Finding square area when inscribed circle has known area

[BraindeadX64]

Homework Statement

The circle inside the large square below has an area of 47.4 cm2.

[URL]http://tinyurl.com/PhysicsSquare[/URL]

Calculate the area of the large square. Note that the circle passes right through a number of grid intersection points.Heres the help the professor gave

[URL]http://tinyurl.com/PhysicsSquare2[/URL]

Homework Equations

C^2 = A^2 + B^2

D= sqrt(x^2 + b^2)

D = 2sqrt(A/pi)

The Attempt at a Solution

4a tall and 1a long

1st attempt.

d = sqrt((1a)^2 + (4a)^2)
d = sqrt(1a + 16a)
d= sqrt(17a)

2nd attempt
d=2sqrt(47.4/pi)
d= 7.77

stuck here I am not sure what else to do

 

[BraindeadX64]

any help would be greatly appreciated!

 

[cupid.callin]

What is D?

 

[BraindeadX64]

What is D?

distance formula for diameter

 

[cupid.callin]

well i haven/t checked your eqn but from the 2 results you got,,, equate them

you get 17a = 7.77²
find a

now you know that area of large Square is 81a²

substitute a

 

[BraindeadX64]

well i haven/t checked your eqn but from the 2 results you got,,, equate them

you get 17a = 7.77²
find a

now you know that area of large Square is 81a²

substitute a

well those answers i got were different numbers for radius and it doesn't make any sense to me

 

[BraindeadX64]

the diameter is sqrt(17a), i just can't figure out how to get a by itself so i can do 81 * a^2 for the area of the grid

 

[BraindeadX64]

figured it out

a = 7.77cm / (sqrt (17) = 1.88cm

Area of the box = (9a*9a) =81a^2 =81*(1.88)^2 = 286.3 cm^2

 

[zgozvrm]

figured it out

a = 7.77cm / (sqrt (17) = 1.88cm

Area of the box = (9a*9a) =81a^2 =81*(1.88)^2 = 286.3 cm^2

Uh, no.

Area of a circle is [itex]A = \pi r^2[/tex] and r = D/2 so, [itex]A = \pi D^2/4[/tex]<br /> <br /> but [itex]D^2 = x^2 + y^2 = (4a)^2 + (3a)^2 = 16a^2 + 9a^2 = 25a^2[/tex], therefore<br /> <br /> $$A = \frac{25a^2 \pi}{4} = 47.4 ~ cm^2$$<br /> <br /> Simplifying, we get<br /> <br /> $$a^2 = \frac{4 (47.4 ~ cm^2)}{25 \pi} = \frac{189.6 ~ cm^2}{25 \pi} = \frac{7.584}{\pi} ~ cm^2$$<br /> <br /> Area of the square is then<br /> <br /> $$81a^2 = 81 \cdot \frac{7.584}{\pi} ~ cm^2 = \frac{614.304}{\pi} ~ cm^2 \approx 195.539 ~ cm^2$$[/itex][/itex][/itex]

 

[BraindeadX64]

Uh, no.

Area of a circle is [itex]A = \pi r^2[/tex] and r = D/2 so, [itex]A = \pi D^2/4[/tex]<br /> <br /> but [itex]D^2 = x^2 + y^2 = (4a)^2 + (3a)^2 = 16a^2 + 9a^2 = 25a^2[/tex], therefore<br /> <br /> $$A = \frac{25a^2 \pi}{4} = 47.4 ~ cm^2$$<br /> <br /> Simplifying, we get<br /> <br /> $$a^2 = \frac{4 (47.4 ~ cm^2)}{25 \pi} = \frac{189.6 ~ cm^2}{25 \pi} = \frac{7.584}{\pi} ~ cm^2$$<br /> <br /> Area of the square is then<br /> <br /> $$81a^2 = 81 \cdot \frac{7.584}{\pi} ~ cm^2 = \frac{614.304}{\pi} ~ cm^2 \approx 195.539 ~ cm^2$$[/itex][/itex][/itex]

[itex][itex]$<br /> those numbers are for the example not the circle at the top, my online submittion says it was right soo yeah.$[/itex][/itex]

 

[zgozvrm]

My bad!

I didn't notice the circles were different and focused on the second one.
Your answer is correct.