Question

A circle with a radius of 2 is centered on "O". Square OABC has sides with length of 1. Sides AB and CB extend past B to meet the circle at D and E; respectively. What is the exact area of this region (shown by arrow), which is bound BD, BE and minor arc connecting D and E

Answer 1

See attached image

Answer 2

Find the area using calculus or geometry/trig ?

Answer 3

Hi Ranga... Geometry

Answer 4

Your drawing is better than mine!
Right on point

Answer 5

Okay. The equation for the circle is: x^2 + y^2 = 2^2 = 4.
You can find the coordinates of points B, D and E.
B is (1,1). D is (1, ?) and E is (?, 1). Can you fill in the "?" marks?

Answer 6

Thinking...

Answer 7

x^2 + y^2 = 4
when x = 1: 1 + y^2 = 4. y^2 = 3. y = +\(\sqrt 3\).
So D is (1, \(\sqrt 3\)) and E is (\(\sqrt 3\), 1).

Using the above coordinates you can find the area of the triangle DBE.
What is the rea of the triangle DBE?

Answer 8

y^2 = 3 should give y =\(\pm \sqrt 3\) but we are interested only in the first quadrant and so I chose +\(\sqrt 3\).

Answer 9

is this right? \[\pi/3-\sqrt{3+1}\]

Answer 10

How did you get the above result?

Answer 11

that is what the answer key shows ... but I don't understand why.

Answer 12

I have to log off after this reply but I was thinking of first finding the area of the TRIANGLE DBE (1/2 * base * height) and adding the area of the sector DE above the straight line DE and adding it to the triangle area to find the total area that is asked.

Answer 13

ahhh... I think that is a very good lead.

Answer 14

thanks for much for taking the time to help me.

Answer 15

You are a very good teacher.

Answer 16

Area of triangle DBE = 1/2 * base * height = 1/2 * (\(\sqrt 3 -1\)) * (\(\sqrt 3 -1\)) = 1/2 * (\(\sqrt 3 -1)^2 = 1/2 * (3 + 1 - 2\sqrt 3) = 2 - \sqrt 3\) ------- (1)



Area of sector DE above the line joining D and E = Area of sector ODE - Area of isosceles triangle ODE = 1/2 * r^2 * \((\theta - \sin(\theta))\) ----- (2)
where \(\theta\) is the angle DOE.

We have to find \(\theta\) .
Slope of line OE = (1-0)/(\(\sqrt 3 - 0\)) = 1/\(\sqrt 3\). Therefore, OE makes 30 degrees with the x-axis.
Slope of line OD = (\(\sqrt 3 - 0\))/(1-0) = \(\sqrt 3\). Therefore, OD makes 60 degrees with the x-axis.
Therefore, \(\theta = 30\text{ degrees or }\pi /6 \text{ radians.}\)
(2) becomes: Area of sector DE above the line joining D and E = 1/2 * 4 * \((\pi / 6 - \sin(\pi /6))\) = 2 * \((\pi / 6 - 1/2) = \pi / 3 - 1.\) Add this to (1):
\( 2 - \sqrt 3 + \pi / 3 - 1 = \pi / 3 - \sqrt 3 + 1\).