1. ## I'm stuck with this question about circle

Only (a) I need help with.

Thank you very much .

2. ## Re: I'm stuck with this question about circle

Hint: draw 2 radius lines from center of circle to ends of water line;
then you'll get a 3-7-7 triangle...

3. ## Re: I'm stuck with this question about circle

You can represent the pipe as $\displaystyle x^2+ y^2= 49$ by setting up a coordinate system with origin at the center of the pipe. Since the maximum depth of the water is 2 cm, its surface is given by y= -7+ 2= -5. The lower half of the pipe is given by $\displaystyle y= -\sqrt{49- x^2}$. The surface of the water crosses the circle when $\displaystyle y= -5= -\sqrt{49- x^2}$ so $\displaystyle 25= 49- x^2$, $\displaystyle x^2= 49- 25= 24$. $\displaystyle x= -2\sqrt{6}$ and $\displaystyle x= 2\sqrt{6}$. Integrate $\displaystyle -5- (-\sqrt{49-x^2})= \sqrt{49- x^2}- 5$ from $\displaystyle x= -2\sqrt{6}$ to $\displaystyle x= 2\sqrt{6}$ to find the area.

4. ## Re: I'm stuck with this question about circle

http://mathworld.wolfram.com/CircularSegment.html

Area = $7^2\arccos\left(\dfrac{5}{7}\right)-5\sqrt{24}$