The Sum of the Squares

Geometry Level 2

Consider a version of the Pythagorean theorem with mistakes (where mistakes are in bold):

The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side .

Suppose the constraint of "right triangle" is in place and "square roots" are changed to "squares".

A: Which sides would you pick for "any two sides"?

B: If the measurement of the "remaining side" is 1 meter, what are the measurements of "any two sides"?

Simplify your answer to B with a rationalized denominator.

A: two legs, B:

\sqrt{2}

and

\sqrt{2}

A: a leg and the hypotenuse, B:

1

and

\frac{1}{\sqrt{2}}

A: a leg and the hypotenuse, B:

1

and

\sqrt{2}

A: a leg and the hypotenuse, B:

1

and

\frac{\sqrt{2}}{2}

A: two legs, B:

\frac{1}{\sqrt{2}}

and

\frac{1}{\sqrt{2}}

A: two legs, B:

\frac{\sqrt{2}}{2}

and

\frac{\sqrt{2}}{2}

2 solutions

Joshua Lowrance
May 2, 2019

The Pythagorean Theorem states that $a^2+b^2=c^2$ for any right triangle, if $a$ and $b$ are the two legs, and $c$ is the hypotenuse. Therefore, we know that the answer to A is two legs. For part B, we want the squares of the two numbers to sum to $1$ (as stated by the Pythagorean Theorem). Only two answers fit this requirement:( $\frac{1}{\sqrt{2}} \text{and} \frac{1}{\sqrt{2}}$ ) and ( $\frac{\sqrt{2}}{2} \text{and} \frac{\sqrt{2}}{2}$ ). However, it asks for a rationalized denominator; or in other words, a denominator that does not have a square root. Only $\frac{\sqrt{2}}{2} \text{and} \frac{\sqrt{2}}{2}$ fits this.

David Velasco
May 2, 2019

The hypotenuse is always longest, so the answer of A cannot be "a leg and the hypotenuse."

Right isosceles triangles always have sides of length $(1, 1, \sqrt{2})$ where the last length is the hypotenuse. So, it cannot be $\sqrt{2}$ .

Therefore, the answer of B is $\sqrt{2}$ 's reciprocal.

The Sum of the Squares

2 solutions

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