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

RainbowPanda Nov 29, 2018

#2**+2 **

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

CPhill Nov 29, 2018

#4**+2 **

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

CPhill Nov 29, 2018