Answer :
Certainly! Let's go through the problem step-by-step.
### Step 1: Understanding the Problem
We are dealing with a right triangle. We know that one leg is 4 cm longer than the other leg. We also need the area of the triangle to be greater than or equal to 6 cm².
### Step 2: Define the Variables
Let [tex]\( x \)[/tex] represent the length of the shorter leg of the right triangle (in cm).
Then, the longer leg will be [tex]\( x + 4 \)[/tex] cm.
### Step 3: Area of the Triangle
The area [tex]\( A \)[/tex] of a right triangle is given by:
[tex]\[ A = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
In our case:
- Base = shorter leg = [tex]\( x \)[/tex] cm
- Height = longer leg = [tex]\( x + 4 \)[/tex] cm
So the area can be written as:
[tex]\[ A = \frac{1}{2} \times x \times (x + 4) \][/tex]
### Step 4: Set Up the Inequality
We want the area to be greater than or equal to 6 cm². Thus, we write the inequality:
[tex]\[ \frac{1}{2} x (x + 4) \geq 6 \][/tex]
### Step 5: Simplify the Inequality
First, multiply both sides of the inequality by 2 to eliminate the fraction:
[tex]\[ x (x + 4) \geq 12 \][/tex]
Expanding the left-hand side, we get:
[tex]\[ x^2 + 4x \geq 12 \][/tex]
### Step 6: Move All Terms to One Side
Let's rearrange the inequality to standard quadratic form:
[tex]\[ x^2 + 4x - 12 \geq 0 \][/tex]
### Step 7: Solve the Quadratic Inequality
To solve the quadratic inequality [tex]\( x^2 + 4x - 12 \geq 0 \)[/tex], we first solve the corresponding quadratic equation:
[tex]\[ x^2 + 4x - 12 = 0 \][/tex]
This quadratic equation can be solved using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 1, \quad b = 4, \quad c = -12 \][/tex]
Plug these values into the quadratic formula:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 48}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 8}{2} \][/tex]
Solving these, we get two solutions:
[tex]\[ x = \frac{-4 + 8}{2} = 2 \][/tex]
[tex]\[ x = \frac{-4 - 8}{2} = -6 \][/tex]
### Step 8: Analyze the Solutions
The solutions to the quadratic equation are [tex]\( x = 2 \)[/tex] and [tex]\( x = -6 \)[/tex]. To find the values of [tex]\( x \)[/tex] that satisfy the inequality [tex]\( x^2 + 4x - 12 \geq 0 \)[/tex], we can test intervals determined by these roots:
1. For [tex]\( x \geq 2 \)[/tex]
2. For [tex]\( x \leq -6 \)[/tex]
### Step 9: Final Answer
Thus, the length of the shorter leg [tex]\( x \)[/tex] must satisfy one of the following inequalities:
[tex]\[ x \geq 2 \][/tex]
or
[tex]\[ x \leq -6 \][/tex]
Since a leg length can't be negative, we discard [tex]\( x \leq -6 \)[/tex]. Therefore, to ensure that the area of the triangle is greater than or equal to 6 cm², the length of the shorter leg should be:
[tex]\[ x \geq 2 \][/tex]
So, the shorter leg should be at least 2 cm long.
### Step 1: Understanding the Problem
We are dealing with a right triangle. We know that one leg is 4 cm longer than the other leg. We also need the area of the triangle to be greater than or equal to 6 cm².
### Step 2: Define the Variables
Let [tex]\( x \)[/tex] represent the length of the shorter leg of the right triangle (in cm).
Then, the longer leg will be [tex]\( x + 4 \)[/tex] cm.
### Step 3: Area of the Triangle
The area [tex]\( A \)[/tex] of a right triangle is given by:
[tex]\[ A = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
In our case:
- Base = shorter leg = [tex]\( x \)[/tex] cm
- Height = longer leg = [tex]\( x + 4 \)[/tex] cm
So the area can be written as:
[tex]\[ A = \frac{1}{2} \times x \times (x + 4) \][/tex]
### Step 4: Set Up the Inequality
We want the area to be greater than or equal to 6 cm². Thus, we write the inequality:
[tex]\[ \frac{1}{2} x (x + 4) \geq 6 \][/tex]
### Step 5: Simplify the Inequality
First, multiply both sides of the inequality by 2 to eliminate the fraction:
[tex]\[ x (x + 4) \geq 12 \][/tex]
Expanding the left-hand side, we get:
[tex]\[ x^2 + 4x \geq 12 \][/tex]
### Step 6: Move All Terms to One Side
Let's rearrange the inequality to standard quadratic form:
[tex]\[ x^2 + 4x - 12 \geq 0 \][/tex]
### Step 7: Solve the Quadratic Inequality
To solve the quadratic inequality [tex]\( x^2 + 4x - 12 \geq 0 \)[/tex], we first solve the corresponding quadratic equation:
[tex]\[ x^2 + 4x - 12 = 0 \][/tex]
This quadratic equation can be solved using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 1, \quad b = 4, \quad c = -12 \][/tex]
Plug these values into the quadratic formula:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 48}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 8}{2} \][/tex]
Solving these, we get two solutions:
[tex]\[ x = \frac{-4 + 8}{2} = 2 \][/tex]
[tex]\[ x = \frac{-4 - 8}{2} = -6 \][/tex]
### Step 8: Analyze the Solutions
The solutions to the quadratic equation are [tex]\( x = 2 \)[/tex] and [tex]\( x = -6 \)[/tex]. To find the values of [tex]\( x \)[/tex] that satisfy the inequality [tex]\( x^2 + 4x - 12 \geq 0 \)[/tex], we can test intervals determined by these roots:
1. For [tex]\( x \geq 2 \)[/tex]
2. For [tex]\( x \leq -6 \)[/tex]
### Step 9: Final Answer
Thus, the length of the shorter leg [tex]\( x \)[/tex] must satisfy one of the following inequalities:
[tex]\[ x \geq 2 \][/tex]
or
[tex]\[ x \leq -6 \][/tex]
Since a leg length can't be negative, we discard [tex]\( x \leq -6 \)[/tex]. Therefore, to ensure that the area of the triangle is greater than or equal to 6 cm², the length of the shorter leg should be:
[tex]\[ x \geq 2 \][/tex]
So, the shorter leg should be at least 2 cm long.
