To find the lengths of the legs of the right triangle, we can use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
Given the conditions:
1. The hypotenuse measures 16 cm.
2. One leg (let's call it [tex]\( b \)[/tex]) is 1 cm more than twice the length of the other leg (let's call it [tex]\( a \)[/tex]).
We can set up the relationship as follows:
[tex]\[ b = 2a + 1 \][/tex]
Using the Pythagorean theorem:
[tex]\[ a^2 + b^2 = 16^2 \][/tex]
Substitute [tex]\( b \)[/tex] in the equation:
[tex]\[ a^2 + (2a + 1)^2 = 16^2 \][/tex]
Let's simplify and solve this equation step-by-step:
First, expand [tex]\( (2a + 1)^2 \)[/tex]:
[tex]\[ (2a + 1)^2 = 4a^2 + 4a + 1 \][/tex]
Substitute back into the equation:
[tex]\[ a^2 + 4a^2 + 4a + 1 = 256 \][/tex]
[tex]\[ 5a^2 + 4a + 1 = 256 \][/tex]
Bring 256 to the left side to set the equation to 0:
[tex]\[ 5a^2 + 4a + 1 - 256 = 0 \][/tex]
[tex]\[ 5a^2 + 4a - 255 = 0 \][/tex]
To solve the quadratic equation:
[tex]\[ 5a^2 + 4a - 255 = 0 \][/tex]
Use the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -255 \)[/tex].
Solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 5 \cdot (-255)}}{2 \cdot 5} \][/tex]
[tex]\[ a = \frac{-4 \pm \sqrt{16 + 5100}}{10} \][/tex]
[tex]\[ a = \frac{-4 \pm \sqrt{5116}}{10} \][/tex]
We calculate the discriminant:
[tex]\[ \sqrt{5116} \approx 71.52 \][/tex]
Thus, solving for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{-4 + 71.52}{10} \][/tex]
[tex]\[ a \approx 6.752 \][/tex]
Since we only need a positive value:
[tex]\[ a \approx 6.8 \][/tex]
Finally, substitute [tex]\( a \)[/tex] back to find [tex]\( b \)[/tex]:
[tex]\[ b = 2a + 1 \][/tex]
[tex]\[ b = 2 \cdot 6.8 + 1 \][/tex]
[tex]\[ b \approx 13.6 + 1 \][/tex]
[tex]\[ b \approx 14.5 \][/tex]
Therefore, the lengths of the legs of the triangle are:
- The shortest leg is [tex]\( \boxed{6.8} \)[/tex] cm.
