Answer :
Let's denote the length of the longer leg as x cm.
From the problem statement, we are told:
1. The shorter leg is 5 cm shorter than the longer leg. Hence, the length of the shorter leg will be x - 5 cm.
2. The hypotenuse is 5 cm longer than the longer leg. Hence, the length of the hypotenuse will be x + 5 cm.
Given that this is a right triangle, we can use the Pythagorean theorem to relate the sides. The Pythagorean theorem states that, in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This can be written as:
[tex]\[ \text{Hypotenuse}^2 = \text{Longer leg}^2 + \text{Shorter leg}^2 \][/tex]
Plugging in the values from our problem, we get:
[tex]\[ (x + 5)^2 = x^2 + (x - 5)^2 \][/tex]
Now, let's solve this equation step by step:
1. First, we expand both the squared terms on both sides:
[tex]\[ (x + 5)^2 = x^2 + 2x5 + 5^2 \][/tex]
[tex]\[ (x - 5)^2 = x^2 - 2x5 + 5^2 \][/tex]
2. After simplifying, we get:
[tex]\[ x^2 + 10x + 25 = x^2 + (x^2 - 10x + 25) \][/tex]
3. Combine like terms:
[tex]\[ x^2 + 10x + 25 = 2x^2 - 10x + 25 \][/tex]
4. To solve for x, we'll subtract x^2 from both sides of the equation:
[tex]\[ 10x + 25 = x^2 - 10x + 25 \][/tex]
5. Next, we'll move all terms involving x to one side of the equation:
[tex]\[ 10x + 10x = x^2 - 25 \][/tex]
[tex]\[ 20x = x^2 - 25 \][/tex]
6. We now have a quadratic equation. Let's move all terms to one side for solving it:
[tex]\[ x^2 - 20x - 25 = 0 \][/tex]
7. To solve this quadratic equation, we can use different methods such as factoring, completing the square, or applying the quadratic formula. Factoring this quadratic doesn't yield integer solutions, so we'll use the quadratic formula given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\(a = 1\)[/tex], [tex]\(b = -20\)[/tex], and [tex]\(c = -25\)[/tex].
8. Plugging in the values for a, b, and c:
[tex]\[ x = \frac{-(-20) \pm \sqrt{(-20)^2 - 41(-25)}}{2*1} \][/tex]
[tex]\[ x = \frac{20 \pm \sqrt{400 + 100}}{2} \][/tex]
[tex]\[ x = \frac{20 \pm \sqrt{500}}{2} \][/tex]
[tex]\[ x = \frac{20 \pm 10\sqrt{5}}{2} \][/tex]
[tex]\[ x = 10 \pm 5\sqrt{5} \][/tex]
We discard the negative solution since the length of a side cannot be negative. Thus, we have:
[tex]\[ x = 10 + 5\sqrt{5} \][/tex]
Now we have the length of the longer leg. We can find the shorter leg and the hypotenuse by subtracting and adding 5 cm, respectively:
- Shorter leg: [tex]\( x - 5 = (10 + 5\sqrt{5}) - 5 = 5 + 5\sqrt{5} \)[/tex] cm
- Hypotenuse: [tex]\( x + 5 = (10 + 5\sqrt{5}) + 5 = 15 + 5\sqrt{5} \)[/tex] cm
Thus, the side lengths of the triangle are:
- Longer leg: [tex]\( 10 + 5\sqrt{5} \)[/tex] cm
- Shorter leg: [tex]\( 5 + 5\sqrt{5} \)[/tex] cm
- Hypotenuse: [tex]\( 15 + 5\sqrt{5} \)[/tex] cm
From the problem statement, we are told:
1. The shorter leg is 5 cm shorter than the longer leg. Hence, the length of the shorter leg will be x - 5 cm.
2. The hypotenuse is 5 cm longer than the longer leg. Hence, the length of the hypotenuse will be x + 5 cm.
Given that this is a right triangle, we can use the Pythagorean theorem to relate the sides. The Pythagorean theorem states that, in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This can be written as:
[tex]\[ \text{Hypotenuse}^2 = \text{Longer leg}^2 + \text{Shorter leg}^2 \][/tex]
Plugging in the values from our problem, we get:
[tex]\[ (x + 5)^2 = x^2 + (x - 5)^2 \][/tex]
Now, let's solve this equation step by step:
1. First, we expand both the squared terms on both sides:
[tex]\[ (x + 5)^2 = x^2 + 2x5 + 5^2 \][/tex]
[tex]\[ (x - 5)^2 = x^2 - 2x5 + 5^2 \][/tex]
2. After simplifying, we get:
[tex]\[ x^2 + 10x + 25 = x^2 + (x^2 - 10x + 25) \][/tex]
3. Combine like terms:
[tex]\[ x^2 + 10x + 25 = 2x^2 - 10x + 25 \][/tex]
4. To solve for x, we'll subtract x^2 from both sides of the equation:
[tex]\[ 10x + 25 = x^2 - 10x + 25 \][/tex]
5. Next, we'll move all terms involving x to one side of the equation:
[tex]\[ 10x + 10x = x^2 - 25 \][/tex]
[tex]\[ 20x = x^2 - 25 \][/tex]
6. We now have a quadratic equation. Let's move all terms to one side for solving it:
[tex]\[ x^2 - 20x - 25 = 0 \][/tex]
7. To solve this quadratic equation, we can use different methods such as factoring, completing the square, or applying the quadratic formula. Factoring this quadratic doesn't yield integer solutions, so we'll use the quadratic formula given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\(a = 1\)[/tex], [tex]\(b = -20\)[/tex], and [tex]\(c = -25\)[/tex].
8. Plugging in the values for a, b, and c:
[tex]\[ x = \frac{-(-20) \pm \sqrt{(-20)^2 - 41(-25)}}{2*1} \][/tex]
[tex]\[ x = \frac{20 \pm \sqrt{400 + 100}}{2} \][/tex]
[tex]\[ x = \frac{20 \pm \sqrt{500}}{2} \][/tex]
[tex]\[ x = \frac{20 \pm 10\sqrt{5}}{2} \][/tex]
[tex]\[ x = 10 \pm 5\sqrt{5} \][/tex]
We discard the negative solution since the length of a side cannot be negative. Thus, we have:
[tex]\[ x = 10 + 5\sqrt{5} \][/tex]
Now we have the length of the longer leg. We can find the shorter leg and the hypotenuse by subtracting and adding 5 cm, respectively:
- Shorter leg: [tex]\( x - 5 = (10 + 5\sqrt{5}) - 5 = 5 + 5\sqrt{5} \)[/tex] cm
- Hypotenuse: [tex]\( x + 5 = (10 + 5\sqrt{5}) + 5 = 15 + 5\sqrt{5} \)[/tex] cm
Thus, the side lengths of the triangle are:
- Longer leg: [tex]\( 10 + 5\sqrt{5} \)[/tex] cm
- Shorter leg: [tex]\( 5 + 5\sqrt{5} \)[/tex] cm
- Hypotenuse: [tex]\( 15 + 5\sqrt{5} \)[/tex] cm
