Question

2 Question Geometry Check!! Pls! Will Medal!!

Answer 1

1. is the third one.
2. is correct.

:)

Answer 2

Can you please tell me how you got the third answer for number 1? Thank you.

Answer 3

Anyone is able to intepret

Answer 4

Because the vertices of the rhombus are located at a0 0b -a 0 and 0b and the midpoint would be a/2 , b/2

Answer 5

@Nitroster

Answer 6

you answer Q1 correctly (-a/2, b/2) is in the 2nd quadrant (you want negative x and positive y)

Answer 7

Yeah, @phi makes more sense.

Answer 8

Hm yeah, cause ProMiner isn't exactly showing his work. Thanks to the both of you!

Answer 9

However, you did not get the length of the diagonal of a square

what is the distance from point (0,0) to point (a,a,) (that is the diagonal)
you can use the distance formula if you know it

Answer 10

or you could use pythagoras with two sides = a and you want to find the hypotenuse (the diagonal)

Answer 11

There was also the other diagonal, which I used... Ill show you my work :)
√(a - 0)^2 + (0 - a)^2
√a^2 + a^2
2a

Answer 12

ok, you should get
\[ \sqrt{a^2+a^2}= \sqrt{2a^2} \]
that does not simplify to 2a

Answer 13

you can write it as
\[ \sqrt{2a^2}= \sqrt{2}\cdot \sqrt{a^2} \]

Answer 14

but when a number that is squared (^2) is inside a √... it goes away, doesn't it?

Answer 15

yes, but you don't have 2*2*a*a
you have 2*a*a
in other words, only the a is squared. the 2 is not, so it "stays inside" the square root sign

Answer 16

http://www.wolframalpha.com/input/?i=%E2%88%9A2a%5E2
can verify. Plus, there are no answer choices for what you provided.

Answer 17

\[ \sqrt{4a^2}= \sqrt{2^2 a^2}= 2a \]

Answer 18

wolfram is doing something strange.
if you type in sqrt(2*a^2) it probably will work.

Answer 19

hm.. so what answer do I put?

Answer 20

but it is worth understanding how to simplify
\[ \sqrt{2a^2}\]
you can "pull out" any squares (or pairs of the same thing)
so you get
\[ \sqrt{2}\ a \]
that means sqr(2) times "a"
you can switch the order when you multiply, so you can also write it as
2 sqr(2)

Answer 21

* a sqr(2) (not 2 sqr(2) )

Answer 22

Yes I understand. Thanks for that!! but for this occasion, the answer has to be 2a. :P I put in wolfram exactly what you said but your * and () was a distraction. It separated the answer. When it is straight out just √2a^2 it goes to 2a.

Answer 23

the length of the diagonal is not 2a

if it were then by pythagoras
a^2 + b^2 = c^2

with the first leg= a, the 2nd leg also a, and the hypotenuse 2a , you would get
a^2+a^2 = (2a)^2
2a^2 = 4a^2
which is not correct. (because the hypotenuse is a * sqr(2) )

Answer 24

no other answers besides "a" ; "2√a" or "a√2" :o
Thanks for helping me btw! very much :D

Answer 25

yes, and one of those is the correct answer. You seem to be missing the idea.

Answer 26

mhm, how do I do that LOL... I never worked with letters and factoring type of deal

Answer 27

@phi help pls?

Answer 28

@phi pls?

Answer 29

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