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Simple Physics Homework Help Pleasee

  • #1

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 im not sure what else to do
 
Last edited by a moderator:

Answers and Replies

  • #2
any help would be greatly appreciated!
 
  • #3
1,137
0
What is D?
 
  • #4
  • #5
1,137
0
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
 
  • #6
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 doesnt make any sense to me
 
  • #7
the diameter is sqrt(17a), i jsut cant figure out how to get a by itself so i can do 81 * a^2 for the area of the grid
 
  • #8
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
 
  • #9
754
1
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]

but [itex]D^2 = x^2 + y^2 = (4a)^2 + (3a)^2 = 16a^2 + 9a^2 = 25a^2[/tex], therefore

[tex]A = \frac{25a^2 \pi}{4} = 47.4 ~ cm^2[/tex]

Simplifying, we get

[tex]a^2 = \frac{4 (47.4 ~ cm^2)}{25 \pi} = \frac{189.6 ~ cm^2}{25 \pi} = \frac{7.584}{\pi} ~ cm^2[/tex]

Area of the square is then

[tex]81a^2 = 81 \cdot \frac{7.584}{\pi} ~ cm^2 = \frac{614.304}{\pi} ~ cm^2 \approx 195.539 ~ cm^2[/tex]
 
  • #10
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]

but [itex]D^2 = x^2 + y^2 = (4a)^2 + (3a)^2 = 16a^2 + 9a^2 = 25a^2[/tex], therefore

[tex]A = \frac{25a^2 \pi}{4} = 47.4 ~ cm^2[/tex]

Simplifying, we get

[tex]a^2 = \frac{4 (47.4 ~ cm^2)}{25 \pi} = \frac{189.6 ~ cm^2}{25 \pi} = \frac{7.584}{\pi} ~ cm^2[/tex]

Area of the square is then

[tex]81a^2 = 81 \cdot \frac{7.584}{\pi} ~ cm^2 = \frac{614.304}{\pi} ~ cm^2 \approx 195.539 ~ cm^2[/tex]
those numbers are for the example not the circle at the top, my online submittion says it was right soo yeah.
 
  • #11
754
1
My bad!

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

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