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Chapter 11: Problem 53

Write an equation and solve. One leg of a right triangle is 1 in. more than twice the other leg. The hypotenuse is \(\sqrt{29}\) in. long. Find the lengths of the legs.

Short Answer

Expert verified
The lengths of the legs of the right triangle are 2 inches and 5 inches.

Step by step solution

01

Define the variables

Let's denote the shorter leg as x and the longer leg as y. According to the problem, the longer leg is 1 inch more than twice the shorter leg. This can be written as the equation: \(y = 2x + 1\) The hypotenuse is given as \(\sqrt{29}\) inches.
02

Apply the Pythagorean theorem

According to the Pythagorean theorem, we have: \(x^2 + y^2 = (\sqrt{29})^2\) Since we know the relationship between the legs from Step 1, we can substitute the equation from Step 1 into the Pythagorean theorem: \(x^2 + (2x + 1)^2 = 29\)
03

Expand and simplify the equation

Expand and simplify the equation to solve for x: \(x^2 + (4x^2 + 4x + 1) = 29\) Combine the terms on the left side: \(5x^2 + 4x + 1 = 29\) Subtract 29 from both sides: \(5x^2 + 4x - 28 = 0\)
04

Solve the quadratic equation

To find the value of x, we need to solve the quadratic equation. Since this equation is not easily factorable, we can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Plugging the coefficients from the equation into the formula: \(x = \frac{-4 \pm \sqrt{(-4)^2 - 4(5)(-28)}}{2(5)}\) Simplify: \(x = \frac{-4 \pm \sqrt{16 + 560}}{10}\) \(x = \frac{-4 \pm \sqrt{576}}{10}\) \(x = \frac{-4 \pm 24}{10}\) We get two possible solutions for x: \(x = \frac{20}{10} = 2\) or \(x = \frac{-28}{10} = -2.8\) Since the length of a leg cannot be negative, the correct solution for x is 2 inches.
05

Find the length of the longer leg

Now that we have the length of the shorter leg, we can find the length of the longer leg using the equation from Step 1: \(y = 2x + 1\) \(y = 2(2) + 1\) \(y = 4 + 1\) \(y = 5\) So, the length of the longer leg is 5 inches. The lengths of the legs of the right triangle are 2 inches and 5 inches.

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