Thread: Area of a circle joining 2 connected lines

1. Area of a circle joining 2 connected lines

2 lines of length L1 and L2 (not equal) are at right angles. At the ends furthest from the right angle, a curve, a section of a circle of radius r, joins the ends to form a closed figure. What is the method in terms of L1, L2 and r to determine the area of the figure?

3. Re: Area of a circle joining 2 connected lines

Originally Posted by jeffhobson
2 lines of length L1 and L2 (not equal) are at right angles. At the ends furthest from the right angle, a curve, a section of a circle of radius r, joins the ends to form a closed figure. What is the method in terms of L1, L2 and r to determine the area of the figure?
There isn't any method. You only can determine the length of x (if the lengthes of $L_1$ and $L_2$ are known!) and that the $r\geq x$.

4. Re: Area of a circle joining 2 connected lines

Two obvious points: the area of the triangle is $(1/2)L_1L_2$ and the length of the hypotenuse is $h= \sqrt{L_1^2+ L_2^2}$. So the rest of the problem is to find the area of a portion of a circle of radius r, outside a chord of length h.

One way to analyze that is to use the "cosine law". The hypotenuse of the right triangle and two radii of the circle form an isosceles triangle with two sides of length r and one side of length x. According to the cosine law, $x^2= 2r^2(1- cos(\theta))$ where $\theta$ is the central angle. That tells us that the central angle is given by $\theta= arccos(1- \frac{x^2}{r^2})$. The area of a "sector" of a circle of radius r and central angle $\theta$ is $\frac{1}{2}\theta r^2$ so the area of the sector here is $\frac{1}{2}r^2 arccos(1- \frac{x^2}{r^2})$. That isosceles triangle, having two sides of length r and one of length x, can be divided into two right triangles each having hypotenuse r and one leg of length x/2 so the other leg has length $\sqrt{r^2- x^2/4}$. The area of that isosceles triangle, then, is $x\sqrt{r^2- x^2/4}$ and so the area of the circular sector beyond the chord has length $\frac{1}{2}r^2arccos(1- \frac{x^2}{r^2})- x\sqrt{r^2- x^2/4}$.