Question

the perimeter of an isosceles right triangle is ( 62 + 62sqrt{2} ) inches. what is the length, in inches, of the hypotenuse of this triangle?
a 31
b ( 31sqrt{2} )
c 62
d ( 62sqrt{2} )

Explanation:

Step1: Recall the side - length ratio of an isosceles right triangle

In an isosceles right triangle, if the length of each of the two equal sides (legs) is \(a\), then the length of the hypotenuse \(c\) is \(a\sqrt{2}\) (by the Pythagorean theorem \(c=\sqrt{a^{2}+a^{2}}=\sqrt{2a^{2}} = a\sqrt{2}\)). The perimeter \(P=a + a+ a\sqrt{2}=2a + a\sqrt{2}\).

Step2: Set up an equation for the perimeter

We are given that \(P = 62+62\sqrt{2}\). So, \(2a + a\sqrt{2}=62 + 62\sqrt{2}\). Factor out \(a\) from the left - hand side: \(a(2+\sqrt{2})=62(1 + \sqrt{2})\). Multiply the numerator and denominator of the right - hand side by \((2-\sqrt{2})\) to rationalize. But we can also observe by comparing the coefficients:
If \(a(2+\sqrt{2})=62(1+\sqrt{2})\), we can rewrite \(a(2+\sqrt{2})\) as \(a\sqrt{2}(\sqrt{2}+1)\) (since \(a(2+\sqrt{2})=a\sqrt{2}(\sqrt{2}+1)\) by factoring out \(\sqrt{2}\) from \(2+\sqrt{2}\) as \(2+\sqrt{2}=\sqrt{2}(\sqrt{2} + 1)\)).
Equating \(a\sqrt{2}(\sqrt{2}+1)=62(1+\sqrt{2})\), we can cancel out \((\sqrt{2}+1)\) (since \(\sqrt{2}+1
eq0\)). Then \(a\sqrt{2}=62\).

Answer:

C. 62