Find the Area of the Green Triangle in a Square with a Semicircle | Step-by-Step Explanation

[@mohanramachandran4550]

Very interesting to us, Thanks.

[@PreMath]

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😃

[@bobjordan5231]

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!

[@charlesbromberick4247]

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.

[@partisano75]

I solved by Metric ratios of a right triangle...your method is great too! I like your videos, slow, step by step, eloquent.

[@PreMath]

Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@BraBoyza]

I've never heard of that method please share

[@rajendramisir3530]

Thank you for sharing this geometry question and answer. Clear reasoning, strategy, patient explanation, application of theorems and logical approach.

[@alvarosarmientogomez4996]

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.

[@alvarosarmientogomez4996]

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.

[@alvarosarmientogomez4996]

@@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. .😢

[@tombratcher6938]

Brilliant!

[@shoilee2000]

Spot on.

[@nagarajant3404]

Nice one Mr Gomez ! T Nagarajan from Goa, India

[@1Maklak]

It's pretty cool that half a circle and an Egyptian triangle (6,8,10) fit inside a square.

[@tomcruise6738]

Solved this question verbally! Love and Prayers from India! 🇮🇳❤️

[@PreMath]

Bravo! Very smart. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃 Greetings from the USA!

[@deyomash]

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

[@PreMath]

Bravo Derryk! Very smart. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃 Greetings from the USA!

[@murdock5537]

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 🙂

[@j.r.1210]

I enjoyed this problem. Not too difficult, but interesting because it involved the two-tangent theorem applied in two different places.

[@PreMath]

Thanks J R for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@Mathematician6124]

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.

[@batchrocketproject4720]

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.

[@rclrd1]

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

[@PreMath]

Thanks Eric for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@spafon7799]

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.

[@kujmous]

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.

[@PreMath]

Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@edsonstarteri6313]

It is necessary to determine that AF=2.

[@williamjones2407]

Another brilliant problem using geometric principles. Great problem solving

[@PreMath]

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😃

[@py8554]

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.

[@PreMath]

Thanks for sharing Patrick Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@pedromartins1986]

Muito bom o canal. Apresenta desafios matemáticos interessantes.

[@flaviusnita6008]

Define the center of segment AD as G. Draw lines EG, CG and FG. Now it is simple!

[@PreMath]

Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@JLvatron]

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.

[@nigelsw55]

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.

[@PreMath]

Thanks Williamson for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@faizs9676]

👍👍👍 ALSO By two tangent theorem EC=CD=8

[@charger121212]

Excellent approach

[@PreMath]

Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@ebi2ch]

It was similar to a question that appeared in PreMath a while ago, so I was able to solve it.

[@PreMath]

Great job as usual! Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@sleevman]

Thanks for the visit

[@242math]

very well done, great job

[@PreMath]

Thank you! Cheers! Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@magnetshorts8165]

Hello sir, ur video is awesome. Kindly tell me which software do u use for video recording and writing?

[@PreMath]

Dear Javed, we use Camtasia TechSmith utility for our vids. Thanks for asking. Keep smiling😊 Enjoy every moment of your life 🌻

[@susennath6035]

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

[@Allangulon]

After I realized this arrangement produced a 345 triangle the answer was obvious with a simple calculation!

[@PreMath]

Cool Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from Arizona, USA!

[@Allangulon]

@@PreMath G'day from Australia!

[@PreMath]

@@Allangulon You are so generous!

[@MathZoneKH]

Good job sir

[@PreMath]

Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@MathZoneKH]

Thank you sir! God blessed you

[@EnnioPiovesan]

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.

[@edsonstarteri6313]

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.

[@krislegends]

Where are you getting these problems from? Is there a book you can refer me to? Thanks in advance.

[@PreMath]

Dear Kris, I don't know any book. Thanks for asking dear. Take care dear and stay blessed😃

[@AmirgabYT2185]

S=24🔥

[@CARAMENGHITUNG]

Thanks👍👍👍

[@PreMath]

Thank you too Cara. Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@umapathygovindaswamy8622]

Thank you sir 🙏

[@PreMath]

Most welcome Umapathy. Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@fernandoleon4204]

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.

[@annatygrys9043]

I solved it the same way. Greetings.

[@yvessioui2716]

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.

[@bea-pr3ei]

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

[@PreMath]

Thanks for sharing Bea. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃 Greetings from the USA!

[@phungpham1725]

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.

