## Geometry Lesson 1.1 Practice A Answers

Geometry can be a challenging subject for many students, especially when it comes to solving practice problems. Lesson 1.1 in geometry often focuses on basic concepts such as points, lines, and planes. In this article, we will provide you with the answers to Practice A in Lesson 1.1, helping you understand and master the fundamental principles of geometry.

### Exercise 1

For exercise 1, we are given a line segment AB and a point C between A and B. The task is to find the length of AC and CB. To solve this problem, we can use the Segment Addition Postulate, which states that if we have three points A, B, and C on a line, then AB + BC = AC.

Using the given information, we can substitute the values into the equation. If AB = 8 units and AC = 3 units, we can find the length of CB by subtracting AC from AB. Therefore, CB = AB - AC = 8 - 3 = 5 units. So, the length of AC is 3 units, and the length of CB is 5 units.

### Exercise 2

In exercise 2, we are given two points, P and Q, on a line. The task is to find the length of PQ. This problem can be easily solved using the Distance Formula, which states that the distance between two points (x1, y1) and (x2, y2) is given by the formula: √((x2 - x1)^2 + (y2 - y1)^2).

Let's say point P has coordinates (2, 3) and point Q has coordinates (5, 7). We can substitute these values into the Distance Formula to find the length of PQ. Therefore, PQ = √((5 - 2)^2 + (7 - 3)^2) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5 units. So, the length of PQ is 5 units.

### Exercise 3

Exercise 3 presents us with three points, A, B, and C, on a line. We are given that AB = 4x and BC = 7x. The task is to find the value of x and the lengths of AB and BC. To solve this problem, we can use the Segment Addition Postulate again.

Using the information given, we can set up the equation AB + BC = AC. Substituting the values, we have 4x + 7x = AC. Combining like terms, we get 11x = AC. Since AB = 4x and BC = 7x, we can substitute these values into the equation to find AC. Therefore, AC = 4x + 7x = 11x. So, the length of AC is 11x units.

Now, we need to find the value of x. If we know that AB = 4x, and given that AB = 20 units, we can set up the equation 4x = 20. Solving for x, we divide both sides of the equation by 4, and we get x = 5. So, the value of x is 5.

Finally, we can find the lengths of AB and BC by substituting the value of x back into the equations. AB = 4x = 4(5) = 20 units, and BC = 7x = 7(5) = 35 units. Therefore, the length of AB is 20 units, and the length of BC is 35 units.

### Exercise 4

In exercise 4, we are given a line segment PQ with a point R between P and Q. The task is to find the length of PR and QR. This problem can be solved using the Segment Addition Postulate.

Given that PQ = 28 units, PR = 9 units, and QR = 5 units, we can substitute these values into the equation PR + QR = PQ. Therefore, 9 + 5 = PQ. Simplifying, we get 14 = PQ. So, the length of PQ is 14 units.

To find the length of PR, we subtract QR from PQ. Therefore, PR = PQ - QR = 14 - 5 = 9 units. So, the length of PR is 9 units.

Similarly, to find the length of QR, we subtract PR from PQ. Therefore, QR = PQ - PR = 14 - 9 = 5 units. So, the length of QR is 5 units.

### Exercise 5

Exercise 5 involves three collinear points, A, B, and C. We are given that AB = 4x, BC = 3x, and AC = 28 units. The task is to find the value of x and the lengths of AB and BC.

Using the Segment Addition Postulate, we can set up the equation AB + BC = AC. Substituting the values, we have 4x + 3x = 28. Combining like terms, we get 7x = 28. To find the value of x, we divide both sides of the equation by 7. Therefore, x = 4. So, the value of x is 4.

Now, we can find the lengths of AB and BC by substituting the value of x back into the equations. AB = 4x = 4(4) = 16 units, and BC = 3x = 3(4) = 12 units. Therefore, the length of AB is 16 units, and the length of BC is 12 units.

### Exercise 6

In exercise 6, we are given three non-collinear points, P, Q, and R. The task is to find the length of PR and QR. This problem can be solved using the Distance Formula.

Let's say point P has coordinates (2, 3) and point Q has coordinates (5, 7). To find the length of PQ, we can use the Distance Formula: √((x2 - x1)^2 + (y2 - y1)^2). Substituting the values, we have PQ = √((5 - 2)^2 + (7 - 3)^2) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5 units. So, the length of PQ is 5 units.

Now, let's say point R has coordinates (8, 10). To find the length of PR, we use the Distance Formula again: √((x2 - x1)^2 + (y2 - y1)^2). Substituting the values, we have PR = √((8 - 2)^2 + (10 - 3)^2) = √(6^2 + 7^2) = √(36 + 49) = √85 units. So, the length of PR is √85 units.

Similarly, to find the length of QR, we use the Distance Formula: √((x2 - x1)^2 + (y2 - y1)^2). Substituting the values, we have QR = √((8 - 5)^2 + (10 - 7)^2) = √(3^2 + 3^2) = √(9 + 9) = √18 units. So, the length of QR is √18 units.

### Exercise 7

Exercise 7 presents us with three non-collinear points, P, Q, and R. We are given that PQ = 6x, QR = 10x, and PR = 16 units. The task is to find the value of x and the lengths of PQ and QR.

Using the Segment Addition Postulate, we can set up the equation PQ + QR = PR. Substituting the values, we have 6x + 10x = 16. Combining like terms, we get 16x = 16. To find the value of x, we divide both sides of the equation by 16. Therefore, x = 1. So, the value of x is 1.

Now, we can find the lengths of PQ and QR by substituting the value of x back into the equations. PQ = 6x = 6(1) = 6 units, and QR = 10x = 10(1) = 10 units. Therefore, the length of PQ is 6 units, and the length of QR is 10 units.

### Exercise 8

In exercise 8, we are given three non-collinear points, A, B, and C. The task is to find the length of AC and AB. This problem can be solved using the Distance Formula.

Let's say point A has coordinates (2, 3) and point C has coordinates (8, 10). To find the length of AC, we can use the Distance Formula: √((x