Geometry Assignment (Math Problem Sample)

Geometry assignment - 6 questions
Question 10.
Solutions:
PR = RQSegment Bisector Theorem
8x – 5 = 7x – 4 Substitute the value of PR and RQ
8x – 7x – 5 = 7x – 7x – 4 Simplify
x – 5 = -4
x – 5 + 5 = -4 + 5
x = 1
PR = 8x – 5 RQ = 7x – 4 Substitute the value of x to solve for
= 8(1) – 5 = 7(1) – 4 PR and RQ and simplify
= 8 – 5 = 7 – 4
PR = 3 RQ = 3
PQ = PR + RQ Segment Addition Postulate
= 3 + 3Substitute the value of PR and RQ
PQ = 6Final Answer
Question 11.
Solutions:
Using points A(5,-1) and M(1,6) to find the point B.
Since, M is the midpoint of AB, we used the midpoint formula to find the point B.
Md = x1 + x22,y1 + y22Midpoint Formula
(1,6) = 5 + x22,-1 + y22Substitute the value of points A and M
1 = 5 + x22 6 = -1 + y22Solve for the value of x and y coordinates and simplify.
2 = 5 + x2 12 = -1 + y2
2 – 5 = 5 – 5 + x212 + 1 = -1 + 1 + y2
-3 = x2 13 = y2
Therefore, the coordinates of Point B is (-3,13)Final Answer
Question 12.
Solution:
By inspection, you can determine that Plane P is not the name of a plane shown in the figure.
Question 1.
Solutions:
Since D is the midpoint of CE and E is the midpoint of CF, we can find the value of x using the segment addition postulate and the definition of midpoint.
EF = CD + DE
10x + 27 = (8x + 8) + (6x – 5) Substitute the values of EF, CD, and DE
10x + 27 = 14x + 3Simplify
10x – 14x +27 = 14x – 14x + 3
- 4x + 27 – 27 = 3 – 27
-4x-4= -24-4
x = 6Final Answer
2308860206375P
Question 2.
Sol