Question

Suppose a parallelogram is drawn on a coordinate plane. Using vertices with the coordinates (0,0), (2,3), (5,2), and (3,−1), compute the perimeter. Round your answer to the nearest tenth.(1 point)
The perimeter is approximately
units.

Answer 1

The only transformation that changes both a figure's orientation and position is a(1 point)
Responses

dilation.
dilation.

translation.
translation.

rotation.
rotation.

reflection.

Answer 2

Consider ΔABC , with vertices A(0,3) , B(0,1) , and C(−4,2) , and ΔDEF , with vertices D(0,6) , E(0,2) , and F(8,4) . Which transformations could be used to show that the triangles are similar?(1 point)
Responses

ΔABC is dilated by a scale factor of 12 and is reflected across the x-axis.
triangle upper A upper B upper C is dilated by a scale factor of Start Fraction 1 over 2 End Fraction and is reflected across the x -axis.

ΔABC is dilated by a scale factor of 2 and is reflected across the y-axis.
triangle upper A upper B upper C is dilated by a scale factor of 2 and is reflected across the y -axis.

ΔABC is dilated by a scale factor of −2 and is reflected across the y-axis.
triangle upper A upper B upper C is dilated by a scale factor of negative 2 and is reflected across the y -axis.

ΔABC is dilated by a scale factor of 14 and is reflected across the y-axis.

Answer 3

Jorge draws a rectangle measuring 3 ft. by 2 ft. Then, he dilates the rectangle so it is 7.5 ft. by 5 ft. If x represents the area of the original rectangle, which expression can he use to discover the area of the dilated rectangle?(1 point)
Responses

x(2.5)2
x left parenthesis 2.5 right parenthesis squared

x(1.5)2
x left parenthesis 1.5 right parenthesis squared

x+(2.5)2
x plus left parenthesis 2.5 right parenthesis squared

x(2.5)

Answer 4

A laundry soap company decides it wants to shrink the size of its box to save on cost. The surface area of the original box is 112 cm2 . If it creates a new box using a scale factor of 34 , determine the surface area of the new box. Round to the nearest tenth if necessary.(1 point)

Answer 5

A right triangle has a segment from the right angle that meets the hypotenuse at a right angle at point p. Point p splits the hypotenuse into two segments whose lengths are 16 units and 4 units. Find the distance from the perpendicular line to the right angle.(1 point)
units

Answer 6

To find the perimeter of the parallelogram defined by the vertices \((0,0)\), \((2,3)\), \((5,2)\), and \((3,-1)\), we will first compute the lengths of the sides using the distance formula.

### Distance Formula
The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

### Calculate the lengths of the sides:

1. Distance between \((0,0)\) and \((2,3)\):
\[
d_1 = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}
\]

2. Distance between \((2,3)\) and \((5,2)\):
\[
d_2 = \sqrt{(5-2)^2 + (2-3)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}
\]

3. Distance between \((5,2)\) and \((3,-1)\):
\[
d_3 = \sqrt{(3-5)^2 + (-1-2)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}
\]

4. Distance between \((3,-1)\) and \((0,0)\):
\[
d_4 = \sqrt{(0-3)^2 + (0+1)^2} = \sqrt{(-3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10}
\]

### Perimeter Calculation
Since \(d_1\) and \(d_3\) are equal (both \(\sqrt{13}\)), and \(d_2\) and \(d_4\) are equal (both \(\sqrt{10}\)), the perimeter \(P\) of the parallelogram is given by:
\[
P = 2(d_1 + d_2) = 2(\sqrt{13} + \sqrt{10})
\]

### Numerical Values
Now we evaluate these square roots:

\[
\sqrt{13} \approx 3.60555
\]
\[
\sqrt{10} \approx 3.16228
\]

Adding these values:
\[
\sqrt{13} + \sqrt{10} \approx 3.60555 + 3.16228 \approx 6.76783
\]

Now we can find the perimeter:
\[
P \approx 2 \times 6.76783 \approx 13.53566
\]

### Rounding
Rounding to the nearest tenth, the perimeter is approximately:
\[
\boxed{13.5} \text{ units.}
\]

Answer 7

The only transformation that changes both a figure's orientation and position is a **reflection**.

