# Challenging "Find the Hashed Area" Question

• Jul 12th 2012, 10:16 AM
danielmc
Challenging "Find the Hashed Area" Question
See attached question. It asks: "A circle is inside a square, which is also found inside another square with with 6 meter sides. Find the hashed area".

Is the correct answer: 9pi/2 meters^2 or 9pi meters^2?

Can you please systematically outline the steps that you took to arrive at the answer?

Thank you for your help and explanation.
• Jul 12th 2012, 10:20 AM
richard1234
Re: Challenging "Find the Hashed Area" Question
The side length of the smaller square is $3 \sqrt{2}$ (do you see why?).

Therefore the diameter of the circle is $3 \sqrt{2}$, so the radius is $\frac{3 \sqrt{2}}{2}$. The area of the circle is $(\frac{3 \sqrt{2}}{2})^2 \pi = \frac{9 \pi}{2}$.
• Jul 12th 2012, 10:33 AM
danielmc
Re: Challenging "Find the Hashed Area" Question
Hi Richard,

Thanks for the quick response. I do not see how you got the side length of the smaller square to be http://latex.codecogs.com/png.latex?3%20\sqrt{2}.

Can you please explain how you got to that step?

Thanks a million!
• Jul 12th 2012, 10:34 AM
richard1234
Re: Challenging "Find the Hashed Area" Question
Hint: The side length of the larger square is the diagonal of the smaller square.
• Jul 12th 2012, 10:43 AM
danielmc
Re: Challenging "Find the Hashed Area" Question
I understand the rest of it but just can't the length of the inner square. I'm sorry my brain isn't working today. I know it has something to do with the Phytagoreans Theorem. Can you please tell me how you got the side length of the inner square.
• Jul 12th 2012, 10:55 AM
richard1234
Re: Challenging "Find the Hashed Area" Question
A square is composed of two 45-45-90 triangles. The side length of the large square is the diagonal of the small square, so the diagonal of the small square is 6. The diagonal is like the "hypotenuse" of the 45-45-90 triangles, you should be able to find the side length now.
• Jul 12th 2012, 11:01 AM
danielmc
Re: Challenging "Find the Hashed Area" Question
Thank you Richard for the step-wise explanation. I now understand. Since the diagonal of the small square is 6; it would be b^2 + b^2 = 6^2; so solving for b, you would take square root of 36 so 6, then another square root of 6 so "b" (the side of the square) would be 3 sqrt 2. From here, we know that the diameter of the circle is 3 sqrt 2 and we can find radius and then area as you had outlined. Thanks for your help!
• Jul 12th 2012, 12:17 PM
Soroban
Re: Challenging "Find the Hashed Area" Question
Hello, danielmc!

Quote:

A circle is inside a square, which is also found inside another square with with 6 m sides.
Code:

                        E                         o                   3  *  *  3                     *      *                   *          *             A  *              *  B               o-------* * *-------o             * |  *:::::::::::*  | *       3  *  | *:::::::::::::::* |  *  3         *    |*:::::::::::::::::*|    *       *      |:::::::::::::::::::|      *     *        *:::::::::::::::::::*        * H o          *:::::::::o:::::::::*          o F     *        *:::::::::::::::::::*        *       *      |:::::::::::::::::::|      *       3 *    |*:::::::::::::::::*|    * 3           *  | *:::::::::::::::* |  *             * |  *:::::::::::*  | *               o-------* * *-------o             D  *              *  C                   *          *                   3 *      * 3                       *  *                         o                         G
Is the correct answer: $\tfrac{9\pi}{2}\:m^2\,\text{ or }\,9\pi\:m^2\,?$

I'd like to know how you got those two answers . . .

The circle is inscribed in square $ABCD$
. . which is inscribed in square $EFGH$ as shown.

The top triangle $ABE$ is an isosceles right triangle with legs of length 3.
. . Hence: $AB = 3\sqrt{2}$

But $AB$ is the diameter of the circle.
. . Hence, the radius is: . $r \:=\:\frac{3\sqrt{2}}{2}$

Now you can find the area of the circle . . . right?