Can't solve this elementary school problem involving area of circles

[shirozack]

Homework Statement

See attach

Relevant Equations

none

I can find the area of e and f but otherwise am stuck with many simultaneous equations. i don't think my approach is correct though considering it is an elementary school problem. it should be simpler?

1 semicircle area = π(52)/2 = 12.5π = 39.25 cm2

Square area = 10x10 = 100cm2

Square – 2 semicircle = 2 edge = 21.5, so 1 edge = 10.75

so 2 e + edge = semicircle

e = (39.25 – 10.75) / 2 = 14.25 cm2

so f = 39.25 – 14.25 = 25 cm2

area of quad = 1/4 * pi * 102 = 25pi = 78.5 cm2

c+d = quad – (e+f) = 39.25 cm2

b+c = semi – e = 25cm2

also, 39.25 – c = b+e = b + 14.25

so c = 25 - b

get rid of c by setting them equal so, 39.25 – d = 25 - b

so d = 14.25 + b

a + b + quad = square

a + b =21.5 cm2

a+d + b+c+e+f = square

a+d +64.25 = 100, a+d = 35.75

sub d in, a+14.25+b = 35.75, so a+b =

 

[docnet]

I'm getting $$\text{area}=200 \arcsin\left(\frac{4}{5}\right) + 50 \arcsin\left(\frac{3}{5}\right) - 125\pi + 200$$ as the exact solution using integration. I don't know if there is any simple trick.

 

[PeroK]

It looks tricky. One approach is to note that, with $R = 10cm$:
$$a = R^2 - (b + c + e) - (d+f+e + c) + (e + c)$$$$= R^2 - \frac{\pi R^2}{8} - \frac{\pi R^2}{4} + (e + c)$$The problem is then to calculate the area $(e + c)$. I've seen this before:



This gives a different expression from the one @docnet gave, but it's the same numerical answer.

 

[PeroK]

PS this shows that there is a difference between "elementary" and "simple"!

 

[FactChecker]

This is an elementary-school problem?!? Not in my book.

 

[Vanadium 50]

There are a lot of seemingly extraneous elements on this figure. Perhaps they are hints. Perhaps they are distractions. A high priority would be to try and figure this out. Personally - and I admit this is style - I would start with the only relevant elements and if I found a needed an auxiliary element, I'd add it.

I would not go charging in with numbers,'

I would sat it up as L = length of the small square (10 cm). For heavens sake, I would not call it R!

Then the shaded area is the area of the square minus the area of the big circle, minus the area of the small circle, plus the area of overlap (I double counted). It is probably worth noticing that the overlap region is two segments, back to back.

 

[PeroK]

I would sat it up as L = length of the small square (10 cm). For heavens sake, I would not call it R!

Whoever heard of the area of a circle being $\pi L^2$?

 

[Vanadium 50]

Whoever heard of an area of a square being R²?

 

[Lnewqban]

I can find the area of e and f but otherwise am stuck with many simultaneous equations.

You don't even need to calculate those areas, which are only distractions produced by the irrelevant vertical semicircle.

By using only trigonometry, I would calculate the areas of the two triangles formed (orange and cyan colored) and containing the area of interest.
Note the 1:2 proportion of the radii.

Then, would subtract the areas of the two circular segments.
Please, see:
https://en.wikipedia.org/wiki/Circular_segment

 

[Vanadium 50]

Then the shaded area is the area of the square minus the area of the big circle, minus the area of the small circle, plus the area of overlap (I double counted). It is probably worth noticing that the overlap region is two segments, back to back.

It may (or may not) be incrementally easier to work with triangles and sectors instead of segments.

Now that I have worked it through to the end, I do not see a simple way. There may be some room for cleverness in determining whether you write the answer in terms of arcsines, arcsines or arctangents.

 

[anuttarasammyak]

One approach is to note that, with R=10cm:
a=R2−(b+c+e)−(d+f+e+c)+(e+c)=R2−πR28−πR24+(e+c)The problem is then to calculate the area (e+c).

To go further trigonometry seems required.

 

[PhDeezNutz]

Formatting got messed up when I edited my post, gonna retype it

 

[PhDeezNutz]

$\text{Area} \left(E\right) = \text{Area} \left( \text{Square}\right) - \text{Area} \left( \text{semi-circle}\right) - \text{Area} \left(\text{quarter-circle} \right) + \text{Area} \left( B \right) + \text{Area} \left( C \right)$

Recall the area of a segment of a circle is $\frac{1}{2} r^2 \left( \theta - \sin \theta \right)$ so the above becomes

$\text{Area} \left(E\right) = 100 - \frac{25 \pi}{2} - 25 \pi + 50 \left(\theta_1 - \sin \theta_1 \right) + \frac{25}{2} \left( \theta_2 - \sin \theta_2 \right)$

So we have to solve for $\theta_1$ and $\theta_2$ we have the following system of linear equations (which doesn't have a singular solution which is why I'll use the law of Sines and half angle identities later). First equation is the sum of the angles of a quadrilateral. The next two equations are the sums of the angles of triangles. The fourth equation is a statement that $\theta_3$ and $\theta_4$ are compliments. The 5th and last equation can be derived from the other 4.



