Let's Think outside the Box! | Calculate the length X | (Nice explanation) |

Learn how to calculate the length X. Important Geometry skills are also explained: intersecting chords theorem; Pythagorean Theorem; Thales' theorem. Step-by-step tutorial by PreMath.com
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Let's Think outside the Box! | Calculate the length X | (Nice explanation) | #math #maths
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#FindX #CalculateLengthX #ThalesTheorem #Radius #GeometryMath #PythagoreanTheorem #Semicircle #IntersectingChordsTheorem
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[@bigm383]

Professor, you have such a lovely turn of phrase and lilt to your voice that it is a delight just to listen!

[@PreMath]

Thank you kindly! ❤️ You are awesome. Keep it up 👍

[@bigm383]

@@PreMath❤😀🥂

[@monroeclewis1973]

Another way: as the professor, I used the Pythagorean theorem to find the radius,15. Then, I constructed a perpendicular bisector from O to the chord CB at point Q. I found the length of that bisector using similar triangles: AC/ AB :: OQ/ OB, 4x sq rt 14/30 = OQ/15. Therefore, OQ =2x sq rt 14. I found QP by adding 8 and 18 to get 26. 1/2 26 = 13. (Or, just note OQ bisects CB to get 13.) 13- 8 =5 which is QP. OQ and QP form the 2 smaller sides of rt triangle OQP, with OP or “x” the hypotenuse. Next I used the Pythagorean theorem, adding the squares of 2x sq rt 14 and 5 to get 81. The sq rt 9 is “x.”

[@ronaldreygalos]

i love this❤️❤️❤️

[@eduardoteixeira869]

the interesting about this solution is that you do not need to find the radius = 15 since the two triangles are similar and the similar ratio is 2 since one side of the large triangle is the diameter and the correspondent side of the smaller triangle is the radius. I love this one too. Thanks

[@halkagros3829]

I solved this way

[@phungpham1725]

From O draw OH perpendicular to the chord CB. We have OH//= 1/2AC= 2sqrt14 and HP= 13-8=5 ----> sq x= 4. 14 +25= 81-----> x= 9 units

[@gnpatterson]

Yes this method is quicker and easier than method shown in video

[@marioalb9726]

Pytagorean theorem: (2R)²=CB² + AC² 4.R² = (18+8)² + (4√14)² 4R² = 676 + 224 R= √225 = 15 cm Cosine rule: x² = a² + b² - 2 ab cos α x² = R² + 18² - 2.R.18.(26/2R) x² = R² + 324 - 468 x² = R² - 144 x = 9 cm ( Solved √ )

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@spafon7799]

Another way to solve. First, as in the solution given, you use pythagorean to find that the diameter AB is 30) Then draw a line segment from O to chord BC that intersects it at a right angle. Call the intersection point between them D. Triangle ODB is similar with triangle ACB, and its dimensions are 1/2 of those of ACB. Thus BD is 1/2 of BC, i.e. 13. Therefore PD is 5. Also OD is 1/2 of AC, so OD is 2sqrt(14). Since you have PD and OD you can then apply pythagorean on triangle OPD to solve for OP, which gives the answer of x=9.

[@marioalb9726]

Pytagorean theorem: (2R)²=CB² + AC² 4.R² = (18+8)² + (4√14)² 4R² = 676 + 224 R= √225 = 15 cm Intersecting chords theorem: (R-x)(R+x) = 18 . 8 R² - x² = 144 x² = R² - 144 x = 9 cm ( Solved √ )

[@marioalb9726]

Right triangle, leg 8+18=26cm tan α = 4√14 / 26. --> α= 29,9264° R = (26/2) / cos α R = 15 cm Stewart theorem for isosceles triangle: x² = R² - 8•18 x² = 15² - 144 x= 9 cm. ( Solved √ )

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@richard6216]

Alternatively, you can extend a perpendicular bisector from center O to chord BC. Solve for the square of the perpendicular bisector using the Pythagoras theorem, which turns out to be 56.Then use the Pythagoras theorem again on the smaller triangle with OP as its hypotenuse, to solve for 'x', which will give you 9.

[@PreMath]

Great! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@Abby-hi4sf]

Will you explain how you get 56? Thank you

[@richard6216]

@@Abby-hi4sf let's say OD is the perpendicular bisector of chord BC(bisects the chord at point D - perpendicular from the center of a circle to any chord bisects the chord). Then, BD =13. Consider the right triangle ODB. Then, apply Pythagoras theorem and you will get 56 as the square of OD.

