Asked by Kenny
Let ABCD be a square, and EFGH be a square whose area is twice that of ABCD. Prove that the ratio of AB to AC is equal to the ratio of AB to EF
Answers
Answered by
bobpursley
if the second square is twice the area, then the sides for the large square must be sqrt2 times as large.
EF=AB*sqrt2
EF/AB =sqrt2
Now consider the smaller square. ABC forms a right triangle, sides AB, CD, and AC
but CD=AB so AC by Pythagorean theorem is AC=sqrt(AB^2+AB^2)=sqrt(2AB^2
=AB*sqrt2
AC/AB = sqrt2
so the Ration of AB to AC is the same as ratio of AB to AC
EF=AB*sqrt2
EF/AB =sqrt2
Now consider the smaller square. ABC forms a right triangle, sides AB, CD, and AC
but CD=AB so AC by Pythagorean theorem is AC=sqrt(AB^2+AB^2)=sqrt(2AB^2
=AB*sqrt2
AC/AB = sqrt2
so the Ration of AB to AC is the same as ratio of AB to AC
Answered by
Reiny
let the side of square ABCD be x
then area = x^2
area of square EFGH = 2x^2
side of square EFGH = √2 x
EF = √2 x
AC^2 = x^2 + x^2 = 2x^2
AC = √2 x
AB/AC = x/√2x = 1/√2
AB/EF = x/√2x = 1/√2
thus AB/AC = AB/EF
then area = x^2
area of square EFGH = 2x^2
side of square EFGH = √2 x
EF = √2 x
AC^2 = x^2 + x^2 = 2x^2
AC = √2 x
AB/AC = x/√2x = 1/√2
AB/EF = x/√2x = 1/√2
thus AB/AC = AB/EF
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