Find the Area of the Green Triangle in a Square with a Semicircle | Step-by-Step Explanation
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- เผยแพร่เมื่อ 10 ก.พ. 2025
- Learn how to find the area of a triangle in a square with a semicircle by using the two-tangent theorem and the pythagorean theorem. Quick and easy tutorial by PreMath.com
Very interesting to us, Thanks.
You are very welcome Mohan!👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and all the best😃
Sure enjoy these. I'm a professional engineer and these are refreshing to do myself (if I can) and to listen to your description of the solution. A favorite channel now!
THE TWO TANGENT THEOREM - so very powerful, and the key to this problem (which I couldn´t come up with without your help). Thanks Profe.
I solved by Metric ratios of a right triangle...your method is great too! I like your videos, slow, step by step, eloquent.
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I've never heard of that method please share
This can also be solved without so much complication.
Let's draw the center of the semicircle, which we will call O, in the middle of the AD side. The radius of this semicircle = OA = OE = 4.
After, and by elementary geometry, it's clear that the polygons OAFE and OECD are similar. Therefore DC/DO = OA/AF and, substituting for their values, 8/4 = 4/AF. So, AF = (4×4)/8 = 16/8 = 2 and the base of the triangle FBC = 8 - 2 = 6.
Finally, the green shaded area ( triangle FBC ) = (6×8)/2 = 24.
Greetings from Spain. Bye.
But man...if that's geometry for kids. I shouldn't, but I do.
Only. ..draw a point O in the middle of DA. Then join O with E. I imagine you will see two polygons : OAFE and ODCE . Why are they similar ?.
Well.
Can you see that CD=CE=8 ?
Can you see that OA=OE=4 ?
And can you see that the OEC and ODC angles ( in the larger polygon ) are right, as are the OAF and OEF angles in the smaller polygon as well ?.
Can you really see it? . I think so.
But then the conclusion is forced. All the angles of both polygons are equal. ..and EF = AF = 2.
They are similar. ..whether you like it or not. Not only similar, but also that the sides of the larger polygon are twice the respective sides of the smaller one.
I hope that is clear.
By the way. ..I smoke more than one pack a day. I know. ..bad habit. ..but one must die of something. ..😢.
As "the teacher" says. ..stay blessed. Bye.
@@rohitmadashri7250 The demonstration you are asking me for is already published and I hope you understand it better now. Similar doesn't mean equal, as is obvius, but congruent.
Dejan Potocnik responds in the same way, although more abbreviated.
As "the teacher" says. ..stay blessed. Bye.
Ah...finally I repeat that I use to smoke more than a pack for a day. It's a bad habit. ..I know. ..but you must die of something. .😢
Brilliant!
Spot on.
Nice one Mr Gomez ! T Nagarajan from Goa, India
Thank you for sharing this geometry question and answer. Clear reasoning, strategy, patient explanation, application of theorems and logical approach.
Solved this question verbally!
Love and Prayers from India! 🇮🇳❤️
Bravo! Very smart.
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Greetings from the USA!
It's pretty cool that half a circle and an Egyptian triangle (6,8,10) fit inside a square.
I enjoyed this problem. Not too difficult, but interesting because it involved the two-tangent theorem applied in two different places.
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Cant believe i actually did it the same way and i used the two tangent theorem for thr first time without cheating on the video. Thanks, i make these with breakfast every morninf haha
Bravo Derryk! Very smart.
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Greetings from the USA!
Interesting question, excellent solving, many thanks!
or:
tan(φ) = 4/8 = 1/2 → tan(2φ) = 2tan(φ)/(1 - tan^2(φ)) = 1/(3/4) = 4/3 = 8/k →
k = 6 → 6(8)/2 = 24 square units 🙂
Another brilliant problem using geometric principles. Great problem solving
Thanks William for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Even after playing the video as far as the two-tangent clue I was stumped by this one (failing to apply the two-tangent rule to AFE). Really pleasing solution. Thank you.