[@Abby-hi4sf]

Thank you@@richard6216

[@fikirfikir3039]

بعد حساب الوتر والزاوية B نستطيع استخدام نظرية (الكاشي) لإيجاد قيمة X

[@DB-lg5sq]

Bravo

[@PreMath]

❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@fikirfikir3039]

@@PreMath Thank you

[@user-my7ki4it3s]

found diameter of the semi-circle using Pythagorean theorem as well, but to calculate x I used law of cosines. still obtained 9. thanks for explanation

[@PreMath]

Very cool! ❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@Copernicusfreud]

That is how I did it too. The law of cosines seemed much easier. Cosine (ABC) = 26/30 = 13/15. It was a simple "plug-and-chug" at that point.

[@marioalb9726]

A simple plug and chug??? Who.thought outside the box ? Stewart theorem for isosceles triangle: x² = R² - 8•18 !!!!! 😅

[@DanielNeedham2500]

I used trigonometry to find the angles of the 90 triangle then used the cosine rule to find x. The funny thing is I used intersecting chord theorem on one of the last semi-circle problems finding the length of the chord AB

[@johankotze42]

Draw the perpendicular from O to BC, lable it D. By proportionality, OD = (1/2)AC = 2*Root(14) Again by proportionality, D is midway of BC, therefore DP =5. By Pythagoras x = root(5^2 + (2*root(14)^2) = 9 The value of AB is not needed.

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@paulscheele623]

Alternatively, use the law of cosines on OPB, where x^2 = 8^2 + 15^2 - 2*8*15*cos(angle B). But because ABC is a right triangle, cos(B) is 26/30, so x^2 = 81, so x=9

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@pushkarjakhar9943]

Draw perpendicular to chord BC from centre O and name OZ as perpendicular from the centre bisects chord then OZ =1/2 BC = 1/2(26) = 13 . PZ = PC - CZ = 18 - 13 = 5 . Also by using mid point theorem OP = 1/2 AC = 1/2(4√14) = 2√14. Using pythagors in ∆OPZ :- OP²= PZ² + OZ² x² = 25 + 56 x² = 81 x = 9 This is the easiest and fastest method. Sir !

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@mohabatkhanmalak1161]

Wonderfull, I managed to get this one although had to scratch my head for a while. I knew the solution lay within that circle and just went through all the circle theorems. Thanks proffessor, I am 65 and enjoying math!☘🌵🌿

[@PreMath]

Glad it helped! ❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Stay blessed, Mohabat ji 👍

[@susanwang797]

From O, draw a line parallel to AC which intersects AB at E. Since triangle ABC is similar to OBE, It is easy to find OE =2•sqrt (14), PE= 5, so x = 9 by pythagorean theorem.

[@devondevon4366]

x=9 Since a triangle inside a semi-circle is a right triangle Using Pythagorean (4 sqrt 14 )^2 + 26^2 = c^2 c = 30 = diamter Hence the radius = 15 Using Thales theorem the chord intersect at 15 - x , and x+ 15 , as well 18 and 8 Hence 15-x * 15 + x = 8 *26 225 - x ^2 = 144 x ^2 = 81 x = 9 Can use trigometry to find x , since know all three sides and this is 30-60-90. To find x one can use 15, 30 degrees and 8 t

[@PreMath]

❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@montynorth3009]

Angle ACB = 90 degrees within semi-circle. Pythagoras, AB^2 = (16*14) + 26^2. AB^2 = 224 + 676 = 900. AB = 30. OB = 15. Cos ABC = CB/AB = 26/30 = 13/15. Triangle OBP cosine rule. x^2 = PB^2 + OB^2 - 2 * PB * OB * cos ABC. x^2 = 8^2 + 15^2 - 2 * 8 * 15 * 13/15. x^2 = 64 + 225 - 208. x^2 = 81.

[@ВерцинГеториг-ч5ь]

С центра О проводится прямая параллельная АС , которая перпендикулярная СВ , как и АС , и является срединой линией треугольника АВС , следовательно равна 1/2 АС=2\|14 , второй катет образовавшегося треугольника =(18+8)/2-8=5 . По Пифагору - Х*2=(2\|14)*2 + 5*2=81 , Х=\|81=9 .

[@Stanislaw2344]

Moc pěkný příklad. s elegantním řešením.

[@uwelinzbauer3973]

Hello friends of math and geometry. I used different steps, I first calculated angle OBC by inv tan, then used the cosine rule to find value for x. Worked as well, got x=9. Best greetings/wishes!

