Question

the ratio of the areas of two squares one having its diagonal double that of the other is

Answer 1

AREA OF A SQUARE
$\frac{ {(diagonal)}^{2} }{2}$
let diagonal of 1st square be x
so diagonal of the 2nd square be 2x
area of 1st square be
$= \frac{ {x}^{2} }{2}$
area of 2nd square be
$\frac{ ({2x})^{2} }{2} = \frac{4 {x}^{2} }{2} = 2 {x}^{2}$
so the ratio of the areas of the two squares be
$\frac{ {x}^{2} }{2} : \: 2 {x}^{2} = \frac{1}{2} : \: 2 = 1: \: 4$

Answer 2

Answer:

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Step-by-step explanation:

let diagonals be 2d and d.

then, A1/A2=1-2 (2d)^2/1-2d^2

A1/A2=4d^2/d^2

A1/A2=4/1

A1:A2=4:1

the ratio is 4:1

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