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# I need help with a small portion of this question

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It's a bit messy but im sure you get what im goin at here. This is what i've got so far:

1. Given

2. AB= \(3^2+1^2=c^2\)

9+1=c^2

sqrt(10)=sqrt(c^2)

c=sqrt(10)

BC=\(3^2+1^2=c^2\)

9+1=c^2

sqrt(10)=sqrt(c^2)

c=sqrt(10)

CD=\(3^2+1^2=c^2\)

9+1=c^2

sqrt(10)=sqrt(c^2)

c=sqrt(10)

DA=\(3^2+1^2=c^2\)

9+1=c^2

sqrt(10)=sqrt(c^2)

c=sqrt(10)

3. \(AB\cong BC\cong CD \cong DA\)

4. I'm stuck here >.<

I know that they are congruent but how do i prove that and i know that it is obviously a square

Nov 29, 2018

#1
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Uhh is it

AC=1/3

BD=3/-1=-3/1

1/3*-3/1=-3/3=-1

Nov 29, 2018
#2
+103148
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Here's the way I might do it, RP

1. Find the side lengths

2. Show that the diagonals have the same length

3. Show that the length  of any side =

diagonal length / √2

3. This will mean that we have a  square because the sides are congruent,  the diagonals are congruent and because of the proof of  (3),   they will meet at right angles

Nov 29, 2018
edited by CPhill  Nov 29, 2018
#3
+2448
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And how do I do that...

RainbowPanda  Nov 29, 2018
#4
+103148
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Example .....to find a side length.....use the distance formula

Are you familiar with this???

Distance from A to B   =

sqrt  [ ( 1 - 2)^2 + (3 - 0)^2 ]  = sqrt [ (-1)^2 + 3^2] = sqrt [ 1 + 9 ] =  sqrt (10)

Then find the distances between B,C     C,D      and      D,A

They should all be =  sqrt (10)

Let me know if you have trouble

Nov 29, 2018