Answer :
To prove that in a \( 45^\circ-45^\circ-90^\circ \) triangle the hypotenuse is \(\sqrt{2}\) times the length of each leg, we are given that the triangle is an isosceles right triangle where the legs have equal length \(a\). The proof should proceed by using the Pythagorean theorem and then finding the relationship between the hypotenuse \(c\) and the legs \(a\).
Given:
[tex]\[ a^2 + a^2 = c^2 \][/tex]
Combine like terms on the left side:
[tex]\[ 2a^2 = c^2 \][/tex]
We need to isolate \(c\). For this, we take the principal (positive) square root of both sides of the equation:
[tex]\[ \sqrt{2a^2} = \sqrt{c^2} \][/tex]
Since \(\sqrt{c^2} = c\) and \(\sqrt{2a^2} = \sqrt{2} \cdot \sqrt{a^2} = \sqrt{2} \cdot a\):
[tex]\[ c = \sqrt{2} \cdot a \][/tex]
This final step shows that the length of the hypotenuse \(c\) is \(\sqrt{2}\) times the length of each leg \(a\). Hence, the correct step is to:
Determine the principal square root of both sides of the equation.
Given:
[tex]\[ a^2 + a^2 = c^2 \][/tex]
Combine like terms on the left side:
[tex]\[ 2a^2 = c^2 \][/tex]
We need to isolate \(c\). For this, we take the principal (positive) square root of both sides of the equation:
[tex]\[ \sqrt{2a^2} = \sqrt{c^2} \][/tex]
Since \(\sqrt{c^2} = c\) and \(\sqrt{2a^2} = \sqrt{2} \cdot \sqrt{a^2} = \sqrt{2} \cdot a\):
[tex]\[ c = \sqrt{2} \cdot a \][/tex]
This final step shows that the length of the hypotenuse \(c\) is \(\sqrt{2}\) times the length of each leg \(a\). Hence, the correct step is to:
Determine the principal square root of both sides of the equation.
