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## #1 2012-03-04 03:08:46

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Area above a chord.

Hi;

The question is to find the shaded area between the rectangle and the top part of the circle.

Let's see if Geogebra can help.

1)Draw 4 points (0,0),(0,.53)(1.53,.53)(1.53,0) and call them A,B,C and D.

2)Draw point E at (.765, .35)

3)Use the circle through 3 points tool and draw a circle through A,E and D

4)Draw a line through E and perpendicular to the x axis.

5)Find the point of intersection between the perpendicular line and the circle. Call the bottom point H.

6)Find the midpoint between E and H and call it I. This is the midpoint of the circle.

7)Measure IH with the measuring tool. That is the radius of the circle. You should get 1.01104

8)Measure the angle DIA with the angle tool, you should get 98.33948 degrees.

Now we can use the area above a chord formula

Plugging in r = 1.01104 and θ = 98.33948  we get 0.37153. Now we subtract from .8109 - 0.37153 = 0.439369, which is close. See the drawing to check yours.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #2 2012-03-04 03:13:44

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### Re: Area above a chord.

Hi bobbym

I don't think that's a semicircle.

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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## #3 2012-03-04 03:34:16

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Area above a chord.

Hi;

Yes, I know I have corrected it, thanks.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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