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 equations

Expand the squared terms in the Pythagorean equations. For example, $(x + 1)^{2}=x^{2}+2x + 1$.

Step3: Simplify equations

Combine like - terms to get quadratic equations in standard form $ax^{2}+bx + c = 0$.

Step4: Solve quadratic equations

Use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or simple algebraic manipulation to solve for $x$. Reject negative values of $x$ as side - lengths of a triangle cannot be negative.

Step5: Find side - lengths

Substitute the valid value of $x$ back into the expressions for the sides of the triangle to find their lengths.

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 through 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 $a = 1$, $b = 1$, $c=-24$. So $x=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-24)}}{2\times1}=\frac{-1\pm\sqrt{1 + 96}}{2}=\frac{-1\pm\sqrt{97}}{2}$. Since side - length 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 $7$.

1. For the second triangle with sides $x$ and $20$ and hypotenuse $x + 4$:

• Apply the Pythagorean theorem: $x^{2}+20^{2}=(x + 4)^{2}$. Expand to get $x^{2}+400=x^{2}+8x + 16$. Subtract $x^{2}$ from both sides: $400=8x + 16$. Subtract $16$ from both sides: $384=8x$. Solve for $x$: $x = 48$. The sides are $48$, $20$ and $x + 4=52$.

1. For the third triangle with sides $x$ and $10$ and hypotenuse $x + 6$:

• By the Pythagorean theorem, $x^{2}+10^{2}=(x + 6)^{2}$. Expand: $x^{2}+100=x^{2}+12x+36$. Subtract $x^{2}$ from both sides: $100=12x + 36$. Subtract $36$ from both sides: $64=12x$. Solve for $x$: $x=\frac{16}{3}\approx5.33$. The sides are $x\approx5.33$, $10$ and $x + 6\approx11.33$.

1. For the fourth triangle with sides $x - 1$ and $x + 8$ and hypotenuse $12$:

• Apply the Pythagorean theorem: $(x - 1)^{2}+(x + 8)^{2}=12^{2}$. Expand: $x^{2}-2x + 1+x^{2}+16x + 64 = 144$. Combine like - terms: $2x^{2}+14x+65 - 144=0$, so $2x^{2}+14x - 79 = 0$. Using the quadratic formula with $a = 2$, $b = 14$, $c=-79$, $x=\frac{-14\pm\sqrt{14^{2}-4\times2\times(-79)}}{2\times2}=\frac{-14\pm\sqrt{196 + 632}}{4}=\frac{-14\pm\sqrt{828}}{4}=\frac{-14\pm6\sqrt{23}}{4}=\frac{-7\pm3\sqrt{23}}{2}$. Since side - length cannot be negative, we take the positive root. $x=\frac{-7 + 3\sqrt{23}}{2}\approx\frac{-7+3\times4.796}{2}=\frac{-7 + 14.388}{2}=3.694$. The sides are $x - 1\approx2.694$, $x + 8\approx11.694$ and $12$.