I did it slightly differently. If you define the center of the circle as O: Draw lines OC and OF. Note that the angle AOE (call it alpha) is the same angle as the angle DCE. Furthermore, the angle DCO is alpha/2 since line CO runs down the middle between CD and CE. The angle AOF is also alpha/2 since it similarly splits between OA and OF. Thus triangles CDO and OAF are similar. Consequently CD/DO=OA/AF. Thus 8/4=4/AF, hence AF=2, hence FE=6, and the green area is (6*8)/2=24. As I look at other people's solutions I see that other people solved it this way also.
Even though I had forgotten the two tangent theorem, it took me little time to figure out there are two identical triangles sitting within the polygon formed by B, D, E and the midpoint of AB. And the same goes for the polygon formed by A, F, E and the midpoint of AB.
Thanks for sharing Patrick
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I did it in 2 ways. First, using the tangent theory as you did. Secondly, using the kite concept and using a little bit of trigonometry with it.
Muito bom o canal. Apresenta desafios matemáticos interessantes.
Construct G as midpoint of AD. Segment GE divides white space into similar right angle kites. CDGE is 8×4=32; GAFE is 4×2=8; ABCD = 64. Subtract the kites.
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It is necessary to determine that AF=2.
Define the center of segment AD as G. Draw lines EG, CG and FG. Now it is simple!
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It was similar to a question that appeared in PreMath a while ago, so I was able to solve it.
Great job as usual!
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I also formed 2 tangent triangles by joining the circle centre to C & E. Then I trig'd their angles at point C.
Subtract 2 of these angles from 90, gave angle BCF, which allowed to trig side BF.
Quick summary:
By two-tangents theorem
EC = DC = 8 and
AF = FE = y
Therefore
FB = AB - AF = 8 - y
FC = EC + FE = 8 + y
By Pythogoras
(FC)^2 = (FB)^2 + (BC)^2
(8 + y)^2 = (8 - y)^2 + 8^2
Hence
y = 2
Area of triangle FBC is therefore
FB.BC/2 = (8 - y).4 = 24
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Thanks for the visit
Excellent approach
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By joining the centre of AD to the points E and C you form two similar triangles so EC = 8. Then use Pythagoras to determine length of EF = AF.
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👍👍👍 ALSO By two tangent theorem EC=CD=8
very well done, great job
Thank you! Cheers!
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If O is the centre of the semicircle and CFB=α we can write:
EFA=180°-α so EFO=90°-α/2, but EOF is the complementary of EFO so EOF=α/2.
Now FCD=CFB=α (alternate interior angles) so ECO=α/2.
As a consequence the triangles EOC and EFO are similar so: EO:EF=CE:EO --> 4:EF=8:4 --> EF=2.
But AF=EF=2 so FB=8-2=6. And the green area is 6*8/2=24.
Thanks👍👍👍
Thank you too Cara.
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After I realized this arrangement produced a 345 triangle the answer was obvious with a simple calculation!
Cool
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Love and prayers from Arizona, USA!
@@PreMath
G'day from Australia!
@@Allangulon
You are so generous!
I solved it the same way. Greetings.
S=24🔥
If O is the center of the circle, angles DCO and OCE are equal to alpha, and tan(alpha)=0.5. Since angle BCF = beta = pi/2 - 2*alpha, FB = 8 * tan(beta) and tan(beta) = cotan(2*alpha) = 0.75.
Please mention DC =CE, as common point tangent to same circle.
So they are equal.
It will help us to understand quickly why FC=8+y
Good job sir
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Thank you sir! God blessed you
Hello sir, ur video is awesome. Kindly tell me which software do u use for video recording and writing?
Dear Javed, we use Camtasia TechSmith utility for our vids. Thanks for asking. Keep smiling😊 Enjoy every moment of your life 🌻
I got 17.76 m2. I used similar triangles concept and solved.