- **Dilation** changes the size of the figure but not its orientation.
- **Translation** changes the position of a figure but does not change its orientation.
- **Rotation** changes the orientation of the figure around a point but does not change its position.

Thus, the correct answer is **reflection**.

Answer 8

To determine which transformations can show that triangles \( \Delta ABC \) and \( \Delta DEF \) are similar, we need to analyze the vertices of both triangles and see how we can manipulate one to get the other.

### Vertices:
- \( \Delta ABC: A(0,3), B(0,1), C(-4,2) \)
- \( \Delta DEF: D(0,6), E(0,2), F(8,4) \)

### Step 1: Compare Side Lengths
First, let's calculate the lengths of the sides of both triangles.

#### Triangle \( \Delta ABC \):
1. \( AB \) (between points \( A(0,3) \) and \( B(0,1) \)):
\[
AB = \sqrt{(0-0)^2 + (3-1)^2} = \sqrt{0 + 4} = 2
\]

2. \( BC \) (between points \( B(0,1) \) and \( C(-4,2) \)):
\[
BC = \sqrt{(0 - (-4))^2 + (1-2)^2} = \sqrt{(4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}
\]

3. \( CA \) (between points \( C(-4,2) \) and \( A(0,3) \)):
\[
CA = \sqrt{(0 - (-4))^2 + (3-2)^2} = \sqrt{(4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17}
\]

#### Triangle \( \Delta DEF \):
1. \( DE \) (between points \( D(0,6) \) and \( E(0,2) \)):
\[
DE = \sqrt{(0-0)^2 + (6-2)^2} = \sqrt{0 + 16} = 4
\]

2. \( EF \) (between points \( E(0,2) \) and \( F(8,4) \)):
\[
EF = \sqrt{(0 - 8)^2 + (2-4)^2} = \sqrt{(-8)^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68}
\]

3. \( FD \) (between points \( F(8,4) \) and \( D(0,6) \)):
\[
FD = \sqrt{(0 - 8)^2 + (6-4)^2} = \sqrt{(-8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68}
\]

### Step 2: Compare Ratios of Side Lengths
Now we check the ratios of corresponding sides:

For \( \Delta ABC \):
- \( AB = 2 \)
- \( BC = \sqrt{17} \)
- \( CA = \sqrt{17} \)

For \( \Delta DEF \):
- \( DE = 4 \)
- \( EF = \sqrt{68} \)
- \( FD = \sqrt{68} \)

### Scale Factor Calculation
1. Compare \( AB \) and \( DE \):
\[
\text{Scale Factor} = \frac{ DE }{ AB } = \frac{ 4 }{ 2 } = 2
\]

2. Compare \( BC \) and \( EF \):
\[
\text{Scale Factor} = \frac{ EF }{ BC } = \frac{ \sqrt{68} }{ \sqrt{17} } = \sqrt{ \frac{68}{17} } = \sqrt{4} = 2
\]

3. Compare \( CA \) and \( FD \):
\[
\text{Scale Factor} = \frac{ FD }{ CA } = \frac{ \sqrt{68} }{ \sqrt{17} } = \sqrt{ \frac{68}{17} } = \sqrt{4} = 2
\]

### Conclusion
All sides are in proportion. Thus, the triangles are similar. A similarity transformation can be a dilation by a factor of 2, keeping orientation or a dilation with a reflection across the y-axis.

