uestion 1

Question

A)
Use the image to answer the question.

Triangle upper J upper K upper L is divided into triangle upper P upper K upper Q and quadrilateral upper J upper P upper Q upper L by vertical line upper P upper Q. Point upper P is on line upper J upper K. Point upper Q is on diagonal upper L upper K.

△JKL∼△PKQ, and KP¯¯¯¯¯¯¯¯=14 m, JK¯¯¯¯¯¯¯¯=56 m, and KQ¯¯¯¯¯¯¯¯=21 m. What is the length of KL¯¯¯¯¯¯¯¯?

(1 point)
Responses

84 m
84 m

5.25 m
5.25 m

72 m
72 m

34 m
34 m
Question 2
A)
Use the image to answer the question.

Triangle upper J upper K upper L is divided into triangle upper P upper K upper Q and quadrilateral upper J upper P upper Q upper L by vertical line upper P upper Q. Point upper P is on line upper J upper K. Point upper Q is on diagonal upper L upper K.

The two triangles are right triangles. If △JKL∼△PKQ, and PQ¯¯¯¯¯¯¯¯=7, KQ¯¯¯¯¯¯¯¯=25 km, and QL¯¯¯¯¯¯¯=175 km, then what is the measure of JK¯¯¯¯¯¯¯¯?

(1 point)
Responses

189 km
189 km

197 km
197 km

200 km
200 km

192 km
192 km
Question 3
A)
Use the image to answer the question.

Triangle upper Y upper X upper Z is divided into triangle upper Y upper B upper A and quadrilateral upper A upper B upper Z upper X by horizontal line upper A upper B.

If △XYZ∼△AYB, and XY¯¯¯¯¯¯¯¯=24 yards, YZ¯¯¯¯¯¯¯=30 yards, and YA¯¯¯¯¯¯¯¯=8 yards, then what is the length of YB¯¯¯¯¯¯¯¯?

(1 point)
Responses

10 yards
10 yards

25 yards
25 yards

15 yards
15 yards

12 yards
12 yards
Question 4
A)Myk is 5 feet and 2 inches tall, and their shadow is 84 inches tall. The flagpole next to them has a shadow that is 31 feet long. Myk uses similar triangles to find the height of the flagpole. Can you calculate the height of the flagpole to the nearest foot?(1 point)
Responses

23 feet
23 feet

374 feet
374 feet

23 inches
23 inches

17 feet
17 feet
Question 5
A)If △ABC∼△LMN and AB¯¯¯¯¯¯¯¯=5, AC¯¯¯¯¯¯¯¯=4, BC¯¯¯¯¯¯¯¯=3, and LM¯¯¯¯¯¯¯¯¯=7.5, then how long is LN¯¯¯¯¯¯¯¯?(1 point)
Responses

6
6

7
7

4.5

Answer 1

Let's go through each question one by one:

Question 1

Given:

• Similar triangles \( \triangle JKL \sim \triangle PKQ \)
• \( KP = 14 \) m
• \( JK = 56 \) m
• \( KQ = 21 \) m

Using the property of similar triangles: \[ \frac{JK}{KP} = \frac{KL}{KQ} \] Substituting the known values: \[ \frac{56}{14} = \frac{KL}{21} \] \[ 4 = \frac{KL}{21} \] Now, cross-multiply: \[ KL = 4 \times 21 = 84 \text{ m} \] Answer: 84 m

Question 2

Given:

• Similar triangles \( \triangle JKL \sim \triangle PKQ \)
• \( PQ = 7 \)
• \( KQ = 25 \) km
• \( QL = 175 \) km

Using the properties of similar triangles: \[ \frac{JK}{PQ} = \frac{KL}{KQ} \] First, we need to calculate \( KL \): \[ \frac{JK}{7} = \frac{KL}{25} \] Using the triangle similarity in \( QL \): \[ \frac{JK}{PQ} = \frac{JL}{QL} \] Thus: \[ \frac{JK}{7} = \frac{JK + KL}{175} \] To find \( JK \): Assuming \( JK + KL = JL \).

Furthermore, we know: Let's assume \( JK = x \), then \( KL = (x/7) \times 25 \): \[ x + \frac{25x}{7} = 175 \] Multiplying through by 7: \[ 7x + 25x = 1225 \] \[ 32x = 1225 \] \[ x = \frac{1225}{32} \] This calculation gives us approximately \( JK = 192.5 \approx 192 \) km. Answer: 192 km

Question 3

Given:

• \( XY = 24 \) yards
• \( YZ = 30 \) yards
• \( YA = 8 \) yards

Using the property of similar triangles: \[ \frac{XY}{YA} = \frac{YZ}{YB} \] This means: \[ \frac{24}{8} = \frac{30}{YB} \] So: \[ 3 = \frac{30}{YB} \] Cross-multiplying gives: \[ 3YB = 30 \implies YB = 10 \text{ yards} \] Answer: 10 yards

Question 4

Myk's height: \(5 \text{ ft} 2 \text{ in} = 62 \text{ in} \) (since 5 ft = 60 in and 2 in = 2 in). Myk's shadow: \(84 \text{ in}\) Flagpole shadow: \(31 \text{ ft} = 372 \text{ in}\)

Using the triangles' similarities: \[ \frac{\text{height of flagpole}}{372} = \frac{62}{84} \] Cross-multiplying: \[ \text{height of flagpole} = 372 \times \frac{62}{84} \approx 284 \text{ in} \approx 23.5 \text{ ft} \] Rounding gives approximately \(23\) ft. Answer: 23 feet

Question 5

Given:

• Triangles \( \triangle ABC \sim \triangle LMN \)
• \( AB = 5 \)
• \( AC = 4 \)
• \( BC = 3 \)
• \( LM = 7.5 \)

Using the property of similar triangles: To find \( LN \): \[ \frac{AB}{LM} = \frac{AC}{LN} \] Setting proportions: \[ \frac{5}{7.5} = \frac{4}{LN} \] Cross-multiplying gives: \[ 5 \cdot LN = 4 \cdot 7.5 \implies LN = \frac{30}{5} = 6 \] Answer: 6

For your review:

1. 84 m
2. 192 km
3. 10 yards
4. 23 feet
5. 6