In the figure below, what is the perimeter of square ABCD? (The answer is 8√2, but I don't understand why)

Question

............................|...........................
............................|...........................
...........................B...........................
............................|...........................
__.__.__A__.__|__.__C__.__.__
............................|..........................
...........................D..........................
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............................|...........................

Just to make it easier to understand...
These are the points:
A (-2, 0)
B (0, 2)
C (2, 0)
D (0, -2)

And I thought the answer to this was 16,
But the correct answer is 8√2
And I don't understand why.

Oh yeah... I forgot to add...
This isn't for my homework.
It's from the practice booklet for my PSATs

Answer 1

Distance AB=sqrt{(0+2)^2+(2-0)^2)
=sqrt8=2 sqrt2 units
BC=sqrt{(2-0)^2+(0-2)^2
=sqrt8=2sqrt2 units
Similarly,you would find CA and AD aksi equal to 2 sqrt2 units each
Therefore the perimetr is
4*2sqrt2 units
=8sqrt2 units
[04]

Answer 2

The diagonals of the square are 4 units long. That means if you turn the square into four triangles by cutting along the lines x=0 and y=0, each of those right triangles has short sides 2 units long.

The square of the long side (which is one side of the square) is equal to the sum of the squares of the other two sides. Well the other two sides are each 2, so you square that, and they're each 4. Add them, and they're 8. Take the square root, and you end up with √8 which simplifies to 2√2.

Your square consists of 4 sides like that, so you multiply by four to get 8√2.

Answer 3

For the purpose of this exercise, let me refer to the origin (0,0) as O. To find the perimeter of a square you need only find the length of one side and then multiply that by 4, since all sides have the same length by definition.

This gives you a right triangle OBC that has leg length of BO = 2 and OC = 2. The hypotenuse of OBC (BC) is the lenght of the sides of the square. According to the pythagorean theorem, the lenght of the hypotenuse is sqrt (2^2+2^2) = sqrt (8) = 2*sqrt(2). 4 times this side length (the perimeter of the square) is therefore 4*2*sqrt(2) = 8*sqrt(2)... Solved...

Answer 4

HI, ok what you have to do is make a right triangle by connecting points A and B. You have to find the line that connect A and B it's not straight it's skewed. When you make a right triangle, your hypotenus is AB the other two sides are 2 so the hyp. is 2rut2 then you multiply that by 4 and you get the answer. Hope I explained it well.

Answer 5

AB² = 2² + 2² = 8
AB = √8 = 2√2
BC = CD = DA = 2√2 also
P = 4 x ( 2√2 )
P = 8√2

Answer 6

If you take a close look at the values you have given, it is not a square but a parallelogram with the two long/horizontal sides (4) and the two short/vertical sides (0.4).

So, you have (2 x 4) + (2 x 0.4) = 8.8

Hope this helps to visualize the problem pictorially.

Answer 7

from A to midpoint is 2, from midpoint to d is 2 forms a special triangle. the distance from a to d is 2√2, times 4 to get p which is 8√2

Answer 8

Take one fourth of the shape - say, AB, and calculate it's length -
sqrt( (XlengthofAB^2)+(YlenghtofAB^2)) or:
sqrt((2^2)+(2^2)) = sqrt(4+4)=sqrt(8)= sqrt(4*2) = 2*sqrt(2).
Then multiply that one side by four, and viola - 8*sqrt(2)

Answer 9

lol you ppl with yall homework ....LOL xD