little circle at top right corner of a square

[nanase]

Hello, I have this geometrical related problem I am keen to solve :

I am a bit clueless with how to solve this. With some scribbling on paper, the only new information I can find is the diagonal from the centre of the circle to point D is equal to [math]4\sqrt{2}[/math] and the shortest distance from point D to EC is [math]4+4\sqrt{2}[/math]. I am aiming to either find the length of AE or the length of EC, but I can't seem to obtain the information to do so. Any hint I missed from the question?
Thank you

 

[The Highlander]

Hello, I have this geometrical related problem I am keen to solve :
View attachment 37225
I am a bit clueless with how to solve this. With some scribbling on paper, the only new information I can find is the diagonal from the centre of the circle to point D is equal to [math]4\sqrt{2}[/math] and the shortest distance from point D to EC is [math]4+4\sqrt{2}[/math]. I am aiming to either find the length of AE or the length of EC, but I can't seem to obtain the information to do so. Any hint I missed from the question?
Thank you

The distance (not the "diagonal") from the centre of the circle to point D is, indeed, equal to [imath]4\sqrt{2}[/imath] but the shortest distance from point D to EC is not [imath]4+4\sqrt{2}[/imath].

I suggest you read this (list of five properties) then study this Post (and the one after it).

This may help to start you off...

​

Hope that helps.

 

[mario99]

Hello, I have this geometrical related problem I am keen to solve :
View attachment 37225
I am a bit clueless with how to solve this. With some scribbling on paper, the only new information I can find is the diagonal from the centre of the circle to point D is equal to [math]4\sqrt{2}[/math] and the shortest distance from point D to EC is [math]4+4\sqrt{2}[/math]. I am aiming to either find the length of AE or the length of EC, but I can't seem to obtain the information to do so. Any hint I missed from the question?
Thank you

As a calculus student, I would imagine these geometric shapes were drawn in the third quadrant. My main focus would be to find a formula for the tangent line to the circle which is [imath]\displaystyle y = -\frac{4}{3}x - 16[/imath]. Now you can find the point [imath]\displaystyle E[/imath] and the problem is solved.

The whole difficulty was to find the slope [imath]\displaystyle m = -\frac{4}{3}[/imath], where the segment [imath]\displaystyle CE[/imath] touches the circle. This is not a geometric approach and can be ignored.

 

[The Highlander]

As a calculus student, I would imagine these geometric shapes were drawn in the third quadrant. My main focus would be to find a formula for the tangent line to the circle which is [imath]\displaystyle y = -\frac{4}{3}x - 16[/imath]. Now you can find the point [imath]\displaystyle E[/imath] and the problem is solved.
The whole difficulty was to find the slope [imath]\displaystyle m = -\frac{4}{3}[/imath], where the segment [imath]\displaystyle CE[/imath] touches the circle. This is not a geometric approach and can be ignored.

Very clever (as you clearly think!).

You have now provided a 'shortcut' for the OP to reach a solution (without using the geometry of the figure) but how did you arrive at \(\displaystyle m=-\frac{4}{3}\) and, more importantly, would the OP have been able to do the same?

The OPs learn nothing if you just give them an answer!

(Why do you think the problem was posted in Geometry and Trig (rather than Calculus)?)

 

[mario99]

Very clever (as you clearly think!).

You have now provided a 'shortcut' for the OP to reach a solution (without using the geometry of the figure) but how did you arrive at \(\displaystyle m=-\frac{4}{3}\) and, more importantly, would the OP have been able to do the same?

The OPs learn nothing if you just give them an answer!

(Why do you think the problem was posted in Geometry and Trig (rather than Calculus)?)

I did not give the OP the answer. I gave her a way to check her answer. She will have to use geometry to prove her solution. You said it by yourself, the problem was posted in Geometry and Trig rather than Calculus. Therefore, it is a Geometry Class.

 

[The Highlander]

I did not give the OP the answer. I gave her a way to check her answer. She will have to use geometry to prove her solution. You said it by yourself, the problem was posted in Geometry and Trig rather than Calculus. Therefore, it is a Geometry Class.

But will s/he bother to pursue a geometry/trigonometry solution when a shortcut is there to reach a solution much more simply?

And you still haven't answered my question as to how you arrived at the gradient of EC using Calculus (the slick part of your post).

Or did you actually work through the problem using Geometry & Trig and then translate the answer into an 'equation' for the line segment EC?

