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Sagot :
To determine which statements are true, let's analyze the given conditions and equations step-by-step.
### Step 1: Determine equations for the rectangle
1. Let [tex]\( x \)[/tex] be the length of the rectangle.
2. The width of the rectangle, given that it is 3 inches less than its length, is [tex]\( x - 3 \)[/tex].
### Step 2: Calculate the area of the rectangle
1. The area of the rectangle is given by [tex]\( \text{length} \times \text{width} \)[/tex].
2. This can be written as [tex]\( x \times (x - 3) \)[/tex].
3. Each right triangle formed by cutting the rectangle along the diagonal has an area of 44 square inches, so the area of the entire rectangle is [tex]\( 2 \times 44 = 88 \)[/tex] square inches.
Thus, the first statement:
The area of the rectangle is 88 square inches.
is True.
### Step 3: Form the equation from the triangle definition
1. From the area of one of the right triangles, we have:
[tex]\[ \frac{1}{2} \times \text{base} \times \text{height} = 44 \][/tex]
2. Since the base and height are the dimensions of the rectangle, we get:
[tex]\[ x \times (x - 3) = 88 \][/tex]
3. Simplifying this equation, we get:
[tex]\[ x(x - 3) = 44 \][/tex]
Thus, the second statement:
The equation [tex]\( x(x-3)=44 \)[/tex] can be used to solve for the dimensions of the triangle.
is False.
### Step 4: Form the quadratic equation for the length
1. Using the area equation [tex]\( x(x - 3) = 88 \)[/tex], expand and simplify:
[tex]\[ x^2 - 3x = 88 \][/tex]
2. Rearrange it to form a quadratic equation:
[tex]\[ x^2 - 3x - 88 = 0 \][/tex]
Thus, the third statement:
The equation [tex]\( x^2-3x-88=0 \)[/tex] can be used to solve for the length of the rectangle.
is True.
### Step 5: Verify the triangle dimensions
1. Let's solve for [tex]\( x \)[/tex] using the quadratic equation:
[tex]\[ x^2 - 3x - 88 = 0 \][/tex]
2. The factors of -88 that add up to -3 are examined. Solving this, we get possible values of [tex]\( x \)[/tex]:
[tex]\[ x = 11 \quad \text{or} \quad x = -8 \quad \text{(ignore negative value as a dimension cannot be negative)} \][/tex]
3. If [tex]\( x = 11 \)[/tex], then the width [tex]\( x - 3 = 8 \)[/tex].
4. Verify the area of the triangle given the base and height are 11 and 8 respectively:
[tex]\[ \frac{1}{2} \times 11 \times 8 = 44 \][/tex]
Thus, the fourth statement:
The triangle has a base of 11 inches and a height of 8 inches.
is True.
### Step 6: Verify the width of the rectangle
1. We need to check if the width can be 4 inches:
2. From the quadratic equation, possible values of [tex]\( x \)[/tex] were 11 or -8 (ignore -8).
3. When [tex]\( x = 11 \)[/tex], the width is [tex]\( x - 3 = 11 - 3 = 8 \)[/tex], not 4.
Thus, the fifth statement:
The rectangle has a width of 4 inches.
is False.
### Conclusion:
The true statements are:
1. The area of the rectangle is 88 square inches.
2. The equation [tex]\( x^2 - 3x - 88 = 0 \)[/tex] can be used to solve for the length of the rectangle.
3. The triangle has a base of 11 inches and a height of 8 inches.
### Step 1: Determine equations for the rectangle
1. Let [tex]\( x \)[/tex] be the length of the rectangle.
2. The width of the rectangle, given that it is 3 inches less than its length, is [tex]\( x - 3 \)[/tex].
### Step 2: Calculate the area of the rectangle
1. The area of the rectangle is given by [tex]\( \text{length} \times \text{width} \)[/tex].
2. This can be written as [tex]\( x \times (x - 3) \)[/tex].
3. Each right triangle formed by cutting the rectangle along the diagonal has an area of 44 square inches, so the area of the entire rectangle is [tex]\( 2 \times 44 = 88 \)[/tex] square inches.
Thus, the first statement:
The area of the rectangle is 88 square inches.
is True.
### Step 3: Form the equation from the triangle definition
1. From the area of one of the right triangles, we have:
[tex]\[ \frac{1}{2} \times \text{base} \times \text{height} = 44 \][/tex]
2. Since the base and height are the dimensions of the rectangle, we get:
[tex]\[ x \times (x - 3) = 88 \][/tex]
3. Simplifying this equation, we get:
[tex]\[ x(x - 3) = 44 \][/tex]
Thus, the second statement:
The equation [tex]\( x(x-3)=44 \)[/tex] can be used to solve for the dimensions of the triangle.
is False.
### Step 4: Form the quadratic equation for the length
1. Using the area equation [tex]\( x(x - 3) = 88 \)[/tex], expand and simplify:
[tex]\[ x^2 - 3x = 88 \][/tex]
2. Rearrange it to form a quadratic equation:
[tex]\[ x^2 - 3x - 88 = 0 \][/tex]
Thus, the third statement:
The equation [tex]\( x^2-3x-88=0 \)[/tex] can be used to solve for the length of the rectangle.
is True.
### Step 5: Verify the triangle dimensions
1. Let's solve for [tex]\( x \)[/tex] using the quadratic equation:
[tex]\[ x^2 - 3x - 88 = 0 \][/tex]
2. The factors of -88 that add up to -3 are examined. Solving this, we get possible values of [tex]\( x \)[/tex]:
[tex]\[ x = 11 \quad \text{or} \quad x = -8 \quad \text{(ignore negative value as a dimension cannot be negative)} \][/tex]
3. If [tex]\( x = 11 \)[/tex], then the width [tex]\( x - 3 = 8 \)[/tex].
4. Verify the area of the triangle given the base and height are 11 and 8 respectively:
[tex]\[ \frac{1}{2} \times 11 \times 8 = 44 \][/tex]
Thus, the fourth statement:
The triangle has a base of 11 inches and a height of 8 inches.
is True.
### Step 6: Verify the width of the rectangle
1. We need to check if the width can be 4 inches:
2. From the quadratic equation, possible values of [tex]\( x \)[/tex] were 11 or -8 (ignore -8).
3. When [tex]\( x = 11 \)[/tex], the width is [tex]\( x - 3 = 11 - 3 = 8 \)[/tex], not 4.
Thus, the fifth statement:
The rectangle has a width of 4 inches.
is False.
### Conclusion:
The true statements are:
1. The area of the rectangle is 88 square inches.
2. The equation [tex]\( x^2 - 3x - 88 = 0 \)[/tex] can be used to solve for the length of the rectangle.
3. The triangle has a base of 11 inches and a height of 8 inches.
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