A semicircle and a tangent line in a square

Question

In the following square, a semi-circle and a tangent line to it is drawn. What is the area of the shaded region?

I solved it as follows:

We have $\tan\alpha=1/2$, and area of the shaded triangle is equal to

$$\dfrac12\times10\times10\times\sin(90^{\circ}-2\alpha)=50\sin(90^{\circ}-2\alpha)$$

From $\tan\alpha=1/2$, we have $\tan(2\alpha)=\dfrac{1}{1-(1/2)^2}$ and $\tan(90^{\circ}-2\alpha)=1-(1/2)^2=3/4$.

Now, drawing the $3-4-5$ triangle, we can see that $\sin(90^{\circ}-2\alpha)=3/5$ and the area of the blue triangle is $30 \text{cm}^2$.

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I am looking for other methods to solve this problem without trigonometry.

Answer 1

Here's a simpler dissection approach, using $10$ sub-triangles of equal area ($8$ congruent red, $2$ congruent white; the dotted gray lines are guides).

Three of these fill the target triangle, so the target's area is $3/10$-ths the square's. $\square$

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Here's a (refinement of my previous) fun way ...

The square can be subdivided into $20$ sub-triangles: $16$ congruent "$1\times 2$" right triangles (shaded red), and $4$ congruent obtuse triangles (shaded blue). These triangles all have the same area. (Compare the ones marked with dots.)

Six sub-triangles fill the target triangle, so the ratio of the target's area to the square's is $6:20 = 3:10$. $\square$

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Here's an alternative, also based on consideration of "$1\times2$" right triangles:

$$\begin{align} \frac{|\triangle A'BC|}{|\triangle ABC|}&=\frac{|A'M|}{|AM|}=\frac{3}{5} \\[4pt] \to\qquad |\triangle A'BC|&=\frac35|\triangle ABC|=\frac35\cdot\frac12|\square ABCD|=\frac{3}{10}|\square ABCD| \end{align}$$

Answer 2

Bisector $CE$ (see figure below) cuts $FD$ at its midpoint $H$. Triangles $CFE$ and $CFH$ are similar and we thus get $FH=2\sqrt5$ and $FA=\sqrt{AD^2-FD^2}=2\sqrt5$.

$AG$ is the same as the altitude of triangle $DFA$, hence $AG=4$ and $BG=6$. The area of triangle $BCF$ can then be computed as ${1\over2}BC\cdot BG=30$.

Answer 3

Maybe not the most elegant but a little coordinate geometry never hurts. Set the origin at the center of the left edge of the square. We will try to find the coordinates $(x,y)$ of the intersection of the circle and triangle. At this point, the circle has tangent $(y, -x)$. A point on the tangent is also $(10,5)$ so we have for some $c \in \mathbb{R}$ that \begin{align} x+cy &= 10 \\ y-cx &= 5\\ x^2+y^2 &=25 \end{align} which is easily seen to have $(x,y) = (4,-3)$ as the only reasonable solution. Thus, the answer is simply $$\frac{1}{2} \times 10 \times (10-4) = 30$$ as desired.

Answer 4

$\triangle ABE\cong\triangle BEF$

$S_{ABFE}=2S_{ABE}=\frac{5\times 10}2=50$

$FG=\frac12 EF$

$EF=\frac 12 BF$

$\Rightarrow S_{EFG}=\frac14S_{BEF}=\frac{25}4$

$S_{EDGF}=2\times S_{EFG}=\frac {25}2$

$S_{BGC}=100-(50+12.5)=37.5$

$\triangle GFC=\triangle GTC$

$\triangle FMT=\triangle GFC$

$S_{M'O'F}=S_{MTC}$

$\Rightarrow S_{M'O'MCF}=2S_{GFC}$

$x=S_{BFC}=4S_{GFC}$

$S_{BMO'}=S_{GFC}\Rightarrow x=4S_{GFC}$

that is we have:

$x+\frac x 4=\frac54 x=37.5=4\times 7.5=30 m^2$

Answer 5

Another way is by adding height segments (may not be the shortest). In the figure $A'$ is the symmetric of $A$ with respect to $(OB)$. We find $A'M= 8$ then by Pythagoras $MB=6$ and the area is deduced.

The height values can be found by taking the areas of triangles in two different ways. For example, $5\sqrt{5}AD=50$, $DB=4\sqrt{5}$. $10MA'=4\sqrt{5}\times 4\sqrt{5}=80$. So $MA'=8$.

Answer 6

Extend $CF$ to meet $AB$ at $H$.

$\triangle{DEC}$ is similar to $\triangle{AHE}$. Therefore, $AH = HF = \frac12 AE = \frac14 AB = \frac14 CF$. $$S_{\triangle{BHC}} = \frac12 \times BC \times \frac34 AB = \frac38 S_{\text{square}}$$ $$S_{\triangle{CBF}} = \frac{CF}{CH} \times S_{\triangle{BHC}} = \frac45 \times \frac38 S_{\text{square}} = \frac{3}{10} S_{\text{square}}$$