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. Part B: Explain how you know the triangles from Part A are similar. Part C: If EG = 2 and EF = 8, find the length of segment ED. Show your work.
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if you draw it out it helps alot
i still dont understand it
Okay well do you at leastget why I made the drawing
yes i do
i just dont know the answers to the questions
Okay can you explain why you know the drawing is correct?
the drawing is here on my paper too, and u drew exactly whats on the paper so ofc its correct but how do i find the pair of similar triangles
You can identify a similar triangle when it looks alike What are two angle measures that look alike? *hint: we split two trangles in half with the line DG*
whew im super bad at this.. since its split in half would that make DE and DF or am i thinking about it wrong
A triangle has the name DEF: which means the corners of the triangles land on the D, E, and F. do u understand?
yes so overall what should i put for part A
Part A: I will give you three choices a)EDF and DGF b)DEF and FED c)DEG and DGF d)EDG and DEF
D?
No, they do not have the same size.
egd and fgd
Yes correct!
is part c 4?
or no?
Moving onto Part b)
What do you think?
Okay, i know they are the same size
which makes them similar
?
Yes
okay and then for part c i came up with 4
for the length of ED
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is that 64?
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so that would make part c answer..
What do you think lol?
honestly im not sure..lol i think 4
?
No lol where did you get four from
idek im super confused
out of what numbers would it be
2 sqrt(2)
\[2\sqrt{2}\]
6?
\(\color{#0cbb34}{\text{Originally Posted by}}\) @lowkey \[2\sqrt{2}\] \(\color{#0cbb34}{\text{End of Quote}}\)
2.8......
yes lmao
okay could you help me with one more that im going to post? i have to be finished soon lol
Sorry I gtg lol <3 @dude
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