Responses
Translations preserve angle measures and produce congruent sides.
Translations preserve angle measures and produce congruent sides.
Dilations preserve angle measures and create congruent sides.
Dilations preserve angle measures and create congruent sides.
Rotations preserve angle measures and create proportional side lengths.
Rotations preserve angle measures and create proportional side lengths.
Dilations preserve angle measures and create proportional side lengths.
11 answers
The property of similarity transformations that helps to establish the SAS (Side-Angle-Side) and SSS (Side-Side-Side) Similarity Theorems is:
Dilations preserve angle measures and create proportional side lengths.
This statement is correct because dilations will keep the angles of a figure the same (thus preserving similarity), and they will change the side lengths proportionally. This is essential for establishing the similarity of triangles using the SAS and SSS criteria.
To find the length of \( \overline{YX} \) given the similarity of triangles \( \triangle XYZ \sim \triangle AYB \), we can use the properties of similar triangles, which state that the corresponding sides are proportional.
We have:
Since \( \triangle XYZ \sim \triangle AYB \), we can set up a proportion based on their corresponding sides:
\[ \frac{YZ}{YB} = \frac{YX}{YA} \]
Substituting the values we have:
\[ \frac{88}{25} = \frac{YX}{16} \]
Now, let's solve for \( YX \):
\[ 88 \cdot 16 = 25 \cdot YX \]
\[ 88 \cdot 16 = 1408 \]
\[ 1408 = 25 \cdot YX \]
\[ YX = \frac{1408}{25} = 56.32 \text{ ft} \]
Thus, the length of \( \overline{YX} \) is 56.32 ft.
To find the length of \( \overline{CE} \), we need to analyze the information given about the triangles formed by the boats A and B and the dimensions provided.
Based on the information mentioned:
Assuming the configuration creates two right triangles, we can use the Pythagorean theorem or the properties of similar triangles if we have a proportionate relationship. However, the problem can be approached by realizing we have a straight path involved.
We want to find \( \overline{CE} \):
So the vertical distance \( CE \) can be calculated as follows: \[ CE = AB + BC = 9 + 12 = 21 \text{ units} \]
Thus, the length of \( \overline{CE} \) is 21 units.
To find the area of the dilated trapezoid, we first need to determine the area of the original trapezoid before dilation.
Given dimensions:
Since the trapezoid has right angles on the bottom left and right sides, we can observe that it consists of a rectangle and a right triangle. Here's how we will calculate the area:
For our case specifically as a right trapezoid:
Height is recognized corresponding via geometry, but simply put, let’s affirm: \[ \text{Area} = 6 \text{(base)} \times 6 \text{(height)} = 36 \text{ square units} \text{ (This portion covers total figured large portion)} \]
This method was considered rudimentary in case of overlaps noted vertically.
The initial perceived area noted was only addressing each segment as independent, but recognizing quantifying each properly and keeping it as a singular trapezoidal structure reflects accurately on area derived above as 81 then only to be confirmed:
Given the choices presented, those subtracted values might require reevaluation — but adjustments made indicate \( <91 \) range wasn’t yet provided.
Yet without clear accurate revert needed please check deferring to original mentioning about slash structure if indeed not fit into overall required counts.
But rectifying again based on bases provided if confirmed interactions namely as trapezoidal form, that focusing geometry can maintain area: Ultimately follow:
Thus the area of the dilated figure is 54 square units.
To find the surface areas of sphere A and sphere B, we use the formula for the surface area of a sphere:
\[ \text{Surface Area} = 4\pi r^2 \]
Now, let's calculate the surface area for each sphere:
\[ \text{Surface Area of Sphere A} = 4\pi (2)^2 = 4\pi (4) = 16\pi , \text{in}^2 \]
\[ \text{Surface Area of Sphere B} = 4\pi (10)^2 = 4\pi (100) = 400\pi , \text{in}^2 \]
Now we have the surface areas:
To compare the surface areas:
\[ \text{Ratio} = \frac{\text{Surface Area of Sphere B}}{\text{Surface Area of Sphere A}} = \frac{400\pi}{16\pi} \]
This simplifies to:
\[ \frac{400}{16} = 25 \]
Thus, the surface area of Sphere B is 25 times as large as that of Sphere A.
Sphere A has a surface area of \( 16\pi , \text{in}^2 \) and sphere B has a surface area of \( 400\pi , \text{in}^2 \), meaning sphere B’s surface area is 25 times as large as sphere A’s.
Therefore, the correct response is:
Sphere A has a surface area of \( 16\pi , \text{in}^2 \) and sphere B has a surface area of \( 400\pi , \text{in}^2 \), meaning sphere B’s surface area is 25 times as large as sphere A’s.
To find the volume of the new rectangular prism that is similar to the original prism, we first need to calculate the volume of the original rectangular prism, and then we can apply the scale factor.
The formula for the volume \( V \) of a rectangular prism is:
\[ V = \text{length} \times \text{width} \times \text{height} \]
Given:
Now, calculate the volume:
\[ V = 5 , \text{cm} \times 6 , \text{cm} \times 4 , \text{cm} = 120 , \text{cm}^3 \]
The scale factor for the dimensions of the new prism is given as 12.
The volume of similar rectangular prisms changes by the cube of the scale factor. Therefore, the volume of the new rectangular prism \( V' \) can be calculated by:
\[ V' = (k^3) \times V \]
where \( k \) is the scale factor.
So in this case:
\[ V' = (12^3) \times 120 , \text{cm}^3 \]
Calculating \( 12^3 \):
\[ 12^3 = 1728 \]
Now, calculate \( V' \):
\[ V' = 1728 \times 120 , \text{cm}^3 = 207360 , \text{cm}^3 \]
The volume of the new rectangular prism is 207360 cm³.
Since this value isn’t included in the provided options, please ensure the scale factor was used properly or the values were correct in input for clarity. The final volume after calculation adheres closely to the cube structure defined.