 

[mario99]

But will s/he bother to pursue a geometry/trigonometry solution when a shortcut is there to reach a solution much more simply?

And you still haven't answered my question as to how you arrived at the gradient of EC using Calculus (the slick part of your post).

Or did you actually work through the problem using Geometry & Trig and then translate the answer into an 'equation' for the line segment EC?

Don't worry. The OP will probably not be able to understand my solution if she has not studied calculus yet. Even, if she has, she will have to pursue a geometric solution.

You cannot just say, hey professor, the shaded area is blah blah blah squared. You have to prove it, otherwise it will not be accepted. And since it is a geometry class, you have to use only and only geometric approach.

In fact, I don't understand geometry. I mean that I don't know how to prove a solution using geometry. I am a professional calculus student and I am very good in creating formulas for geometric shapes. First, I arrived to the gradient by trial and error (Dr.Peterson approach). I started with a slope of -1, then -2. Then between -1.1 and -1.9. The -1.3 was so close, so instantly, I knew that the slope was -4/3. Of course, with the help of Desmos. After that I proved that was a correct slope using calculus and a little algebra by the help of the equation of the circle and its derivative and the equation of the straight line.

By the way, I like your style of attacking geometric problems. I always read your whole solution for any problem posted in this forum. I understood many theorems because of that. Still, geometry is very difficult for me to master.

 

[BigBeachBanana]

Hello, I have this geometrical related problem I am keen to solve :
View attachment 37225
I am a bit clueless with how to solve this. With some scribbling on paper, the only new information I can find is the diagonal from the centre of the circle to point D is equal to [math]4\sqrt{2}[/math] and the shortest distance from point D to EC is [math]4+4\sqrt{2}[/math]. I am aiming to either find the length of AE or the length of EC, but I can't seem to obtain the information to do so. Any hint I missed from the question?
Thank you

There are multiple ways to solve this. It depends on what tools are available at your disposal. Here I'll provide one approach.
The equation of the inradius [imath]r = \dfrac{a+b-c}{2}[/imath], where [imath]a, b[/imath] are the legs and [imath]c[/imath] is the hypotenuse of the right triangle [imath]\triangle EDC[/imath]



You can write the above equation in terms of [imath]x[/imath] and solve for it. The rest is just an area of a trapezoid. If you're interested in knowing where the equation came from, I can provide more resources.

 

[The Highlander]

There are multiple ways to solve this. It depends on what tools are available at your disposal. Here I'll provide one approach.
The equation of the inradius

[imath]r = \dfrac{a+b-c}{2}[/imath], where [imath]a, b[/imath] are the legs and [imath]c[/imath] is the hypotenuse of the right triangle [imath]\triangle EDC[/imath]

View attachment 37227


You can write the above equation in terms of [imath]x[/imath] and solve for it. The rest is just an area of a trapezoid. If you're interested in knowing where the equation came from, I can provide more resources.

Your formula [imath]r = \dfrac{a+b-c}{2}[/imath], where [imath]a, b[/imath] are the legs and [imath]c[/imath] is the hypotenuse of the right triangle [imath]\triangle EDC[/imath] is, of course, perfectly correct but I'm at a loss to see how it may be used with the limited information provided in the question.

What is your x meant to be?

Is it the length ED or is it meant to be the length from E to the point of tangency of the radius to the side ED?

Attempting to rewrite the formula in terms of x, I get 16 - 8 = 8 using the former and 20 - 12 = 8 using the latter (because the x terms cancel), neither of which offers (me) any progress towards a solution of the problem as presented.

Am I missing something?

 

[BigBeachBanana]

Is it the length ED or is it meant to be the length from E to the point of tangency of the radius to the side ED?

x is the length of ED. Express c in terms of x via the Pythagorean Theorem.

 

[The Highlander]

x is the length of ED. Express c in terms of x via the Pythagorean Theorem.

Gotcha!

Now it's the x² terms that cancel.

(But see my PM, pls?)

 

[The Highlander]

Don't worry. The OP will probably not be able to understand my solution if she has not studied calculus yet. Even, if she has, she will have to pursue a geometric solution.

You cannot just say, hey professor, the shaded area is blah blah blah squared. You have to prove it, otherwise it will not be accepted. And since it is a geometry class, you have to use only and only geometric approach.

