A)Correctly label the sides of the triangle shown below(3 points)

Question

Drop target 1
Drop target 2
Drop target 3

answers:
Hypotenuse
leg
B)For the triangle shown above, RP¯¯¯¯¯¯¯¯ =24, and DR¯¯¯¯¯¯¯¯=7
, which of the following the correct way to set up the problem to find DP?(2 points)
Responses

24+7=x
24 plus 7 is equal to x

242+72=x2
24 squared plus 7 squared is equal to x squared

x2+72=242
x squared plus 7 squared is equal to 24 squared

x2+242=72
x squared plus 24 squared is equal to 7 squared
C)What is DP? Round your answer to the nearest tenth if necessary(2 points)
DP =
Question 2
A)For the triangle shown below, correctly label the sides(3 points)
Drop target 1
Drop target 2
Drop target 3
answers
hypotenuse
leg
B)If CE = 29 and AC = 20, which of the following equations can be used to find AE?(2 points)
Responses

202+x2=292
20 squared plus x squared is equal to 29 squared

20+x=29
20 plus x is equal to 29

202+292=x2
20 squared plus 29 squared is equal to x squared

20+29=x
20 plus 29 is equal to x
C)Using the information given above, find AE. Round to the nearest tenth if necessary.(2 points)
AE =
Question 3
A)

For the triangle pictured above, which side is the hypotenuse?

(1 point)
Responses

The side labeled 11.9 km
The side labeled 11.9 km

The side labled 14.7 km
The side labled 14.7 km

We do not have enough information in the given information.
We do not have enough information in the given information.

The side labeled x
The side labeled x
B)What is the value of x? Round to the nearest tenth if necessary. (1 point)
Responses

8.6 km
8.6 km

74.5 km
74.5 km

2.8 km
2.8 km

18.9 km
18.9 km
Question 4
A)

If YS = 6.3 km and YL = 15.4 km, find the length of LS. Round your answer to the nearest tenth if necessary. You must show ALL work in order to receive full credit. You may type your work or use the button to insert a picture of your work.

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(4 points)
0 / 10000 Word Limit
Question 5
A)

If JE = 4.8 and EW = 28.6, find the length of JW. Round your answer to the nearest tenth if necessary.

You must show ALL work in order to receive full credit. You may type your work or use the button to insert a picture of your work.

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(4 points)
0 / 10000 Word Limit
Question 6
A)
If one of the legs of a triangle measures 10 cm and the hypotenuse measures 50.5 cm, find the length of the other leg? Round your answer to the nearest tenth if necessary.





You must show ALL work in order to receive full credit. You may type your work or use the button to insert a picture of your work.

A good example of how work should be shown is

(4 points)

Answer 1

Question 1

A) Correctly label the sides of the triangle shown below:

• Hypotenuse: Drop target 1 (the longest side opposite the right angle)
• Leg: Drop target 2 (one of the two sides that form the right angle)
• Leg: Drop target 3 (the remaining side that forms the right angle)

B) Given RP = 24 and DR = 7, the correct way to set up the problem to find DP is:

• x² + 7² = 24² (This applies the Pythagorean theorem, where x is the length of DP).

C) To find DP:

Using the equation from part (B):

\[ x^2 + 7^2 = 24^2 \]

Calculating:

\[ x^2 + 49 = 576 \] \[ x^2 = 576 - 49 \] \[ x^2 = 527 \] \[ x = \sqrt{527} \] \[ x \approx 22.9 \] (rounded to the nearest tenth)

So, DP ≈ 22.9.

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Question 2

A) For the triangle shown below, correctly label the sides:

• Hypotenuse: Drop target 1 (longest side opposite the right angle)
• Leg: Drop target 2 (one of the sides forming the right angle)
• Leg: Drop target 3 (the remaining side)

B) If CE = 29 and AC = 20, the equation that can be used to find AE is:

• 20² + x² = 29² (applying the Pythagorean theorem).

C) To find AE:

Using the equation from part (B):

\[ 20^2 + x^2 = 29^2 \]

Calculating:

\[ 400 + x^2 = 841 \] \[ x^2 = 841 - 400 \] \[ x^2 = 441 \] \[ x = \sqrt{441} \] \[ x = 21 \]

So, AE = 21.

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Question 3

A) For the triangle pictured above, which side is the hypotenuse?

The side labeled 14.7 km (hypotenuse is the longest side).

B) To find the value of x using the Pythagorean theorem:

1. Assuming the legs of the triangle are labeled as 11.9 km and x, we can set up: \[ 11.9^2 + x^2 = 14.7^2 \]

2. Solving for x: \[ x^2 = 14.7^2 - 11.9^2 \] \[ x^2 = 216.09 - 141.61 \] \[ x^2 = 74.48 \] \[ x = \sqrt{74.48} \] \[ x \approx 8.6 \] km (rounded to the nearest tenth)

So, x ≈ 8.6 km.

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Question 4

A) Using YS = 6.3 km and YL = 15.4 km, to find LS:

Using the Pythagorean theorem:

1. Let LS = x. \[ YS^2 + LS^2 = YL^2 \] \[ (6.3)^2 + x^2 = (15.4)^2 \]

2. Calculating: \[ 39.69 + x^2 = 237.16 \] \[ x^2 = 237.16 - 39.69 \] \[ x^2 = 197.47 \] \[ x = \sqrt{197.47} \] \[ x \approx 14.0 \] km (rounded to the nearest tenth)

So, LS ≈ 14.0 km.

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Question 5

A) Using JE = 4.8 and EW = 28.6 to find JW:

Using the Pythagorean theorem:

1. Let JW = x. \[ JE^2 + JW^2 = EW^2 \] \[ (4.8)^2 + x^2 = (28.6)^2 \]

2. Calculating: \[ 23.04 + x^2 = 817.96 \] \[ x^2 = 817.96 - 23.04 \] \[ x^2 = 794.92 \] \[ x = \sqrt{794.92} \] \[ x \approx 28.2 \] (rounded to the nearest tenth)

So, JW ≈ 28.2.

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Question 6

A) If one of the legs measures 10 cm and the hypotenuse measures 50.5 cm:

1. Let the unknown leg be x. We will use the Pythagorean theorem: \[ 10^2 + x^2 = 50.5^2 \]

2. Calculating: \[ 100 + x^2 = 2550.25 \] \[ x^2 = 2550.25 - 100 \] \[ x^2 = 2450.25 \] \[ x = \sqrt{2450.25} \] \[ x \approx 49.5 \] (rounded to the nearest tenth)

So, the length of the other leg is approximately 49.5 cm.