Considering the options given, the one that matches this understanding is:

**ΔABC is dilated by a scale factor of 2 and is reflected across the y-axis.**

Hence, the correct choice is:

**ΔABC is dilated by a scale factor of 2 and is reflected across the y-axis.**

Answer 9

To solve for the area of the dilated rectangle based on the area of the original rectangle and the scale factor of the dilation, we follow these steps:

1. **Calculate the area of the original rectangle**:
The area \( x \) of the original rectangle measuring 3 ft by 2 ft is:
\[
x = 3 \times 2 = 6 \text{ ft}^2
\]

2. **Determine the scale factor of the dilation**:
The dimensions of the dilated rectangle are 7.5 ft by 5 ft.
To find the scale factor, we can calculate:
- Width scale factor:
\(\frac{7.5 \text{ ft}}{3 \text{ ft}} = 2.5\)
- Height scale factor:
\(\frac{5 \text{ ft}}{2 \text{ ft}} = 2.5\)

Since both dimensions scale by the same factor, the scale factor for the dilation is 2.5.

3. **Calculate the area of the dilated rectangle**:
When a shape is dilated by a scale factor \( k \), the area scales by \( k^2 \). Therefore, the area of the dilated rectangle can be expressed as:
\[
\text{Area of dilated rectangle} = x \cdot (2.5)^2
\]

Thus, the correct expression that Jorge can use to discover the area of the dilated rectangle is:
\[
x(2.5)^2
\]

Therefore, the answer is:
**x(2.5)²**

Answer 10

To determine the surface area of the new box after applying a scale factor, we use the following relationship:

When the dimensions of a three-dimensional object are scaled by a factor of \(k\), the surface area is scaled by a factor of \(k^2\).

### Given:
- Surface area of the original box: \(112 \, \text{cm}^2\)
- Scale factor: \(k = \frac{3}{4} = 0.75\)

### Step 1: Calculate the Scale Factor for Surface Area
The scale factor for the surface area is the square of the linear scale factor:
\[
k^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}
\]

### Step 2: Calculate the Surface Area of the New Box
Now, we can find the surface area of the new box \(A_{new}\) by multiplying the original surface area by the scale factor for the surface area:
\[
A_{new} = A_{original} \cdot k^2 = 112 \, \text{cm}^2 \cdot \frac{9}{16}
\]

### Step 3: Perform the Calculation
Calculating this gives:
\[
A_{new} = 112 \cdot \frac{9}{16} = 112 \cdot 0.5625 = 63 \, \text{cm}^2
\]

### Final Answer
Thus, the surface area of the new box is:
\[
\boxed{63 \, \text{cm}^2}
\]

Answer 11

Let's denote the vertices of the right triangle as \(A\), \(B\), and \(C\), where \(C\) is the right angle. Let \(P\) be the point where the perpendicular from \(C\) meets the hypotenuse \(AB\).

Given that segment \(AP\) (the portion of the hypotenuse from \(A\) to \(P\)) is 16 units long and segment \(PB\) (the portion of the hypotenuse from \(P\) to \(B\)) is 4 units long, the entire length of the hypotenuse \(AB\) is:
\[
AB = AP + PB = 16 + 4 = 20 \text{ units.}
\]

### Step 1: Use Similar Triangles
The triangles \(ACP\) and \(BPC\) are similar to the original triangle \(ABC\) because they share angles with it, and they both have a right angle at \(C\).

Let:
- \(h\) be the distance from the right angle at \(C\) to the hypotenuse \(AB\).

### Step 2: Apply the relationships for the altitude to the hypotenuse
The length of the segment from point \(C\) to point \(P\) (the altitude \(h\)) can be found using the formula for the area of the triangle in two different ways.

The area of triangle \(ABC\) can be expressed in two ways:
1. Using the legs \(a\) and \(b\):
\[
\text{Area} = \frac{1}{2} \cdot a \cdot b
\]

2. Using the hypotenuse and the altitude to the hypotenuse:
\[
\text{Area} = \frac{1}{2} \cdot AB \cdot h
\]

### Step 3: Relate the segments to find h
There is also a relationship between the segments formed by the foot of the altitude on the hypotenuse and the altitude itself, given by:
\[
h^2 = AP \cdot PB
\]

Substituting the lengths of the segments:
\[
h^2 = 16 \cdot 4 = 64
\]
Taking the square root:
\[
h = \sqrt{64} = 8
\]

### Conclusion
Thus, the distance from the right angle to the hypotenuse is:
\[
\boxed{8} \text{ units.}
\]