Here is an old geometry/measurement chestnut for you.

[Whateverist]

My take-

---

---

Not convinced that point "D" is required.

Yeah, I think you're right.  When I was writing it (I didn't do a first draft) I thought I might want it to refer to the part I wanted the area of but then I realized the "part of the second circle not inside the quarter circle" would get it done.

Edited to say I think there is an issue with your final answer but the wifey just got back from her trip so I can't figure out what it is right now, but I bet you can.

[GrandizerII]

---

sqrt(2) is the measure of the length of AB (Pythagoras' theorem)

The area of the circle with diameter AB is pi*r^2 = pi*(sqrt(2)/2)^2 = pi*(2/4) = 1/2 pi

The area of the quarter circle = pi*r^2/4 = pi*1^2/4 = 1/4 pi

The area to be calculated = the area of the circle with diameter AB - the area of the quarter circle = 1/2 pi - 1/4 pi = 1/4 pi

---

D'oh, I did the same mistake as another poster here. Yeah, polymath's answer makes sense.

[Fireball]

My take-

---

---

Not convinced that point "D" is required.

Yeah, I think you're right.  When I was writing it (I didn't do a first draft) I thought I might want it to refer to the part I wanted the area of but then I realized the "part of the second circle not inside the quarter circle" would get it done.

Edited to say I think there is an issue with your final answer but the wifey just got back from her trip so I can't figure out what it is right now, but I bet you can.

Yeah, I took the original diameter wrong. My answer should be twice what I originally came up with.

[polymath257]

An old problem of Hipparchus.

---

The area of the half-circle ABD is the same as the area of the quarter circle. Subtract the area outside of the triangle and the quarter circle from both.
The answer is the same as the area of the triangle OAB where O is the center of the quarter circle. In other words, 1/2.

---

Thanks for the Hipparchus reference.  I didn't know that and will have to look into who that was.  I don't recall where I encountered it but for whatever reason I woke up too early this morning thinking about it and had to recreate it.

I love the elegance of the solution though and how unlikely it seems when you consider the shape whose area we must find in itself.  Very satisfying I think.

There is a generalization due to al-Haytham. Take a large half-circle with diameter AB. Inscribe a triangle ABC where C is on the half-circle. Now, draw two half circles with diameters AC and BC outside of the triangle.


The question concerns the area outside the original circle and inside the two smaller circles.  

The Hipparchus figure is where the point C is on a perpendicular bisector of AB.

---

The area of the two small half-circles is the same as the area of the large half circle because the triangle is a right triangle. So the area of the two 'lunes' is the same as the area of the triangle ABC.

---

[I_am_not_mafia]

Draw a quarter circle and label the center point C and the points on either end of its arc, A and B.

Let the length of the quarter circle's radius = 1 unit 

Now draw the circle with diameter AB, and label another point on this circle D such that point D is not inside the original quarter circle CAB.

Find the area of circle ABD outside of quarter circle CAB.  Place answer inside hide tags and justify your result.

What's in it for me?

[GrandizerII]

Draw a quarter circle and label the center point C and the points on either end of its arc, A and B.

Let the length of the quarter circle's radius = 1 unit 

Now draw the circle with diameter AB, and label another point on this circle D such that point D is not inside the original quarter circle CAB.

Find the area of circle ABD outside of quarter circle CAB.  Place answer inside hide tags and justify your result.

What's in it for me?

Half a quarter of pie?

[Whateverist]

Thanks for the Hipparchus reference.  I didn't know that and will have to look into who that was.  I don't recall where I encountered it but for whatever reason I woke up too early this morning thinking about it and had to recreate it.

I love the elegance of the solution though and how unlikely it seems when you consider the shape whose area we must find in itself.  Very satisfying I think.

There is a generalization due to al-Haytham. Take a large half-circle with diameter AB. Inscribe a triangle ABC where C is on the half-circle. Now, draw two half circles with diameters AC and BC outside of the triangle.


The question concerns the area outside the original circle and inside the two smaller circles.  

The Hipparchus figure is where the point C is on a perpendicular bisector of AB.

Oh goody.  We're getting ready to go to the Y but I should have time for it in a few hours.  

To really make the Hipparchus figure fit this generalization we just need to put two of the original quarter circles back to so that solution to that problem is double what we had before.  That solution should then hold even if the half circles are not congruent.

First thoughts.

---

The triangle inscribed must be a right triangle where the diameter is its hypotenuse.  So the area of a square drawn on the hypotenuse must equal in area the sum of the squares on the other two sides.  But it can be shown that the area of a semicircle drawn on the hypotenuse must also equal in area the sum of the semicircles drawn on the other two sides. So we just need to show that the two semicircles drawn on the legs have the same area as the area of the inscribed triangle.

---

---

Yeah, I think you're right.  When I was writing it (I didn't do a first draft) I thought I might want it to refer to the part I wanted the area of but then I realized the "part of the second circle not inside the quarter circle" would get it done.

Edited to say I think there is an issue with your final answer but the wifey just got back from her trip so I can't figure out what it is right now, but I bet you can.

Yeah, I took the original diameter wrong. My answer should be twice what I originally came up with.

Yup.

---

Draw a quarter circle and label the center point C and the points on either end of its arc, A and B.

Let the length of the quarter circle's radius = 1 unit 

Now draw the circle with diameter AB, and label another point on this circle D such that point D is not inside the original quarter circle CAB.

Find the area of circle ABD outside of quarter circle CAB.  Place answer inside hide tags and justify your result.

What's in it for me?

Well, a chestnut .. obviously.  

[Whateverist]

Well I misspoke in my first thoughts on the more general problem.

---

What I should have said needed to be shown is that the sum of the semicircles was equal in area to the semicircle containing the inscribed triangle.

But that semicircle is just the reflection across the diameter of the semicircle on the inscribed circle's diameter, so I guess my work is done, since then the area of interest can be shown to be equal to the area of the triangle.

---

This more general problem would seem considerably more complex without having considered the easier form first.