Как в театре не бывает маленьких ролей, так и в математике - маленьких касательных!☝👀
Let's define a point "G" right at the center of AD and draw the lines CG, EG and FG. You can see that triangle CDG is identical to triangle CEG, triangle AFG is identical to triangle EFG and all four triangles are similar. Therefore AF/AG=DG/CD, AF=4/8*4, AF=2. That makes BF=8-2, BF=6 and the area of BCF=6*8/2=24. I think this way is easier.
I understand YT format can be a drag to whatever info we want to put out there. Nevertheles, why not go a little furtheror deeper? On the conjectural side, which can be demonstrated parallel to your actual demonstratoin, for such a given construction, whatever the lenghts given : (1) the riangle will always be a 3-4-5 format rectangle triangle and (2) length FB will always be 3/4th of the square side. And why making this situation a real life application one. Juggling with formulas and numbers is great,being able to see this haibiity to some use is also appealing.
Thank you sir 🙏
Most welcome Umapathy.
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Another method, you can call STATISTIC PITAGORA:
Pitagora's theorem is ALWAYS on (3;4;5) right triangles in 99,9999% exercises. Thus, if one side is 8, then other sides are 6 and 10 of course 😅😅😅
Let's say O is the center of the semicircle. Connect OF, OC; they are the bisectors of the angles AOE and DOE. Because the angle DOA (the diameter) is 180 degrees so the two bisectors form a 90 degrees angle, so we come to the conclusion that the two triangles OAF and CDO are similar so AF/AO =OD/CD ---> AF/4=4/8---.> AE=2---> FB=6.---> The triangle area is 1/2x6x8= 24 square units.
Easier way:
1) Point O is the center of the semicircle
2) Two tangent theory EC = DC
3) OD = OE = 4 (both radii of circle, diameter of circle =8)
4) Triangles OEC and ODC are both right angled triangles with non-hypotenuse sides of 4 and 8 meaning that larger non-hypotenuse angle is 60 degrees (the tan(60) = 2= 8/4.)
5 Angle ODC + OEC + OAE = 180 (straight line) which means OAE = 60 since the other two angles add to 120.
6) AE = AF (two tangent theorem)
7) Triangles OEF and OAF are both right angled triangles whose hypotenuses and one other side are equal, therefore they are similar.
8) therefore EOF =AOF = 30 degrees (since they are each half of AOE which is 60 degrees.
Therefore AF = AO tan 30 degrees = 4 * 1/2 = 2
9) BF = 8 - AF = 6
10) Area = BF * BC/2 = 24
good decision. 👍can I send one interesting geometry problem ( task)?
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Please share your geometry problem.
@@PreMath thank you. where can I send (I mean email address)
@@levanjgarkava9992
premathchannel@gmail.com
@@PreMath thank you 👍
No piroblem nodklek mama dodo yes yeyyýydim (¥)???
Here is how I have solved it:
I have built a similar square ADGH on the side AD and made another semicircle on AD. Also I have built HK so it is tangent to circle (2 semicircles on a single side are a circle) and DG contains K.
Notice that HFCK is parallelogram (AH = KC as sides of a similar squares and AH || KC because AD is perpendicular to HB (HF) and also perpendicular to GC (KC)) and since the circle is inscribed into HFCK , it means that HF + KC = HK + FC (Pitot theorem).
But KC = HF and HK = FC. It means that all sides of HFCK are equal and it is rhombus. If AF = x, FC = 8 + x = HF.
Using Pythagoras theorem for triangle FBC, I got following equation: FC² = FB² + BC². (8 + x)² = 8² + (8 - x)². Solving this equation, I got x = 2.
Since AB = 8 and AF = x, FB = 8 - x = 8 - 2 = 6. Since triangle FBC is right, S(FBC) = FB × BC / 2 = 6 × 8 / 2 = 24. And that's the answer
Wow, amazing.
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Angle between base(X) and height(8) of the green∆is θ. Then
x/8=tanθ=¾
Area(green)=½.x.8=24 &c.
