Question 16

A)
Use the image to answer the question.

A rectangle measures 9 units long and 5 units wide.

Lewis draws a rectangle with dimensions 9 units by 5 units, then dilates it by a scale factor of 3. How many times greater is the area of the dilated rectangle than the area of the original rectangle?

(1 point)
Responses

9 times greater
9 times greater

27 times greater
27 times greater

3 times greater
3 times greater

45 times greater
45 times greater
Question 17
A)A square has an area of 16 square units. Jessica dilates the square using a scale factor of 0.5. What is the area of the dilated square?(1 point)
Responses

2 square units
2 square units

32 square units
32 square units

4 square units
4 square units

8 square units
8 square units
Question 18
A)Calculate and compare the surface area of sphere A, which has a radius of 5 in., and sphere B, which has a radius of 10 in. The formula for the surface area of a sphere is 4πr2.(1 point)
Responses

Sphere A has a surface area of 5π in.2 and sphere B has a surface area of 10π in.2, meaning sphere B’s surface area is 2 times as large as sphere A’s.
Sphere upper A has a surface area of 5 pi in. squared and sphere upper B has a surface area of 10 pi in. squared , meaning sphere upper B ’s surface area is 2 times as large as sphere upper A ’s.

Sphere A has a surface area of 100π in.2 and sphere B has a surface area of 400π in.2, meaning sphere B’s surface area is 4 times as large as sphere A’s.
Sphere upper A has a surface area of 100 pi in. squared and sphere upper B has a surface area of 400 pi in. squared , meaning sphere upper B ’s surface area is 4 times as large as sphere upper A ’s.

Sphere A has a surface area of 25π in.2 and sphere B has a surface area of 100π in.2, meaning sphere B’s surface area is 4 times as large as sphere A’s.
Sphere upper A has a surface area of 25 pi in. squared and sphere upper B has a surface area of 100 pi in. squared , meaning sphere upper B ’s surface area is 4 times as large as sphere upper A ’s.

Sphere A has a surface area of 20π in.2 and sphere B has a surface area of 40π in.2, meaning sphere B’s surface area is 2 times as large as sphere A’s.
Sphere upper A has a surface area of 20 pi in. squared and sphere upper B has a surface area of 40 pi in. squared , meaning sphere upper B ’s surface area is 2 times as large as sphere upper A ’s.
Question 19
A)
Use the image to answer the question.

A rectangular prism is shown with length 4.4 meters, height 10.2 meters, and width 3.6 meters.

If this rectangular prism is dilated by a scale factor of 12
, what would the surface area of the new figure be?

(1 point)
Responses

779.52 cm2
1,536 cm squared

48.72 cm2
48 point 7 2 cm squared

389.76 cm2
389 point 7 6 cm squared

97.44 cm2
97 point 4 4 cm squared
Question 20
A)The dimensions of a rectangular shipping crate are 2.5 ft., 2.5 ft., and 4 ft. If there were a similar shipping crate with a scale factor of 3, what would be the volume of the new shipping crate?(1 point)
Responses

225 ft.3
225 ft. cubed

25 ft.3
25 ft. cubed

675 ft.3
675 ft. cubed

75 ft.3
75 ft. cubed
Question 21
A)
Use the image to answer the question.

A square pyramid measures 8 inches in width, 8 inches in length, and 6 inches in height.

If the square-base pyramid were dilated using a scale factor of 4, what would be the new volume? The vertical height of the pyramid is 6 inches.

(1 point)
Responses

8,192 inches3
8,192 inches cubed

2,048 inches3
2,048 inches cubed

32 inches3
32 inches cubed

512 inches3
512 inches cubed
Question 22
A)
Use the image to answer the question.

On right triangle upper U upper V upper T, the altitude of vertex upper U is drawn as a solid line that extends from upper U to the hypotenuse, upper T upper V. The intersection of the altitude and upper T upper V is labeled upper O.

Given is the right triangle TUV with altitude UO. Apply the Pythagorean Theorem and similar triangles to find the length of OV¯¯¯¯¯¯¯¯.