$\theta_1 + \theta_2 + 2 \theta_3 + 2 \theta_4 = 2 \pi$
$\theta_1 + 2 \theta_3 = \pi$
$\theta_2 + 2 \theta_4 = \pi$
$\theta_3 + \theta_4 = \frac{\pi}{2}$
$\theta_1 + \theta_2 = \pi$

Use Law of Sines

$\frac{ \sin \theta_1}{S} = \frac{\sin \theta_3}{10} \Rightarrow S = \frac{10 \sin \theta_1}{\sin \theta_3}$
$\frac{ \sin \theta_2}{S} = \frac{\sin \theta_4}{5} \Rightarrow S = \frac{5 \sin \theta_2}{\sin \theta_4}$

Setting $S = S$

$\frac{10 \sin \theta_1}{\sin \theta_3} = \frac{5 \sin \theta_2}{\sin \theta_4}$

$\frac{2 \sin \theta_1}{\sin \theta_3} = \frac{\sin \theta_2}{\sin\theta_4}$

Recall from the linear systems of equations that $\theta_3 = \frac{\pi - \theta_1}{2}$ and $\theta_4 = \frac{\pi - \theta_2}{2}$ so....

$\frac{2 \sin \theta_1}{\sin \left(\frac{\pi - \theta_1}{2} \right) } = \frac{\sin \theta_2}{\sin \left( \frac{\pi - \theta_2}{2}\right)}$

Now use half angle identity for sine in both denominators

$\frac{2 \sin \theta_1}{\sqrt{\frac{1 - \cos \left(\pi - \theta_1 \right)}{2}}} = \frac{\sin \theta_2}{\sqrt{\frac{1 - \cos \left(\pi - \theta_2 \right)}{2}}}$

Of course $-\cos \left(\pi - \theta_1 \right) = \cos \left(\theta_1 \right)$ and $-\cos \left(\pi - \theta_2 \right) = \cos \left(\theta_2 \right)$

$\frac{2 \sin \theta_1}{\sqrt{\frac{1 + \cos \theta_1}{2}}} = \frac{ \sin \theta_2}{\sqrt{\frac{1 + \cos \theta_2}{2}}}$

Recall $\theta_2 = \pi - \theta_1$

$\frac{2 \sin \theta_1}{\sqrt{\frac{1 + \cos \theta_1}{2}}} = \frac{ \sin \left(\pi - \theta_1 \right)}{\sqrt{\frac{1 + \cos \left(\pi - \theta_1 \right)}{2}}}$



Of course $\cos \left( \pi - \theta_1 \right) = - \cos \theta_1$

$\frac{2 \sin \theta_1}{\sqrt{\frac{1 + \cos \theta_1}{2}}} = \frac{ \sin \theta_1}{\sqrt{\frac{1 - \cos \theta_1}{2}}}$

Cancelling out like factors

$\frac{2}{\sqrt{1 + \cos \theta_1}} = \frac{1}{\sqrt{1- \cos \theta_1}}$

Squaring both sides and moving the cosines to one side and constants to the other we get $\cos \theta_1 = \frac{3}{5} \Rightarrow \theta_1 = \arccos \left(\frac{3}{5} \right) \Rightarrow \theta_1 = \arcsin \left(\frac{4}{5} \right)$

$\theta_1 = \arcsin \left(\frac{4}{5} \right)$

$\theta_2 = \pi - \arcsin \left(\frac{4}{5} \right)$

Plugging the two angles above into the original expression for $\text{Area} \left( E \right)$ we get

$\text{Area} \left(E \right) = 100 - \frac{25 \pi}{2} - 25 \pi + 50 \left(\theta_1 - \sin \theta_1 \right) + \frac{25}{2} \left( \theta_2 - \sin \theta_2 \right)$

$\text{Area} \left(E \right) = 100 - \frac{25 \pi}{2} - 25 \pi + 50 \left(\arcsin \left( \frac{4}{5} \right) - \sin \arcsin \left( \frac{4}{5} \right) \right) + \frac{25}{2} \left( \pi - \arcsin \left( \frac{4}{5} \right) - \sin \left(\pi -\arcsin \left( \frac{4}{5} \right) \right) \right)$

Simplifying all of this

$\text{Area} \left(E\right) = 50 - 25 \pi + \frac{75}{2} \arcsin \left( \frac{4}{5}\right)$

Multiplying this by $4$ we get the answer @docnet got in #2

 

[PhDeezNutz]

$\text{Area} \left(E\right) = 50 + 25 \pi + \frac{75}{2} \arcsin \left( \frac{4}{5}\right)$

Should have been

$\text{Area} \left(E\right) = 50 - 25 \pi + \frac{75}{2} \arcsin \left( \frac{4}{5}\right)$

fixed now

 

[anuttarasammyak]

θ1=arcsin⁡(4/5)

θ2=π−arcsin⁡(4/5)

BTW from the Figure I see
$$\theta_1=2 \tan^{-1}\frac{1}{2}$$
According to your result
$$\theta_1=\tan^{-1}\frac{4}{3}$$
They equal to 0.9272... Is the equality of these inverse trigonometry functions obvious?

 

[docnet]

BTW from the Figure I see
$$\theta_1=2 \tan^{-1}\frac{1}{2}$$
According to your result
$$\theta_1=\tan^{-1}\frac{4}{3}$$
They equal to 0.9272... Is the equality of these inverse trigonometry functions obvious?

Maybe not obvious, but definitely derivable.

Start with the double angle identity:

$$\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}.$$

Let $\theta = \arctan\left(\frac{1}{2}\right)$. Then, $\tan(\theta) = \frac{1}{2}$.

Substituting into the double-angle identity:

$$\tan(2\theta) = \frac{2 \times \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}.$$

So,

$$\tan(2\arctan\left(\frac{1}{2}\right)) = \frac{4}{3}$$

and

$$2\arctan\left(\frac{1}{2}\right) = \arctan\left(\frac{4}{3}\right).$$

 

[anuttarasammyak]