Question

find the lengths of the sides of each right triangle.
i can use the pythagorean theorem.
$x^{2}+(x + 1)^{2}=49$
$x^{2}+x^{2}+2x + 1=49$
$2x^{2}+2x-48 = 0$

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle with sides \(a\), \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\).

Step2: Expand the equations

Expand \((x + n)^{2}=x^{2}+2nx + n^{2}\) for each case.

Step3: Simplify to quadratic form

Combine like terms to get \(ax^{2}+bx + c = 0\) form.

Step4: Solve quadratic equations

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) and choose the non - negative solution for the length of the side of the triangle.

Answer:

1. For the first triangle with sides \(x\), \(x + 1\) and hypotenuse \(7\):

• First, solve the quadratic equation \(2x^{2}+2x - 48=0\). Divide the entire equation by \(2\) to get \(x^{2}+x - 24=0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\), here \(a = 1\), \(b=1\) and \(c=-24\). Then \(x=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-24)}}{2\times1}=\frac{-1\pm\sqrt{1 + 96}}{2}=\frac{-1\pm\sqrt{97}}{2}\). Since the length of a side cannot be negative, \(x=\frac{-1+\sqrt{97}}{2}\approx\frac{-1 + 9.85}{2}=4.425\). The sides are \(x\approx4.425\), \(x + 1\approx5.425\) and the hypotenuse is \(7\).

1. For the second triangle with sides \(x\), \(x + 4\) and hypotenuse \(20\):

• By the Pythagorean theorem, \(x^{2}+(x + 4)^{2}=400\). Expand to get \(x^{2}+x^{2}+8x + 16=400\), then \(2x^{2}+8x-384 = 0\). Divide by \(2\): \(x^{2}+4x - 192=0\). Factor the quadratic equation: \((x + 16)(x - 12)=0\). So \(x=-16\) or \(x = 12\). Since length cannot be negative, \(x = 12\). The sides are \(x = 12\), \(x + 4=16\) and the hypotenuse is \(20\).

1. For the third triangle with sides \(x\), \(x + 6\) and hypotenuse \(10\):

• By the Pythagorean theorem, \(x^{2}+(x + 6)^{2}=100\). Expand: \(x^{2}+x^{2}+12x+36 = 100\), then \(2x^{2}+12x - 64=0\). Divide by \(2\): \(x^{2}+6x - 32=0\). Using the quadratic formula with \(a = 1\), \(b = 6\) and \(c=-32\), \(x=\frac{-6\pm\sqrt{6^{2}-4\times1\times(-32)}}{2\times1}=\frac{-6\pm\sqrt{36 + 128}}{2}=\frac{-6\pm\sqrt{164}}{2}=\frac{-6\pm2\sqrt{41}}{2}=-3\pm\sqrt{41}\). Since length cannot be negative, \(x=-3+\sqrt{41}\approx-3 + 6.4=3.4\). The sides are \(x\approx3.4\), \(x + 6\approx9.4\) and the hypotenuse is \(10\).

1. For the fourth triangle with sides \(x-1\), \(x + 8\) and hypotenuse \(12\):

• By the Pythagorean theorem, \((x - 1)^{2}+(x + 8)^{2}=144\). Expand: \(x^{2}-2x + 1+x^{2}+16x+64=144\), then \(2x^{2}+14x - 79=0\). Using the quadratic formula with \(a = 2\), \(b = 14\) and \(c=-79\), \(x=\frac{-14\pm\sqrt{14^{2}-4\times2\times(-79)}}{2\times2}=\frac{-14\pm\sqrt{196 + 632}}{4}=\frac{-14\pm\sqrt{828}}{4}=\frac{-14\pm6\sqrt{23}}{4}=\frac{-7\pm3\sqrt{23}}{2}\). Since length cannot be negative, \(x=\frac{-7 + 3\sqrt{23}}{2}\approx\frac{-7+3\times4.796}{2}=\frac{-7 + 14.388}{2}=3.694\). The sides are \(x-1\approx2.694\), \(x + 8\approx11.694\) and the hypotenuse is \(12\).