(1 point)
Responses

15
15

16
16

15−−√
Start Root 15 End Root

415−−√
4 Start Root 15 End Root
Question 23
A)
Use the image to answer the question.

A line, labeled x, divides a right triangle from the vertex opposite the hypotenuse to the hypotenuse. The line forms a second right angle, marked by a small square, where it meets the hypotenuse, and it divides the hypotenuse into 32 inches and 46 inches.

Rico is building a corner shelf in the shape of a right triangle for his bedroom. The image displays his dimensions. Use the geometric mean to find the distance from the corner of the wall to the end of the shelf that is perpendicular to the hypotenuse of the shelf.

(1 point)
Responses

14 inches
14 inches

38.4 inches
38.4 inches

78 inches

1 answer

Let's go through the questions one by one:

Question 16

Original Rectangle Area

  • Length = 9 units
  • Width = 5 units
  • Area = Length × Width = 9 × 5 = 45 square units

Dilated Rectangle Area

  • Scale factor = 3
  • New Length = 9 × 3 = 27 units
  • New Width = 5 × 3 = 15 units
  • New Area = 27 × 15 = 405 square units

Comparison of Areas

  • Ratio of areas = 405 / 45 = 9

Response: 9 times greater


Question 17

Original Square Area

  • Area = 16 square units

Dilated Square Area

  • Scale factor = 0.5
  • New area (scaled) = (0.5)^2 × Original area = 0.25 × 16 = 4 square units

Response: 4 square units


Question 18

Surface Area of Sphere A

  • Radius = 5 in
  • Surface Area = 4πr² = 4π(5)² = 4π(25) = 100π in²

Surface Area of Sphere B

  • Radius = 10 in
  • Surface Area = 4π(10)² = 4π(100) = 400π in²

Comparison of Surface Areas

  • Ratio of surface areas = 400π / 100π = 4

Response: Sphere A has a surface area of 100π in² and sphere B has a surface area of 400π in², meaning sphere B’s surface area is 4 times as large as sphere A’s.


Question 19

Original Surface Area

The formula for the surface area of a rectangular prism is: \[ \text{Surface Area} = 2lw + 2lh + 2wh \]

Given:

  • Length (l) = 4.4 m
  • Height (h) = 10.2 m
  • Width (w) = 3.6 m

Substituting values:

  1. 2(4.4)(10.2) + 2(4.4)(3.6) + 2(10.2)(3.6)
  2. 2(44.88) + 2(15.84) + 2(36.72)
  3. 89.76 + 31.68 + 73.44 = 195.88 m²

Dilated Surface Area

  • Scale factor = 12
  • New surface area = Original surface area × (scale factor)²
  • New surface area = 195.88 × 144 = 28127.52 cm²

Response: 1,536 cm² does not match my calculations. Please double-check dimensions and units as they might have been mischaracterized.


Question 20

Original Volume

  • Volume of a rectangular prism = Length × Width × Height = 2.5 ft × 2.5 ft × 4 ft = 25 ft³

Dilated Volume

  • Scale factor = 3
  • New Volume = (scale factor)³ × Original Volume = 3³ × 25 = 27 × 25 = 675 ft³

Response: 675 ft³


Question 21

Original Pyramid Volume

  • Volume = (1/3) × base area × height
  • Base Area = length × width = 8 in × 8 in = 64 in²
  • Volume = (1/3) × 64 × 6 = 128 in³

Dilated Volume

  • Scale factor = 4
  • New Volume = (scale factor)³ × Original Volume = 4³ × 128 = 64 × 128 = 8192 in³

Response: 8,192 inches³


Question 22

This question requires an image to accurately apply the Pythagorean theorem and similar triangles to find OV. Please provide the specifics from the scenario to continue with this.


Question 23

To find the distance perpendicular to the hypotenuse, we can use the geometric mean: Let one segment of the hypotenuse be \( a = 32 \) inches and the other segment be \( b = 46 \) inches.

Using the geometric mean: \[ h = \sqrt{a \cdot b} \]

Calculating: \[ h = \sqrt{32 \times 46} \approx 38.4 \text{ inches} \]

Response: 38.4 inches


If you need any further calculations or assistance, feel free to ask!

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