# How can I find the area of the shaded region given by two semicircle and a line segment? [closed]

I don't know how to start a problem of this type. How can I find the area of the shaded region?

## closed as off-topic by verret, Jean-Claude Arbaut, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Paul FrostApr 17 at 15:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – verret, José Carlos Santos, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.

• How are $A, B$ connected? another semi-circle? straight line? the picture is unclear. – Yuval Gat Apr 14 at 11:25
• A,B are connected by straight line – sharan kumar Apr 14 at 11:33
• Please write in your post the informations on the question and your attempted solution. Furthermore, I don't think that I have enough information to be able to solve it. Do you have more information about the straight line or the points $A$ and $B$? I also suggest that you take a look at the related questions that appear on the right side of the site for inspiration on the solution. – Ertxiem Apr 14 at 11:59
• @Ertxiem We do have enough information. – Rhys Hughes Apr 14 at 12:02
• @Blue I'm sorry, you're right. I missed that. I'll delete my comment to avoid confusion. – Ertxiem Apr 14 at 12:33

Here's a general solution. I've exchanged $$A$$ and $$B$$ so that my $$a-b$$ is non-negative; also, I've exaggerated the scale a bit.

Let $$\overline{AB}$$ meet $$\bigcirc{A'}$$ again at $$C$$, and let $$D$$ be the point where the semicircles meet. Then, clearly

$$|\text{shaded region}| = |\square ABB'A'|-|\triangle AA'C|-|\text{sector } CA'D|-|\text{sector } BB'D| \tag{1}$$

As $$\square ABB'A'$$ is a trapezoid, we have $$|\square ABB'A'| = \frac12(|AA'|+|BB'|)|A'B'|=\frac12(a+b)^2 \tag{2}$$ Also, defining $$\theta := \arctan\frac{a-b}{a+b}$$, we have \begin{align} |\triangle A'AC| &= \frac12a^2\sin 2\theta \tag{3}\\[4pt] |\text{sector }CA'D| &= \frac12a^2\angle CA'D = \frac12a^2\left(\frac{\pi}{2}-2\theta \tag{4} \right) \end{align} And, of course, $$|\text{sector }BB'D| = \frac{\pi}{4} b^2 \tag{5}$$ Therefore,

$$|\text{shaded region}| = \frac12\left(\;(a+b)^2-a^2\sin 2\theta-\left(a^2+b^2\right)\frac{\pi}{2} +2a^2\theta\;\right)\tag{6}$$

But we can do a little better, since \begin{align}\sin2\theta &= 2\sin\theta\cos\theta = 2\cdot\frac{a-b}{|AB|}\cdot\frac{a+b}{|AB|} = 2\cdot\frac{a^2-b^2}{|AB|^2}= 2\cdot\frac{a^2-b^2}{(a+b)^2+(a-b)^2} \\ &= \frac{a^2-b^2}{a^2+b^2}\tag{7} \end{align}

This gives, after a little clean-up,

$$|\text{shaded region}| = \frac12\left(\;\frac{b(a+b)(2a^2+ab+b^2)}{a^2+b^2}-\left(a^2+b^2\right)\frac{\pi}{2}-a^2\arcsin\frac{a^2-b^2}{a^2+b^2}\;\right) \tag{8}$$

Specifically, with $$a=5$$ and $$b=3$$, we have

$$|\text{shaded region}| = \frac{444}{17}-\frac{17}{2}\pi+ \frac{25}{2}\arcsin\frac{8}{17} = 5.53858\ldots \tag{9}$$

• Nice solution +1. Small typo: $\frac12a^2\left(\frac{\pi}{\color{red}{2}}-2\theta \right)$, but it was recovered in the following steps. – farruhota Apr 15 at 11:36

Circles are placed on the co-ordinate system as shown. Letting the center of the smaller circle be at O=(0, 0) will shorten the later calculation a bit because A = (0,3).

(1) Setup the equation of the straight line AB.

(2) Setup the equation of the circle (centered at K with radius = 5).

(3) Solving the equations in (1) and (2) will give the coordinates of X.

(4) The coordinates of Y is then known. This means $$\angle PKX$$ can also be known.

(5) The required area = [quad OAXK] – [sector OAP] – [sector KPX].

• what application did you use to draw your diagram? Thanks in advance :) – Fareed AF Apr 14 at 12:45
• @FareedAF I am using Geogebra. Its *.ggb format is not readily accepted by this site. You have to change it to *.png first before uploading. – Mick Apr 14 at 12:46

Connect $$A$$ with the center of the semicircle on the left (say $$O$$) and connect $$B$$ with the center of the semicircle on the right (say $$O'$$)

Now $$ABO'O$$ is a trapezoid (since $$(AO)$$ is parellal to $$(BO')$$)

Area of $$ABO'O =A_{ABO'O}$$

$$A_{ABO'O}=\frac{h(b_1+b_2)}{2}=32$$

Finally, $$A_{\text{shaded part}}=A_{ABO'O} - \frac{1}{4}(A_{C_1} +A_{C_2}) = 32-\frac{1}{4}(9\pi+25\pi)=\frac{64-17\pi}{2}$$

Refer to the figure:

$$\hspace{5cm}$$

Introduce the coordinate system: $$F(0,0),A(-3,3),B(5,5)$$. The equation of the line $$AB$$: $$y=\frac14x+\frac{15}{4}.$$ The coordinate of $$E$$: $$\begin{cases}y=\frac14x+\frac{15}{4}\\ (x-5)^2+y^2=5^2\end{cases} \Rightarrow \begin{cases}x=\frac{45}{17}=FG\Rightarrow DG=5-x=\frac{40}{17}\\ y=\frac{75}{17}=EG\end{cases}.$$

The area of the sector $$EDF$$: $$S_{sector \ EDF}=\frac12\cdot 5^2\cdot \arcsin \frac{15}{17}$$ The area of $$EFG$$: $$S_{EFG}=S_{sector \ EDF}-S_{EDG}$$ Finally, the required area: $$S_{AEF}=S_{ACGE}-S_{sector \ ACF}-S_{EFG}=$$ $$\frac{3+\frac{75}{17}}{2}\cdot (3+\frac{45}{17})-\frac{9\pi}{4}-\left(\frac12\cdot 5^2\cdot \arcsin \frac{15}{17}-\frac12\cdot \frac{75}{17}\cdot \frac{40}{17}\right)\approx 5.53857608.$$

Calculate the area of the whole shape by splitting it into 3 regions, 2 quarter circles on the left and right and the centre portion between A and B. Then subtract the 2 semicircles.

• Bear in mind that $\overline{AB}$ intersects the larger circle, which complicates the solution. – Blue Apr 14 at 12:13
• Oh yeah missed that. You can still do the same thing. First you have to calculate the coordinate of the intersection point which is just solving the quadratic then do the same splitting up idea. You will need to extend the right quarter circle a bit with some more shapes but the principle is the same. – G Aker Apr 14 at 12:18