# Area of square

• Apr 3rd 2009, 06:46 AM
rickymylv
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
• Apr 3rd 2009, 06:51 AM
Quote:

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 ! (Wait)
• Apr 3rd 2009, 07:28 AM
Chop Suey
Given a square with sides of length s, the length of the diagonal of the square is equal to $\displaystyle s\sqrt{2}$

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

$\displaystyle \frac{s^2}{(s\sqrt{2})^2}$
• Apr 3rd 2009, 07:35 AM
Soroban
Hello, rickymylv!

Quote:

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:

. . $\displaystyle (A)\;1:3 \qquad (B)\;3:4 \qquad (C)\;2:3 \qquad (D)\;1:2$

A sketch is essential . . .
Code:

      * - - - - - *       |          |       |          |       |          | s       |          |       |          |       * - - - - - *             s
A square with side $\displaystyle s$ has area $\displaystyle s^2.$

Code:

      * - - - - - *       |  _    * |       |  √2s  *  |       |    *    | s       |  *      |       | *        |       * - - - - - *             s
The diagonal $\displaystyle d$ of the square can be found with Pythagorus:

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

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

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

• Apr 3rd 2009, 09:35 AM
lebanon
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.