Find the value of x-y of the given shape ("In the diagram, CD = CE = 30 and F is the midpoint of CE....")

[John Ode]

In the diagram, CD = CE = 30 and F is the midpoint of CE. Two quarter circles are drawn: one with centre C and passing through D and E, and the other with centre F and passing through E. Let x be the area of the region that is inside rectangle GDCF and outside the larger quarter circle. Let y be the area that is inside the larger quarter circle, outside the smaller quarter circle, and outside rectangle GDCF. Let d be the positive difference between x and y. What is the integer closest
to d?

I tried solving it so the numbers on diagram might be wrong.

 

[Dr.Peterson]

In the diagram, CD = CE = 30 and F is the midpoint of CE. Two quarter circles are drawn: one with centre C and passing through D and E, and the other with centre F and passing through E. Let x be the area of the region that is inside rectangle GDCF and outside the larger quarter circle. Let y be the area that is inside the larger quarter circle, outside the smaller quarter circle, and outside rectangle GDCF. Let d be the positive difference between x and y. What is the integer closest
to d?

I tried solving it so the numbers on diagram might be wrong.

View attachment 37304

Please show the work you did, including a marked up diagram (e.g. an additional radius or two), so we can see where you need help.

 

[John Ode]

Okay, I started the areas of the two shapes, since F is the midpoint of CE we know that EF is 15 and CF is 15

the area of a semi is calculated as 1/4pir^2 the small semicircle has radius 15 so 1/4pi(15)^2 = 56.25pi

thé areà of rectangle is calculated by l X w thé w is CF which is 15 and height is CD which is 30 so 15 x 30 = 450

the big semicircle is calculated by 1/4pi(30)^2 = 225pi

then we know that 225pi - 56.25pi - 450 + X = y one of the missing areas

simplify and you get 80.14 + X = y

The question asks us to find difference of Y - X = d and we found that y = 80.14 + X so we get

80.14 + X -X = d
X - X = 0 so we are left with
80.14 = d
the closet integer to d is 80

therefore the closet positivé intégrer difference is 80 but I’m not sure if I’m right.

 

[The Highlander]

In the diagram, CD = CE = 30 and F is the midpoint of CE. Two quarter circles are drawn: one with centre C and passing through D and E, and the other with centre F and passing through E. Let x be the area of the region that is inside rectangle GDCF and outside the larger quarter circle. Let y be the area that is inside the larger quarter circle, outside the smaller quarter circle, and outside rectangle GDCF. Let d be the positive difference between x and y. What is the integer closest
to d?

I tried solving it so the numbers on diagram might be wrong.

View attachment 37304

Hi @John Ode,

Welcome to FMH.

If you can determine the area of either x or y then the problem becomes one of simple additions & subtractions.

The area I have shaded (in green) on your diagram may be determined by Integration, however, if all you are capable of is finding the area of a rectangle then that may be beyond your skill level.

You say you "tried solving" the problem but the only thing I see on your diagram, other than what was given in the question, is what looks to me like 756 (or 758) but I presume this is meant to be 450 (since that is the correct value for the area of the rectangle GDCF). Could you not even calculate the areas of the quarter circles?

As I say, one way to get the shaded region is to find the Area Under the Curve by Integration and you could use the following equation to do so:-

[math]y=\sqrt{900-x^2}[/math]
That would be the equation for the semi-circle that would be formed if your quarter circle CDE was sitting on the x-axis (with C at the Origin) and reflected in the y-axis. You could then use the boundaries: [-15, 0] or [0, 15]; either set would give the same area.

So you would want to evaluate:-

[math]\int_{-15}^{0}\sqrt{900-x^2}\,dx[/math]

or​

[math]\int_{0}^{15}\sqrt{900-x^2}\,dx[/math]
However, if your capabilities don't run to much more than finding the area of rectangles then...

The area of a quarter circle is given by: \(\displaystyle \frac{1}{4}\pi r^2\)

and if I just tell you that the area of the shared region is: [imath]\left(75\pi+112.5\sqrt{3}\right)[/imath], would that help?
Or would you need to show how you arrived at that result?

Please come back and tell us whether you have now been able to complete the question or need further assistance.

In either case please show us your own working and indicate where you are stuck if you haven't managed to reach a solution.

Hope that helps.

 

[John Ode]

I tried send work but it seems it hasn’t been approved yet