# On Area percentage -3 (with bonus challenge!)

Not go on the title, in this discussion I will not deal with percentage but instead measure of area

In the below diagram $$O$$ is the center of the circle with tangents $$\overline{BD}$$ and $$\overline{AE}$$ The area of $\triangle AOB=\frac{1}{2}\overline{OD}^2\tan(\angle AOB)\sec(\angle AOB)$

Proof:

Let $\overline{OD}=x;\angle AOB=\theta$

In $\triangle AEO$ $\tan(\theta)=\dfrac{\overline{AE}}{x}$ $\overline{AE}=x\tan(\theta)........$ As $\angle AEO=90^\degree$, applying Pythagorus theorem $x^2+\overline{AE}^2=\overline{AO}^2$ $\overline{AO}^2=x^2+x^2\tan^2(\theta)=x^2(1+\tan^2(\theta))=x^2\sec^2(\theta)$ $\overline{AO}=x\sec(\theta)........$

Now in $\triangle AEO$ and $\triangle BDO$ $\angle AEO=90^\degree=\angle BDO$ $\angle AOE=\theta=\angle BOD$ $\overline{OE}=x=\overline{OD}$ From RHS criterion of congruence $\triangle AEO\cong\triangle BDO$ $\therefore \overline{AO}=\overline{BO}........$

Area of $\triangle ABO=\frac{1}{2}\overline{BO}\times\overline{AE}$

From $$ and this

Area of $\triangle ABO=\frac{1}{2}\overline{AO}\times\overline{AE}$

From $,$ and this

Area of $\triangle ABO=\frac{1}{2}x\sec(\theta)\times x\tan(\theta)=\frac{1}{2}x^2\tan(\theta)\sec(\theta)$

Bonus In the above diagram if you know the value of $R_1,R_2$ and $R_3$, then find a formula to find the area of the $\red{red}$ region.

Note:

• I myself don't know the answer to the bonus problem. Note by Zakir Husain
2 months, 4 weeks ago

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Centres of circles are joined,then we can find the area of the formed triangle by heron's formula and then we find each angle by cosine rule and area of each sector.Subtracting area of sectors from area of triangle gives our result.

@Zakir Husain,I will try to find shorter method.

- 2 months, 4 weeks ago

Or try writing it out as one formula and try to factorise & simplify :)

- 2 months, 4 weeks ago

If someone gets the answer with solution of bonus, please share! Thanks!

- 2 months, 4 weeks ago I got it Where,s=r1+r2+r3

- 2 months, 4 weeks ago

Incredibly! long formula

- 2 months, 4 weeks ago

Ya,the earlier solution is longer than it.I'll try to short the formula

- 2 months, 4 weeks ago

Can you describe the parts of this formula? Did you "use Heron's formula and then try to subtract the sectors"... If so I think there is a small mistake in Heron's formula, it's: $\sqrt{s (s-r_{1})(s-r_{2})(s-r_{3})}$. Anyways please describe how you got till this formula. Thanks!

- 2 months, 4 weeks ago

@Mahdi Raza,check my solution

- 2 months, 4 weeks ago

Wait…but the triangle’s sides aren’t $r_1,r_2,r_3$, and it is equilateral (tangent of circle from same point)

- 2 months, 4 weeks ago

It might be the smallest among the radii of the circles, but we need more proof. :/

- 2 months, 4 weeks ago

@Jeff Giff, @Mahdi Raza, @Kriti Kamal - the red area is not a triangle so don't confuse, it is a shape made of three arcs not three sides.

- 2 months, 4 weeks ago

Yes, but I meant the smallest triangle containing the red area :)

- 2 months, 4 weeks ago

I have considered them as arcs not as linesegmets :-)

- 2 months, 4 weeks ago

Oic. My solution is different. I thought about the smallest triangle containing the red area :)

- 2 months, 4 weeks ago

So my solution is $sizeof\triangle$containing red area $-sizeofARCH1,2,3$

- 2 months, 4 weeks ago

Maybe I am wrong :D Sorry for being unhelpful :|

- 2 months, 4 weeks ago

See my solution..

- 2 months, 4 weeks ago  Here,is my solution.Please comment if you find any mistake.Sorry for delay,it happens because tommorow my hands were burn by fire.

- 2 months, 4 weeks ago

The angles can be calculated in sin inverse and cos inverse function

- 2 months, 4 weeks ago

@Zakir Husain,can i send another version of solution

- 2 months, 4 weeks ago

Sure!

- 2 months, 4 weeks ago

The area of the triangle can be found using Zakir’s formula above:

$\frac12r_1^2\tan\angle A\sec\angle A$

Or any $x$ and $\theta$ pair with the relation above.

- 2 months, 4 weeks ago

You are assuming that the perpendicular of $\overline{AB}$ at point $D$ will touch the point $C$, the same assumption you have to make to the perpendicular of $\overline{AC}$ and point $B$. Then only you can apply my formula to get the area of $\triangle ABC$. You first have to prove that these assumptions are true for all cases. Or you can build a new formula for this special case separately.

- 2 months, 4 weeks ago

Aha! @Zakir Husain, your formula can be simplified as $\frac12x^2\sin \theta$, since $\tan \theta \sec \theta=\tan \theta \frac{1}{\cos \theta}=\frac{\frac{\sin \theta}{\cos \theta}}{\cos \theta}=\sin \theta$ :)

- 2 months, 4 weeks ago

$\tan\theta\sec\theta=\dfrac{\sin\theta}{\cos\theta}\times\dfrac{1}{\cos\theta}=\red{\dfrac{\sin\theta}{\cos^2\theta}}$

- 2 months, 4 weeks ago

Oops!

- 2 months, 4 weeks ago

Slight improvement: express angles in radians. Then, the formula becomes $\sqrt{sr_1r_2r_3}-\sum _{cyc}r_1^2\tan^{-1}\sqrt{\frac{r_2r_3}{sr_1}}$

- 2 months, 4 weeks ago

I know it.but generally i use my most of work in degrees

- 2 months, 4 weeks ago