In fact, I don't understand geometry. I mean that I don't know how to prove a solution using geometry. I am a professional calculus student and I am very good in creating formulas for geometric shapes. First, I arrived to the gradient by trial and error (Dr.Peterson approach). I started with a slope of -1, then -2. Then between -1.1 and -1.9. The -1.3 was so close, so instantly, I knew that the slope was -4/3. Of course, with the help of Desmos. After that I proved that was a correct slope using calculus and a little algebra by the help of the equation of the circle and its derivative and the equation of the straight line.

By the way, I like your style of attacking geometric problems. I always read your whole solution for any problem posted in this forum. I understood many theorems because of that. Still, geometry is very difficult for me to master.

I have to say, you appear to be begging the question here.

You say yourself that you guessed the gradient (how can you know "instantly" what a slope is just because an approximation looks right?) and then you seem to have employed other (unspecified) techniques to 'verify' your guess; that's not the same as determining a value for something from first principles based on available information.

Have a look at the two methods demonstrated in the conversation I started with @BigBeachBanana; you might find the Algebra, Geometry & Trigonometry displayed there enlightening.

 

[mario99]

I have to say, you appear to be begging the question here.

You say yourself that you guessed the gradient (how can you know "instantly" what a slope is just because an approximation looks right?) and then you seem to have employed other (unspecified) techniques to 'verify' your guess; that's not the same as determining a value for something from first principles based on available information.

Have a look at the two methods demonstrated in the conversation I started with @BigBeachBanana; you might find the Algebra, Geometry & Trigonometry displayed there enlightening.

Guessing the slope then proving it is correct is a high level math skill. I could have found the slope directly with Calculus, but playing around with Desmos is fun.

Also it is not pure guessing, it is trial and error. I knew instantly it was -4/3 because as I keep adding 3s (-1.33333...) the line approaches the circle. It is one of the main concepts of Calculus. But I didn't trust Desmos that it was the correct slope. The line sometimes touches the circle but when we zoom in, it appears it is not tangent. Therefore, I had to prove it with some calculations. Of course, my attack was the hard way to solve the problem, but you have to remember that I have zero knowledge in geometry.

Although the trigonometric approach was fascinating, I am designed to solve things the hard way!

 

[nanase]

There are multiple ways to solve this. It depends on what tools are available at your disposal. Here I'll provide one approach.
The equation of the inradius [imath]r = \dfrac{a+b-c}{2}[/imath], where [imath]a, b[/imath] are the legs and [imath]c[/imath] is the hypotenuse of the right triangle [imath]\triangle EDC[/imath]
View attachment 37227


You can write the above equation in terms of [imath]x[/imath] and solve for it. The rest is just an area of a trapezoid. If you're interested in knowing where the equation came from, I can provide more resources.

may I ask, when you mention that a and b represent the legs, what do you mean with the legs? which legs?

 

[blamocur]

may I ask, when you mention that a and b represent the legs, what do you mean with the legs? which legs?

legs of a right triangle - Google Search

 

[nanase]

legs of a right triangle - Google Search

I see, base and height

 

[nanase]

x is the length of ED. Express c in terms of x via the Pythagorean Theorem.

Thank you BBB, it's my first time seeing the formula r = (a+b-c)/2, and your post here on how to express c really helps me in working through the problem.
so [math]4=\frac{x+16-c}{2}[/math] then I subs c with[math]\sqrt{x^{2}+16^{2}}[/math]
After squaring and cancelling the x^2 on both sides, I ended and obtained x =12.
from there, I calculated the area of trapezium, which gives 160.
Thank you so much BBB for your help and guiding me with the "right level" Maths for me.
Thank you The Highlander for sharing the resource regarding incircle, I read it through and learn new things as well.

 

[nanase]

There are multiple ways to solve this. It depends on what tools are available at your disposal. Here I'll provide one approach.
The equation of the inradius [imath]r = \dfrac{a+b-c}{2}[/imath], where [imath]a, b[/imath] are the legs and [imath]c[/imath] is the hypotenuse of the right triangle [imath]\triangle EDC[/imath]
View attachment 37227


You can write the above equation in terms of [imath]x[/imath] and solve for it. The rest is just an area of a trapezoid. If you're interested in knowing where the equation came from, I can provide more resources.

I am interested in knowing where the equation come from (If you mean the r = (a+b-c)/2 ). I hope It can add to my understanding of how the geometry works (incircle measurements in new to me)

 

[BigBeachBanana]

I am interested in knowing where the equation come from (If you mean the r = (a+b-c)/2 ). I hope It can add to my understanding of how the geometry works (incircle measurements in new to me)

This site gives a brief explanation.