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@@rohitmadashri7250 because
x=8 tan θ
Area =½x.8.
¥di777777(¥7) dim $$$=3 (ki) 12
@@dhrubajyotidaityari9240 dodopppppppooooo armido
Easy legal method without "y": Mark a point O what is the center of the circle, then triangle CDO = CEO and tan(∠DCO)=4/8=0,5. Then use calculator or double angles formula because ∠BFC =2*∠ DCO . tan(∠ DCO )=0,5 therefore tan(∠BFC ) = (2*0,5)/(1-0,5^2)=1/0,75(as tan of double angles formula says) Then BF=0,75*8=6 (from tan definition) Answer: 6*8/2
I have solved before watching video
The answer is 64 minus 2 similar triangles with 2 equal sides. One with base 8 and height 8 and the other one with base and height 4. So 32 + 8. 64 minus 40 is 24
Thank for premath
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I solved in another way: connect E to the centre of circle "O". Apply Pitagora to ODC you get OC, ODC and OEC triangles are the same and we have their dimensions. From this you get OCE ANGLE is 26,56 deg , and DCE angle the same. So you get FCB angle is (90-26,56-26,56=) 36,87. So FC is 8/cos(36,87) = 10. BF is FC*Sen(36,87)=6. Therefore 6*8/2=24. Like if you got it.
Cool. Thanks for sharing Stefano
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@@PreMath thanks for the nice and challenging trivial, it keeps brain in motion.
Thanks to your comment, I learnt the new PITAGORA theorem.
Where are you getting these problems from? Is there a book you can refer me to? Thanks in advance.
Dear Kris, I don't know any book.
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It’s easy to solve this problem using pythagoras and properties of similar triangles
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Let S be the centre of the semicircle.Then the deltoid ECDS is congruent to FSAE. Hence DS is 2xAF. AF =2 ->FB=6 then the green area is 6x8/2=24
Completely agree. I have argued basically the same....perhaps somewhat more extensively. But 48/2 = 24...not 12...obviously.
@@alvarosarmientogomez4996 Thank you, it was my typo.
Good Question
Area of the green
AF =y = 2, BF = 6
Area of green = 1/2 * 6 * 8
= 24
dang forgot the two chords theorem! great brain teasers.
Oh, I never thought of it with the two-tangent theorem i only used that one for the y one the rest i found out by geometric mean theorem
Thanks for sharing Bea.
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Greetings from the USA!
If it’s a multiple choice question. Since you are already know the radius. Just draw it out and measure. Then you got the answer.
DCxAF= radius²
8 x AF=4x4
AF=2
FB=6
CB=8
area=1/2x6x8= 24😛
You could limit yourself to one variable y.
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This is very difficult way of solving problem....we can easily solve this problem using trignometry.....with a hights and distances model problem
Rohit Madashri, a triangle like that follows the 3-4-5 rule and therefore the sides are 6-8-10.
Area = 1/2*8*6 = 24
A lot easier if you ignore the semicircle 😉
Nice problem. I used my calculator and trigonometry (tan), but it feels like cheating as it would have been possible without a calculator as well!
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a = 6; b = 8; c = 10
Right triangle
#RightTriangle
Right triangle FBC
It doesn't concern as a red half circle,doesn't it.
#Pythagoras #PythagoreanTheorem
or maybe you recognize that if one side is 8 than the others in a right triangle must be 6 and 10. therefore the area is 24
triangle is 3:4:5?
10:8:6?
as I thought!!
CE=8=x
EF =y
CE does not equal minus y
Pythagorean Theorem
A = 24
Awesome Theo!
You are the best👍😀
Square root of 100 = 10
y = 2
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Answer 24
Pythagoras
24
It’s a 345 triangle. proportions are double 6 8 10.6x8= 48 half that is 24
0
6; 8; 10
Another 3-4-5 triangle! What are the odds?
@@wwoods66
With that geometry 1:1!
24