Question 201289
let {{{A[1]}}} = the area of the larger square
Let {{{A[2]}}} = the area of the smaller square
Let {{{x}}} = the side of the larger square
Let {{{y}}} = the side of the smaller square
given:
{{{A[1] / A[2] = 6 / 1}}}
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{{{A[1] = x^2}}} definition
{{{A[2] = y^2}}} definition
Substituting:
{{{x^2 / y^2 = 6 / 1}}}
Take the square roots of both sides
{{{x / y = sqrt(6) / 1}}}
Also by definition,
larger diagonal = {{{sqrt(2)*x}}}
Smaller diagonal = {{{sqrt(2)*y}}}
larger diagonal / smaller diagonal = {{{(sqrt(2)*x) / (sqrt(2)*y)}}}
larger diagonal / smaller diagonal = {{{x/y}}}
larger diagonal / smaller diagonal = {{{sqrt(6)/1}}} answer