Distance Midpoint And Angle Measurement Embedded Assessment 2 Answers

Distance, Midpoint, and Angle Measurement: Embedded Assessment 2 Answers

This comprehensive guide provides detailed answers and explanations for an embedded assessment focusing on distance, midpoint, and angle measurements. We'll cover various mathematical concepts and techniques crucial for understanding and solving these types of problems. Remember, while this guide provides answers, the true learning comes from understanding the why behind each step.

Understanding the Fundamentals: Distance, Midpoint, and Angle

Before diving into specific assessment questions, let's review the fundamental principles:

1. Distance Measurement

Distance measurement involves determining the length between two points. In a coordinate plane (with x and y axes), we often use the distance formula, derived from the Pythagorean theorem:

Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

Where (x₁, y₁) and (x₂, y₂) are the coordinates of the two points. This formula calculates the straight-line distance between the points.

2. Midpoint Calculation

The midpoint is the point that lies exactly halfway between two given points. The midpoint formula allows us to easily calculate its coordinates:

Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)

This formula averages the x-coordinates and the y-coordinates of the two points to find the midpoint's coordinates.

3. Angle Measurement

Angle measurement focuses on determining the size of an angle formed by two intersecting lines or rays. Angles are typically measured in degrees (°) or radians. Several methods exist for determining angle measurements, depending on the context:

• Using a protractor: A physical protractor provides a direct measurement of an angle.
• Using geometry principles: Knowledge of geometric properties (e.g., isosceles triangles, right triangles) and theorems (e.g., Pythagorean theorem, Law of Cosines) can be used to calculate angles indirectly.
• Using trigonometry: Trigonometric functions (sine, cosine, tangent) are powerful tools for calculating angles in triangles, given the lengths of their sides.

Sample Embedded Assessment Questions and Answers

Let's tackle some sample questions that might appear in an embedded assessment on distance, midpoint, and angle measurements. We'll provide detailed solutions to demonstrate the application of the formulas and concepts.

Question 1: Finding the Distance Between Two Points

Find the distance between points A(2, 3) and B(7, 15).

Answer:

We use the distance formula:

Distance = √[(7 - 2)² + (15 - 3)²] = √[5² + 12²] = √(25 + 144) = √169 = 13

Therefore, the distance between points A and B is 13 units.

Question 2: Calculating the Midpoint

Determine the midpoint of the line segment connecting points C(-4, 6) and D(10, -2).

Answer:

We use the midpoint formula:

Midpoint = ((-4 + 10)/2, (6 + (-2))/2) = (6/2, 4/2) = (3, 2)

The midpoint of the line segment CD is (3, 2).

Question 3: Angle Measurement using Trigonometry

A right-angled triangle has a hypotenuse of length 10 and one leg of length 6. Find the measure of the angle opposite the leg of length 6.

Answer:

We can use the trigonometric function sine:

sin(θ) = opposite/hypotenuse = 6/10 = 0.6

To find the angle θ, we use the inverse sine function (arcsin or sin⁻¹):

θ = sin⁻¹(0.6) ≈ 36.87°

Therefore, the measure of the angle is approximately 36.87°.

Question 4: Application of Distance and Midpoint in Geometry

Points P(1, 4), Q(7, 2), and R(x, y) form an isosceles triangle with PQ = PR. If the midpoint of QR is (4, 1), find the coordinates of R (x, y).

Answer: This problem combines distance and midpoint concepts.

1. Find PQ: Using the distance formula, PQ = √[(7 - 1)² + (2 - 4)²] = √(36 + 4) = √40

2. Use the midpoint of QR: The midpoint of QR is given as (4, 1). Let R = (x, y). Using the midpoint formula:

   ( (7 + x)/2, (2 + y)/2 ) = (4, 1)

   This gives us two equations:

   (7 + x)/2 = 4 => 7 + x = 8 => x = 1 (2 + y)/2 = 1 => 2 + y = 2 => y = 0

   Therefore, R = (1, 0).

3. Verify PQ = PR: Using the distance formula, PR = √[(1 - 1)² + (0 - 4)²] = √16 = 4. Note that PQ = √40 and PR = 4. There's an error in the problem statement or a misunderstanding of the isosceles triangle condition (it implies that there are two sides with equal length). This means that we need to re-examine the problem.

Corrected Question 4: Application of Distance and Midpoint in Geometry

Points P(1, 4), Q(7, 2), and R(x, y) form an isosceles triangle with PQ = PR. If the midpoint of QR is (4, 1), find the coordinates of R (x, y). Assume PQ = QR.

Corrected Answer:

1. Find PQ: Using the distance formula, PQ = √[(7 - 1)² + (2 - 4)²] = √40

2. Use the midpoint of QR: The midpoint of QR is given as (4, 1). Let R = (x, y). Using the midpoint formula:

   ( (7 + x)/2, (2 + y)/2 ) = (4, 1)

   This gives us two equations:

   (7 + x)/2 = 4 => 7 + x = 8 => x = 1 (2 + y)/2 = 1 => 2 + y = 2 => y = 0

   Therefore, R = (1, 0). However, This is not a correct solution. We must consider that PQ = QR.

Let's use a different approach. Since the midpoint of QR is (4, 1), we can find the vector from Q to the midpoint: (4-7, 1-2) = (-3, -1). This is half the vector from Q to R. Therefore, the vector from Q to R is twice this: (-6, -2). Adding this to Q's coordinates gives R's coordinates:

R = (7 - 6, 2 - 2) = (1, 0).

Now, let's verify: QR = √[(1-7)² + (0-2)²] = √40. This confirms PQ = QR.

Therefore, the coordinates of R are (1, 0).

Question 5: Real-World Application

A surveyor needs to measure the distance across a river. He places markers at points A and B on one side of the river, 50 meters apart. He then sights a point C on the opposite bank, such that angle BAC is 60° and angle ABC is 75°. Find the width of the river (the distance from B to C).

Answer:

This problem requires using the Law of Sines. Let's denote the width of the river as BC.

1. Find angle ACB: The sum of angles in a triangle is 180°. Therefore, angle ACB = 180° - 60° - 75° = 45°.

2. Apply the Law of Sines: The Law of Sines states that a/sin(A) = b/sin(B) = c/sin(C). In our case:

   BC/sin(60°) = AB/sin(45°)

   BC = AB * sin(60°) / sin(45°) = 50 * (√3/2) / (√2/2) = 50√3/√2 = 25√6

Therefore, the width of the river is approximately 61.24 meters.

These examples showcase the practical application of distance, midpoint, and angle measurements. Remember that understanding the underlying principles and choosing the appropriate formula or technique are essential for accurately solving these types of problems. Practice is key to mastering these concepts and gaining confidence in your ability to solve more complex geometrical problems. Remember to always check your work and ensure your answers are reasonable within the context of the problem.