Question

mothy says that if you double all the sides of a right triangle, the new triangle is obtuse. Use the
Pythagorean Theorem to explain why this statement is incorrect

Answer 1

@ramen

Answer 2

Could you rephrase the question, please? As well as including the full text and/or inag5es.

Answer 3

Let's denote \(a, b,\) and \(c\) as the original sides to the right triangle, whereby \(a\) and \(b\) are legs, and \(c\) is a hypotenuse.
By the Pythagorean Inequality Theorem, a right triangle satisfies
\(a^2+b^2=c^2\)
Similarly, an obtuse triangle satisfies
\(a^2+b^2<c^2\)

If you double all the sides, it would be represented as
\((2a)^2+(2b)^2=(2c)^2\)
Simplified, that is
\(4a^2+4b^2=4c^2\)
\(4(a^2+b^2)=4(c^2)\)
\(a^2+b^2=c^2\)
Hence, the new sides still satisfy the conditions for a right triangle. By doubling the sides, you are not getting \(a^2+b^2<c^2\). By doubling the sides, you are enlarging the triangle by a factor of 4, that's all.

Answer 4

Nah. You coulda fed your stuff into the inequality and confirmed your [correct] bias that way, equally efficiently.

Simple schematic.

Pythagoras tells us: \(c = \sqrt{ a^2 + b^2}\)

Let: \(\bar a = 2 a \qquad \bar b = 2 b\)

Add the squares: \(\sqrt{ \bar a^2 + \bar b^2 }= \sqrt{(2a)^2 + (2b)^2} = \sqrt{ 4 (a^2 + b^2) }\)

\( = 2 \sqrt{a^2 + b^2} = 2 c\)