Find the area bounded by these 4 arcs

[chocopop]

Homework Statement

Four quarter circles were drawn from the four vertices of the square as shown below. If the side length of the square has a measure of 12±0.05, find the area of the shaded region.

Relevant Equations

Find an equilateral triangle in the figure
Can be solved through 3 equations, 3 unknowns

 

[berkeman]

Homework Statement:: Four quarter circles were drawn from the four vertices of the square as shown below. If the side length of the square has a measure of 12±0.05, find the area of the shaded region.
Relevant Equations:: Find an equilateral triangle in the figure
Can be solved through 3 equations, 3 unknowns

View attachment 271062

Welcome to PhysicsForums.

As you know, you are required to show your best efforts to start the problem before we can offer tutorial assistence. How will you approach this problem? What sort of integral will you set up? What coordinate system (rectangular, polar, etc.) would probably be best for this problem?

(Or, it sounds like you have been given a hint in the problem that you can use and not even have to integrate?)

 

[etotheipi]

It does look a little bit fiddly. By equilateral triangle I presume they're referring to the one joining two corners to the intersection of their respective arcs. You can find the area of that triangle, and then also the little segment at the end of the sector. Might also help if you give the 3 different shapes in the figure names like $A_1$, $A_2$, $A_3$ respectively, or something, so that you can start writing some (3) equations in terms of $r$.

[P.S. If you get fed up of that, you can just do an integral$$A_{\text{shaded}} = \int_{1-\frac{\sqrt{3}}{2}}^{\frac{1}{2}} 2\sqrt{1-(x-1)^2} -1 \, dx$$That should yield pretty easily to a sneaky $\sin{u} = (x-1)$ substitution. But technically that's not a hint, because you're supposed so solve it geometrically. So please don't ban me @berkeman ]

 

[chocopop]

It does look a little bit fiddly. By equilateral triangle I presume they're referring to the one joining two corners to the intersection of their respective arcs. You can find the area of that triangle, and then also the little segment at the end of the sector. Might also help if you give the 3 different shapes in the figure names like $A_1$, $A_2$, $A_3$ respectively, or something, so that you can start writing some (3) equations in terms of $r$.

[P.S. If you get fed up of that, you can just do an integral$$A_{\text{shaded}} = \int_{1-\frac{\sqrt{3}}{2}}^{\frac{1}{2}} 2\sqrt{1-(x-1)^2} -1 \, dx$$That should yield pretty easily to a sneaky $\sin{u} = (x-1)$ substitution. But technically that's not a hint, because you're supposed so solve it geometrically. So please don't ban me @berkeman ]

i don't know how to get it

 

[etotheipi]

Okay. Let the shaded square have area $A_1$, let the four arrow-shaped things have areas $A_2$ each, and the four other shapes along the edges have areas $A_3$ each.

The area of the square is $d^2$; what is the area of the square in terms of $A_1$, $A_2$ and $A_3$? Can you then write an equation?

Can you do the same for two other regions of choice, e.g. maybe one of the quarter circles, and one other region?

It's a little bit fiddly, but you'll need to show an attempt.

 

[chocopop]

Okay. Let the shaded square have area $A_1$, let the four arrow-shaped things have areas $A_2$ each, and the four other shapes along the edges have areas $A_3$ each.

The area of the square is $d^2$; what is the area of the square in terms of $A_1$, $A_2$ and $A_3$? Can you then write an equation?

Can you do the same for two other regions of choice, e.g. maybe one of the quarter circles, and one other region?

It's a little bit fiddly, but you'll need to show an attempt.

12(1+0.05)
=12.6
side x side
12.6 x 12.6
=158.76

is it correct, sir?

 

[etotheipi]

No, that's just the area of the large square. I refer you again to the questions posed in #5

 

[berkeman]

Unfortunately the OP has been banned from the PF because of many other thread starts today like this one that showed zero effort, despite repeated warnings from the Mentors. At least he/she showed a little effort in this thread. Thanks for trying to help.