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the hypotenuse of an isosceles right triangle is 8 centimeters longer t…

Question

the hypotenuse of an isosceles right triangle is 8 centimeters longer than either of its legs. find the exact length of each side. (hint: an isosceles right triangle is a right triangle whose legs are the same length.)
the length of one leg is the length of the other leg is and the length of the hypotenuse is
(simplify your answers, including any radicals. use integers or fractions for any numbers in the expressions.)

Explanation:

Step1: Let the length of each leg be \(x\)

Since the hypotenuse \(c\) is \(8\) centimeters longer than either leg, then \(c=x + 8\)

Step2: Apply the Pythagorean theorem

For a right - triangle \(a^{2}+b^{2}=c^{2}\). In an isosceles right - triangle \(a = b=x\) and \(c=x + 8\). So \(x^{2}+x^{2}=(x + 8)^{2}\)

Step3: Simplify the equation

\(2x^{2}=x^{2}+16x + 64\)
\(2x^{2}-x^{2}-16x - 64=0\)
\(x^{2}-16x - 64=0\)

Step4: Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

For the equation \(x^{2}-16x - 64=0\), where \(a = 1\), \(b=-16\), \(c=-64\)
\(x=\frac{16\pm\sqrt{(-16)^{2}-4\times1\times(-64)}}{2\times1}=\frac{16\pm\sqrt{256 + 256}}{2}=\frac{16\pm\sqrt{512}}{2}=\frac{16\pm16\sqrt{2}}{2}=8\pm8\sqrt{2}\)
Since \(x>0\), we take \(x = 8 + 8\sqrt{2}\)

Step5: Find the length of the hypotenuse

\(c=x + 8=(8 + 8\sqrt{2})+8=16 + 8\sqrt{2}\)

Answer:

The length of one leg is \(8 + 8\sqrt{2}\), the length of the other leg is \(8 + 8\sqrt{2}\), and the length of the hypotenuse is \(16 + 8\sqrt{2}\)