math problem?

Question

LOOK AT THE PICTURE

One square is inside another square...the distance from the edge (diagonally) of the outer square to inner square is 2.
The lenght of the side of the outer square is 10...how much is the lenght of the inner square
see picture....
http://tinypic.com/view.php?pic=8ezrfgz&s=1

Answer 1

You have to create a triangle between the diagonal line and the horizontal distance from the small square to the large square. Using Pythagorean Theorem, the distance is sqrt(2).

10 - 2sqrt(2)

about 8.28

Answer 2

The questions is only answerable if we know the angle between the side of length 10 and the diagonal line. If it is 45 degrees then you create a 45-45-90 degree triangle in the corner and use Pythagoras to calculate the other two equal sides.
A^2 + B^2 = C^2 or
if A=B then 2(A^2) =C^2
then If C=2 and C^2=4
then 2(A^2) =2^2=4
Divide both sides by 2
A^2=2 and A=sq rt2
The length of the side of the inner square is
10 - (2 times rt2)
Not a useful number but oh well

Answer 3

The diagonal of the outer square is 10sqrt(2)
Hence the diagonal of the inner square is 10sqrt(2) - 4
The the side of a square equals diagonal/sqrt(2).
Hence side of inner square is 10 -4/sqrt(2)
= 10 - 2sqrt(2)

Answer 4

using the pythagorum theorem in the corner we have x as the distance from one square horizontally to the next so
x^2 + x^2 = 2^2
2x^2=4
x^2=2
x=sqrt(2)
so the width of the inner square =10-2sqrt(2)
10-2(1.414)=7.1716

Answer 5

Find the sides of a right triangle with a hypotenuse of 2... the sides are square root of 2 (1.414).

10 minus 2 * 1.414 will give you the length of the inner square's side. Which is about 7.17

Answer 6

large square diagonal equals 10*sqrt2
small sqaure diagonal equals 10*sqrt2 - 4
x*sqrt2 = 10*sqrt2 - 4
x=(10*sqrt2 - 4)/sqrt2
simplifies to
x=10-2*sqrt2