[@SabatoDeSimone]

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 😅😅😅

[@zawatsky]

Как в театре не бывает маленьких ролей, так и в математике - маленьких касательных!☝👀

[@НоунеймЧел-с7я]

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

[@PreMath]

Wow, amazing. Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@dhrubajyotidaityari9240]

Angle between base(X) and height(8) of the green∆is θ. Then x/8=tanθ=¾ Area(green)=½.x.8=24 &c.

[@PreMath]

Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@dhrubajyotidaityari9240]

@@rohitmadashri7250 because x=8 tan θ Area =½x.8.

[@israfilekiz2288]

¥di777777(¥7) dim $$$=3 (ki) 12

[@israfilekiz2288]

@@dhrubajyotidaityari9240 dodopppppppooooo armido

[@vemuricreativeworks7406]

I got 17.76 m2. I used similar triangles concept and solved.

[@newperve]

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

[@levanjgarkava9992]

good decision. 👍can I send one interesting geometry problem ( task)?

[@PreMath]

Thanks Levan for the feedback. You are awesome 👍 Take care dear and stay blessed😃 Please share your geometry problem.

[@levanjgarkava9992]

@@PreMath thank you. where can I send (I mean email address)

[@PreMath]

@@levanjgarkava9992 [e-mail]

[@levanjgarkava9992]

@@PreMath thank you 👍

[@israfilekiz2288]

No piroblem nodklek mama dodo yes yeyyýydim (¥)???

[@stefanobonzi7332]

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.

[@PreMath]

Cool. Thanks for sharing Stefano Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards

[@stefanobonzi7332]

@@PreMath thanks for the nice and challenging trivial, it keeps brain in motion.

[@inboxingleads6336]

Thanks to your comment, I learnt the new PITAGORA theorem.

[@tahasami597]

Thank for premath

[@PreMath]

You are very welcome Taha! I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@ШуршавийВіталій]

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

[@LarzB]

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

[@dejanpotocnik8407]

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

[@alvarosarmientogomez4996]

Completely agree. I have argued basically the same....perhaps somewhat more extensively. But 48/2 = 24...not 12...obviously.

[@dejanpotocnik8407]

@@alvarosarmientogomez4996 Thank you, it was my typo.

[@ОльгаСоломашенко-ь6ы]

You could limit yourself to one variable y.

[@PreMath]

Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@ramchandrachinchanikar2146]

It’s easy to solve this problem using pythagoras and properties of similar triangles

[@PreMath]

Thanks dear for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@travelerumesh3318]

Good Question

[@DDX01]

I have solved before watching video

[@warchitect73]

dang forgot the two chords theorem! great brain teasers.

[@Su4ji]

Area of the green AF =y = 2, BF = 6 Area of green = 1/2 * 6 * 8 = 24

[@phamcuong9183]

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.

[@shreyanshpatel0707]

DCxAF= radius² 8 x AF=4x4 AF=2 FB=6 CB=8 area=1/2x6x8= 24😛

[@realastrology3976]

This is very difficult way of solving problem....we can easily solve this problem using trignometry.....with a hights and distances model problem

[@kennethweigand4600]

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 😉

[@theophonchana5025]

Right triangle FBC

[@xyz_abc752]

It doesn't concern as a red half circle,doesn't it.

[@theophonchana5025]

a = 6; b = 8; c = 10

[@theophonchana5025]

Right triangle

[@philipkudrna5643]

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!

[@PreMath]

Thanks Philip for sharing your experience. You are awesome 👍 Take care dear and stay blessed😃

[@theophonchana5025]

#Pythagoras #PythagoreanTheorem

[@ブックビッグ]

triangle is 3:4:5? 10:8:6? as I thought!!

[@theophonchana5025]

#RightTriangle

[@bb55555555]

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

[@Cecilia-dg8ol]

CE=8=x EF =y CE does not equal minus y

[@theophonchana5025]

A = 24

[@PreMath]

Awesome Theo! You are the best👍😀

[@theophonchana5025]

Pythagorean Theorem

[@theophonchana5025]

y = 2

[@PreMath]

Good job Theo dear! Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@-basicmaths862]

Answer 24

[@theophonchana5025]

Square root of 100 = 10

[@theophonchana5025]

Pythagoras

[@johnfoggjr2458]

24

[@johnfoggjr2458]

It’s a 345 triangle. proportions are double 6 8 10.6x8= 48 half that is 24

[@theophonchana5025]

6; 8; 10

[@wwoods66]

Another 3-4-5 triangle! What are the odds?

[@Allangulon]

@@wwoods66 With that geometry 1:1!

[@richardcampion4279]

0

[@Mathematician6124]

24