Seth is using the figure shown below to prove the Pythagorean Theorem using triangle similarity: In the given triangle DEF, angle D is 90° and segment DG is perpendicular to segment EF.

The figure shows triangle DEF with right angle at D and segment DG. Point G is on side EF.

Part A: Identify a pair of similar triangles. (2 points)

Part B: Explain how you know the triangles from Part A are similar. (4 points)

Part C: If EG = 2 and EF = 8, find the length of segment ED. Show your work. (4 points)

Okay so i made an extra account just to put this, I answered my own question...

Step-by-step explanation:

Part A — Triangle DGE is similar to DGF.

Part B — Both triangles has a 90-degree angle, with G. The triangles are congruent with each other. So, Side-Angle-Side (SAS) Similarity Postulate, helps me prove the two triangles are similar. 

Part C — The given is EG = 2 and EF = 8. The EF measurement is not needed in this equation because the question is not asking for the larger triangle. So, GD has to be 2, because G is 90-degrees, and so, taking the 2 able doubling it to 4, while remembering that a triangle is 180 degrees, so that would make ED 4 too, because it is the other 90 degrees; they need to be equal. So, ED = 4

OR

Taking the EF = 8 into mind, I think of the fact that D is 90 degrees. So, again, that needs to be doubled to 180 for the full triangle. Both sides should equal to 8, so for the ED to be solved, 8 will be divided in half. The answer is ED = 4

It has not been graded yet so I am not 100 percent sure. If you are gonna use my work, please please please reword it as much as you can. Trust me when I say the plagiarism is caught immediately. Good Luck Guys!!

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.