Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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.
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
