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Sagot :
To determine the lengths of the diagonals of the rhombus, we need to work with the information given:
1. The area of the rhombus is [tex]\(10 \, \text{m}^2\)[/tex].
2. One diagonal is 3 meters less than twice the other diagonal.
Let's denote the lengths of the diagonals as [tex]\(d_1\)[/tex] and [tex]\(d_2\)[/tex].
### Step 1: Setting up the equations
The formula for the area ([tex]\(A\)[/tex]) of a rhombus in terms of its diagonals is:
[tex]\[ A = \frac{1}{2} \times d_1 \times d_2 \][/tex]
Given that the area is [tex]\(10 \, \text{m}^2\)[/tex], we can write:
[tex]\[ \frac{1}{2} \times d_1 \times d_2 = 10 \][/tex]
Multiplying both sides by 2 to eliminate the fraction:
[tex]\[ d_1 \times d_2 = 20 \quad \text{(Equation 1)} \][/tex]
Next, we know that one diagonal is 3 meters less than twice the other. Without loss of generality, assume:
[tex]\[ d_1 = 2 \times d_2 - 3 \quad \text{(Equation 2)} \][/tex]
### Step 2: Solving the system of equations
We have the equations:
[tex]\[ d_1 \times d_2 = 20 \][/tex]
[tex]\[ d_1 = 2 \times d_2 - 3 \][/tex]
Substitute [tex]\(d_1\)[/tex] from Equation 2 into Equation 1:
[tex]\[ (2 \times d_2 - 3) \times d_2 = 20 \][/tex]
Simplifying:
[tex]\[ 2d_2^2 - 3d_2 = 20 \][/tex]
Rearrange into a standard quadratic equation form:
[tex]\[ 2d_2^2 - 3d_2 - 20 = 0 \][/tex]
### Step 3: Solving the quadratic equation
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ d_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -20\)[/tex].
Plugging in the values:
[tex]\[ d_2 = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-20)}}{2 \cdot 2} \][/tex]
[tex]\[ d_2 = \frac{3 \pm \sqrt{9 + 160}}{4} \][/tex]
[tex]\[ d_2 = \frac{3 \pm \sqrt{169}}{4} \][/tex]
[tex]\[ d_2 = \frac{3 \pm 13}{4} \][/tex]
This gives us two potential solutions:
[tex]\[ d_2 = \frac{16}{4} = 4 \][/tex]
[tex]\[ d_2 = \frac{-10}{4} = -2.5 \][/tex]
Since diagonal lengths cannot be negative, we discard [tex]\(d_2 = -2.5\)[/tex]. Thus, [tex]\(d_2 = 4 \, \text{m}\)[/tex].
### Step 4: Finding [tex]\(d_1\)[/tex]
Using Equation 2:
[tex]\[ d_1 = 2 \times d_2 - 3 \][/tex]
[tex]\[ d_1 = 2 \times 4 - 3 \][/tex]
[tex]\[ d_1 = 8 - 3 \][/tex]
[tex]\[ d_1 = 5 \, \text{m} \][/tex]
So, the lengths of the diagonals are:
[tex]\[ d_1 = 5 \, \text{m} \][/tex]
[tex]\[ d_2 = 4 \, \text{m} \][/tex]
Thus, the correct answer is:
(4) 4 m and 5 m
1. The area of the rhombus is [tex]\(10 \, \text{m}^2\)[/tex].
2. One diagonal is 3 meters less than twice the other diagonal.
Let's denote the lengths of the diagonals as [tex]\(d_1\)[/tex] and [tex]\(d_2\)[/tex].
### Step 1: Setting up the equations
The formula for the area ([tex]\(A\)[/tex]) of a rhombus in terms of its diagonals is:
[tex]\[ A = \frac{1}{2} \times d_1 \times d_2 \][/tex]
Given that the area is [tex]\(10 \, \text{m}^2\)[/tex], we can write:
[tex]\[ \frac{1}{2} \times d_1 \times d_2 = 10 \][/tex]
Multiplying both sides by 2 to eliminate the fraction:
[tex]\[ d_1 \times d_2 = 20 \quad \text{(Equation 1)} \][/tex]
Next, we know that one diagonal is 3 meters less than twice the other. Without loss of generality, assume:
[tex]\[ d_1 = 2 \times d_2 - 3 \quad \text{(Equation 2)} \][/tex]
### Step 2: Solving the system of equations
We have the equations:
[tex]\[ d_1 \times d_2 = 20 \][/tex]
[tex]\[ d_1 = 2 \times d_2 - 3 \][/tex]
Substitute [tex]\(d_1\)[/tex] from Equation 2 into Equation 1:
[tex]\[ (2 \times d_2 - 3) \times d_2 = 20 \][/tex]
Simplifying:
[tex]\[ 2d_2^2 - 3d_2 = 20 \][/tex]
Rearrange into a standard quadratic equation form:
[tex]\[ 2d_2^2 - 3d_2 - 20 = 0 \][/tex]
### Step 3: Solving the quadratic equation
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ d_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -20\)[/tex].
Plugging in the values:
[tex]\[ d_2 = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-20)}}{2 \cdot 2} \][/tex]
[tex]\[ d_2 = \frac{3 \pm \sqrt{9 + 160}}{4} \][/tex]
[tex]\[ d_2 = \frac{3 \pm \sqrt{169}}{4} \][/tex]
[tex]\[ d_2 = \frac{3 \pm 13}{4} \][/tex]
This gives us two potential solutions:
[tex]\[ d_2 = \frac{16}{4} = 4 \][/tex]
[tex]\[ d_2 = \frac{-10}{4} = -2.5 \][/tex]
Since diagonal lengths cannot be negative, we discard [tex]\(d_2 = -2.5\)[/tex]. Thus, [tex]\(d_2 = 4 \, \text{m}\)[/tex].
### Step 4: Finding [tex]\(d_1\)[/tex]
Using Equation 2:
[tex]\[ d_1 = 2 \times d_2 - 3 \][/tex]
[tex]\[ d_1 = 2 \times 4 - 3 \][/tex]
[tex]\[ d_1 = 8 - 3 \][/tex]
[tex]\[ d_1 = 5 \, \text{m} \][/tex]
So, the lengths of the diagonals are:
[tex]\[ d_1 = 5 \, \text{m} \][/tex]
[tex]\[ d_2 = 4 \, \text{m} \][/tex]
Thus, the correct answer is:
(4) 4 m and 5 m
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