I recently saw a similar problem online but found the area in a completely different way.


There is a unit square, i.e. a square with side length equal to $\text 1$. Two lines are placed inside the square. Lets call them $\color{red}{\text {Line 1}}$ and $\color{blue}{\text {Line 2}}$.

$\color{red}{\text {Line 1}}$ is straight - It starts at a vertex and intersects the square exactly half way on the opposite edge.

$\color{blue}{\text {Line 2}}$ is curved - It is a quarter circle with radius, $r$, the same length as the square, i.e. $r=1$.

The diagram below shows the representation of the description provided above (I am not a geometric mathematician so forgive my elementary descriptions - please):

enter image description here

The goal is to find the area $A$.

My Attempt:

My intuition was to straight away subtract areas. So find the area of the quarter circle and subtract off the area under the intersection of $\color{red}{\text {Line 1}}$ and $\color{blue}{\text {Line 2}}$. So in the following diagram I find area $A+B+C$ and subtract area $B$ and area $C$:

enter image description here

Now clearly the area $A+B+C$ is simply the area of a quarter circle. This is: $$Area(A+B+C)=\frac {\pi r^2}4=\frac \pi4$$

Next, I find the point of intersection in the Cartesian plane of the $\text 2$ lines (where the red and blue lines intersect). I assume the bottom left point on the square represents the coordinates $(0,0)$.

Therefore the $\color{red}{\text {Line 1}}$ has equation: $$y=-2x+1$$

And $\color{blue}{\text {Line 2}}$ lines has equation: $$y=\sqrt{1-(x-1)^2}$$

The intersection is at: $$-2x+1=\sqrt{1-(x-1)^2} \implies (-2x+1)^2=1-(x-1)^2$$ $$\implies 4x^2-4x+1=-x^2+2x$$ $$\implies 5x^2-6x+1=0 \implies (5x-1)(x-1)=0 \implies x=-\frac 15$$

Then we have $y=-2x+1=-2\frac 15 +1 = \frac 35$ so the coordinates of intersection (inside the square) are $\left(\frac 15, \frac 35 \right)$.

Then the area $B$ is simply the area of a triangle, with height $h$ and base $b$: $$h=\frac 35 \ , \ b=\frac 12 - \frac 15= \frac 3{10}$$ $$\therefore Area(B)=\frac 12 bh= \frac 12 \frac 3{10} \frac 35= \frac 9{100}$$

Now for area $C$, this is simply the area under the curve of a (semi-) circle between the $x$-coordinates $x=0$ and $x=\frac 15$. Clearly this is not the neatest integral, so we just shift the circle coordinates and the area becomes:

$$Area(C)=\int_0^{1/5} \sqrt{1-(x-1)^2}dx$$ $$=\int_{-1}^{-4/5} \sqrt{1-x^2}dx=\left[ \frac 12 \left(x\sqrt{1-x^2}+\sin^{-1}(x) \right) \right]_{-1}^{-4/5}$$ $$=-\frac 6{25}+\frac \pi 4 - \frac 12 \sin^{-1}\left(\frac 45 \right)$$

Putting it all together we get:

$$Area(A)=Area(A+B+C)-Area(B)-Area(C)$$ $$=\frac \pi4- \frac 9{100} - \left[-\frac 6{25}+\frac \pi 4 - \frac 12 \sin^{-1}\left(\frac 45 \right)\right]$$ $$=\frac 3{20}+\frac 12 \sin^{-1}\left(\frac 45 \right)$$ $$\approx 0.61365$$

My Question:

Is this the best approach? Have I done this right? Is there a simpler or neater method? And what would change if we were to calculate any of the other 3 areas?

Thank you in advance.


There is a simpler approach. Consider an extension of the diagram you drew: enter image description here

Note that the desired area "A" is the sum of the areas of the circular sector and the obtuse triangle. As shown in the diagram, the angle subtended by an arc from the vertex on the circumference is half of the angle measured from the center of the circle. Since $\theta = \tan^{-1} \frac{1}{2}$, it follows that $2\theta = 2\tan^{-1} \frac{1}{2}$ and the area of the circular sector is simply $$A_1 = \tan^{-1} \frac{1}{2}.$$ Then the area of the obtuse triangle is $$A_2 = \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot \sin \left(\frac{\pi}{2} - 2\theta\right) = \frac{1}{4} \cos \left(2 \tan^{-1} \frac{1}{2} \right) = \frac{3}{20}.$$ Thus the total area is $$A = \tan^{-1} \frac{1}{2} + \frac{3}{20} \approx 0.613648.$$

  • $\begingroup$ Nice that makes perfect sense and is amazing :) Thank you! $\endgroup$ Jun 16 '18 at 6:18
  • $\begingroup$ Two points for clarification: the obtuse triangle is the smaller triangle in the top square and thus the black length in this triangle is $\frac{1}{2}$. I agree that the evaluation of $\frac{1}{4} \cos (2 \arctan 1/2)$ should be left to the reader. $\endgroup$
    – Toby Mak
    Dec 20 '21 at 2:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.