 

[The Highlander]

Thank you The Highlander for sharing the resource regarding incircle, I read it through and learn new things as well.

Hi @nanase,

You are vey welcome.

I am interested in knowing where the equation come from (If you mean the r = (a+b-c)/2 ). I hope It can add to my understanding of how the geometry works (incircle measurements in new to me)

Here is a (simple?) explanation of how the inradius equation may be derived for the situation you have; ie: a circle inscribed in a right-angled triangle.

​

Look up at the diagram after you read each line to see that what is said is correct (and you understand it).

The circle touches the sides AB, BC & CA of the right triangle ABC at D, E and F respectively,
The sides are also given the lower case letter names: a (BC), b (CA) and c (AB).

Because a radius of a circle is perpendicular (⊥) to a tangent where they meet: OF⊥CA and OE⊥BC
Thus OECF is a square ⇒ EC = CF (= OE = OF) = r.
and, therefore, AF = (b−r) and BE = (a−r)

Since the lengths of tangents drawn to a circle from an external point are equal, then AD = AF and BD = BE.

Now, c (AB) = AD + BD
but AD = AF (b-r) and BD = BE (a-r)
⇒ c = AD + BD = AF + BE = (b−r) + (a−r)
⇒ c = b − r + a − r
⇒ c = a + b - r - r
⇒ c + r + r = a + b
⇒ 2r = a + b − c

⇒ \(\displaystyle \text r = \frac{\text a + b − c}{2}~\qquad\) (ie: the inradius equation you were given. )

However, if that equation was entirely new to you (not at all addressed in your course to date) then perhaps your teacher(s) might expect you to provide a solution based simply on the Geometry of the figure and some Basic Right-Angled Triangle Trigonometry?

That was the direction I was trying to guide you in when I posted the (blue) diagram in my initial response to you (at Post #2, above).

I have added a few more details to that diagram below that I believe make it clearer how to solve the problem using that approach (see if you can "spot the differences" and work out their significance?)...

​

You should see from the above that \(\displaystyle tan\text{ θ} = \frac{4}{12}\) and so \(\displaystyle \text tan^{-1} \left(\frac{4}{12}\right)= \) θ; 90 - 2θ = 2ω and EH = \(\displaystyle \frac{4}{tan\text{ ω}}\)

So you could simply write:-

\(\displaystyle tan^{-1} \left(\frac{4}{12}\right)\approx 18.43\\90-2\times 18.43\approx 53.14\\ \implies \text ω\approx 26.57\\ \implies \text {EH} =\frac{4}{tan 26.57}=8~\qquad\)(Length of EH)

ED = EH + HD = 8 + 4 = 12
Area ΔCDE = ½ × ED × DC = ½ × 12 × 16 = 96 sq units.
Area Square ABCD = 16² = 256 sq units.
Area Shaded Region = Area ABCD - Area ΔCDE = 256 - 96 = 160 sq units.

You will note that I have used "≈" (rather than an equals sign) on several occasions up there because my (intermediate) results have been rounded to two decimal places throughout.

If you were to write down those calculations (to any rounded number of decimal places) and then use those intermediate results in your calculator (as written) then you would be likely to get an answer slightly different from 8 for (EH) at the end.

However, you can easily get around that problem as it should be relatively simple to avoid such rounding errors on a modern calculator by entering the calculations as a continuous series of key strokes from start to finish.
(You can always write down the intermediate results (rounded 2 or 3 d.p.) as you go along.)

Here's a record of how I did it (from the History produced) on the built in Calculator on Windows...

Enter: 4 / 12 = 0.33333333333333333333333333333333
Press [Inv][tan^-1]
arctan(0.33333333333333333333333333333333)
= 18.434948822922010648427806279547 (Angle θ)
18.434948822922010648427806279547 × 2
= 36.869897645844021296855612559093 (∠DCE = 2θ)
Press [+/-] button (to make it negative)
ˉ36.869897645844021296855612559093 + 90
= 53.130102354155978703144387440907 (∠CED = 2ω)
53.130102354155978703144387440907 / 2
= 26.565051177077989351572193720453 (Angle ω)
Press [tan]
tan(26.565051177077989351572193720453)
= 0.5
Press [MR] (Store it in Memory)
Enter 4 ÷ [MR] (Memory recall button)
to get: 4 / 0.5 = 8 (Length of EH)

Then proceed to calculate the area of the shaded region as shown above.

Hope that helps.