse XZ, we can use similar triangles to relate the lengths:
[tex]\frac{XY}{YZ} = \frac{YN}{XZ}[/tex]
Rearrange this proportion:
[tex]YN = (XY \times XZ) \div YZ[/tex]
Now, square both sides of this equation:
[tex]YN^2 = (XY^2 \times XZ^2) \div YZ^2[/tex]
Divide both sides by [tex](XY^2 \times XZ^2 \times YZ^2):[/tex]
[tex]1\div YN^2 = 1\div(XY^2 \times YZ^2)[/tex]
Now, substitute the value of [tex]XY^2 + YZ^2[/tex] from the Pythagorean theorem:
[tex]1/YN^2 = 1/(XZ^2)[/tex]
Since XZ is the hypotenuse, you can rewrite it as [tex]XY^2 + YZ^2:[/tex]
[tex]1\div YN^2 = 1\div(XY^2 + YZ^2)[/tex]
We have successfully proven the equation [tex](\frac{1}{YN})^2 = (\frac{1}{XY})^2 + (\frac{1}{YZ})^2[/tex] for the given right triangle XYZ with the right angle at Y and YN as the perpendicular to XZ.
Answers & Comments
To prove the equation [tex]\frac{1}{YN}^2 = \frac{1}{XY}^2 + \frac{1}{YZ}^2[/tex] for the right triangle XYZ with the right angle at Y and YN as the perpendicular to XZ, you can use the Pythagorean theorem.
According to the Pythagorean theorem, for a right triangle, the sum of the squares of the two shorter sides (legs) is equal to the square of the hypotenuse. In this case, XY and YZ are the legs, and XZ is the hypotenuse.
So, we have:
[tex]XY^2 + YZ^2 = XZ^2[/tex]
Now, since YN is a perpendicular drawn from the right angle vertex Y to the hypotenuse XZ, we can use similar triangles to relate the lengths:
[tex]\frac{XY}{YZ} = \frac{YN}{XZ}[/tex]
Rearrange this proportion:
[tex]YN = (XY \times XZ) \div YZ[/tex]
Now, square both sides of this equation:
[tex]YN^2 = (XY^2 \times XZ^2) \div YZ^2[/tex]
Divide both sides by [tex](XY^2 \times XZ^2 \times YZ^2):[/tex]
[tex]1\div YN^2 = 1\div(XY^2 \times YZ^2)[/tex]
Now, substitute the value of [tex]XY^2 + YZ^2[/tex] from the Pythagorean theorem:
[tex]1/YN^2 = 1/(XZ^2)[/tex]
Since XZ is the hypotenuse, you can rewrite it as [tex]XY^2 + YZ^2:[/tex]
[tex]1\div YN^2 = 1\div(XY^2 + YZ^2)[/tex]
We have successfully proven the equation [tex](\frac{1}{YN})^2 = (\frac{1}{XY})^2 + (\frac{1}{YZ})^2[/tex] for the given right triangle XYZ with the right angle at Y and YN as the perpendicular to XZ.
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