# Geometry Assignment (Math Problem Sample)

geometry assignment with solutions

source..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

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