# Math Help - Area of square

1. ## Area of square

please tell me how to solve this problem

The ratio of the area of square to that of the square drawn on its diagonal is:
(A)1:3
(B)3:4
(C)2:3
(D)1:2

2. Originally Posted by rickymylv
please tell me how to solve this problem

The ratio of the area of square to that of the square drawn on its diagonal is:
(A)1:3
(B)3:4
(C)2:3
(D)1:2

Let x be the side of square (all sides equal)

Length of its diagonal in terms of x = ?? ( Use Pythagoras theorem)
-----------------------------------
Area of square = (side)^2

Now solve it !

3. Given a square with sides of length s, the length of the diagonal of the square is equal to $s\sqrt{2}$

The question asks for the ratio of the area of a square to the area of a square drawn on its diagonal. As in:

$\frac{s^2}{(s\sqrt{2})^2}$

4. Hello, rickymylv!

Please tell me how to solve this problem

The ratio of the area of a square to that of the square drawn on its diagonal is:

. . $(A)\;1:3 \qquad (B)\;3:4 \qquad (C)\;2:3 \qquad (D)\;1:2$
A sketch is essential . . .
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A square with side $s$ has area $s^2.$

Code:
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The diagonal $d$ of the square can be found with Pythagorus:

. . $d^2 \:=\:s^2+s^2 \:=\:2s^2 \quad\Rightarrow\quad d \:=\:\sqrt{2}\,s$

The area of the square of side $d$ is: . $\left(\sqrt{2}\,s\right)^2 \;=\;2s^2$

Therefore, the ratio of the two areas is . . .

5. ## hello

let the side of the square be (a),so the area of the first square is a^2,
then by pythegoras , it's diagonal is a√2,and the area of the second square ia (a√2)^2=2(a^2).
therefore the ratio is:a^2 / 2(a^2)=1/2
D is correct.