SOLUTION: A square with a side length of 2s has a
smaller square inscribed. The vertices of the
smaller square are at the midpoints of the
sides of the larger square.
What is the ratio o
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-> SOLUTION: A square with a side length of 2s has a
smaller square inscribed. The vertices of the
smaller square are at the midpoints of the
sides of the larger square.
What is the ratio o
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Question 1191231: A square with a side length of 2s has a
smaller square inscribed. The vertices of the
smaller square are at the midpoints of the
sides of the larger square.
What is the ratio of the area of the larger square to the area of
the smaller square? Express your answer in
its simplest form Found 2 solutions by Theo, ikleyn:Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! each side of the larger square is equal to 2s.
the area of the larger square is (2s)^2 = 4s^2.
each side of the smaller square is equal to s * sqrt(2).
the area of the smaller squaare is equal to 2s^2.
the radio of the area of the larger square to the area of the smaller square is 4s^2 / 2s^2 = 4/2 = 2/1.
if you let s = 2, then each side of the larger square id equal to 4 and the area of the larger square is equal to 4^2 = 16.
each side of the smaller square is the hypotenuse of the triangle formed by connecting the midpoint of one side of the larger square to the midpoint of the the adjacent side of that larger square.
the length of the hypotenuse is the length of one side of the smaller square.
that would then be sqrt(2^2 + 2^2) = sqrt(8).
the area of the smaller square would be equal to sqrt(8)^2 = 8.
the ratio of the area of the larger square to the area of the smaller square is equal to 16/8 = 2/1.