[@quigonkenny]

As AB is a diameter and C is on the circumference, ∆BCA is a right triangle. Triangle ∆BCA: (4√14)² + (18+8)² = AB² (2r)² = 224 + 26² = 224 + 676 4r² = 900 r = √(900/4) = √225 = 15 Let D be the point on OB where PD is perpendicular to OB. ∆BCA and ∆PDB are similar by SAS at ∠B. Triangle ∆PDB: DB/BP = BC/AB DB/8 = 26/30 = 13/15 DB = 8(13/15) = 104/15 PD/BP = CA/AB PD/8 = 4√14/30 = 2√14/15 PD = 8(2√14/15) = 16√14/15 Triangle ∆ODP: DP² + OD² = OP² (16√14/15)² + (15-104/15)² = x² x² = 256(14)/225 + (121/15)² x² = 3584/225 + 14641/225 x² = 18225/225 = 81 x = √81 = 9

[@reynaldowify]

Thinking out of the box, just mirror the semicircle below itself, . So , you get a rectangle with sides 4 * (sqrt 14) on one side, and 26 on the shorter side. So, the x twice , goes from upper side to lower side. Move the 2 x to the end, and it leaves a base of 10, and hight of 4 * (sqrt 14) By Pythagoras, you get a that 2 x is 18, leaving x as 9 Thanks

[@beta700a]

Really nice solution )) Thanks again for sharing your knowledge 🤩

[@ybodoN]

Generalized: x = ½ √((a − b)² + c²) where BP = a, PC = b and CA = c

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@Valentina-Tina]

Using Thales' theorem is much easier. Draw a parallel Y from the center of the circle to the shorter leg of the triangle.

[@soli9mana-soli4953]

Once known AB with Pythagorean th. you can find the solution applying the intersecting chords theorem

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@hanswust6972]

With radius = 15 and Cosine of angle OBP = 26/30, I used the Law of Cosines in trianhle OBP.

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@misterenter-iz7rz]

r=1/2×sqrt(16×14+26^2)=15, then x=sqrt((18-13)^2+(15^2-13^2)))=sqrt(18^2+13^2-2×13x18+15^2-13^2)=sqrt(18^2+15^2-2×13×18)=9.

[@huntbat]

A box has one too many corners, lets stay inside the triangle XD. Use Pythagorus' theorem to calculate length OB and you'll get 30. Therefore, AO = 15. Calculate angle CAB. Sin^-1(26/30) to get 60.073... . Join points O and C to form another radius, forming iscoceles triangle, AOC. angle CAB = angle CAO, Therefore angle CAO = angle OAC. Subtract angle ACO from 90 to get angle OCP. Use Cosine rule to find x. x = sqrt( 15^2 + 18^2 - 2•15•18•CosOCP)

[@giuseppemalaguti435]

Ipotenusa=2R=30...carnot teorema..x^2=15^2+8^2-2*15*8*cos(arctg(4√14/26))=81

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@stephenbrand5779]

I used the cosine rule to solve this problem. But I prefer your elegant solution.

[@PreMath]

❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@nehronghamil4352]

Alternate Solution: Angle ACB is right angle Call angle CBA angle B AB ^ 2 = 26 ^ 2 + ( 4 * 14 ^.5 ) ^ 2 AB = 30 cos(B) = 26 / AB = 26 / 30 = 13/15 (OB = AB/2) ^ 2 + 8 ^ 2 - 2(OB)(8)cos(B) = x ^ 2 225 + 64 - 240(13/15) = 81 = x ^ 2 x = 9

[@sarantis40kalaitzis48]

i solved it in my mind in less than half minute drawing perpendicular OM to the middle of chord CB,so OM=2*sqrt(14)=sqrt(56) and MP=(26/2)-8=13-8=5 . Then x^2=(sqrt56)^2+5^2=56+25=81 so x=9.

[@MrPaulc222]

Very clean method, thanks. Calculating radius was the easy part. From there I might have been using trig to work out the angle ABC in the right triangle and then, armed with side lengths of 8 and 15 and angle B in a scalene triangle, used the law of cosines to calculate x. I'm sure it would have worked but was much messier than your way.

[@PreMath]

Glad it helped! ❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@Ibrahimfamilyvlog2097l]

Nice sharing 🌹👌❤️ 🌹👌

[@murdock5537]

Nice! ∆ ABC → sin⁡(BCA) = 1 → AC = 4√14; BC = 26 = CP + BP = 18 + 8 → AB = 2r = 30 = AO + BO OP = x = ? cos⁡(β) = 26/30 = 13/15 → x^2 = 225 + 64 - 2(15)(8)cos⁡(β) = 225 - 144 = 81 → x = 9

[@ganymed1236]

Very clever, your method, Sir. I managed it with general cosine law

[@PreMath]

Well done! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@raya.pawley3563]

Thank you

[@PreMath]

You are very welcome! ❤️ You are awesome. Keep it up 👍

[@講場愛丁堡]

Don’t over think Make a mid point of BC as D Then OD=2sqrt14 PD=5 X^2=25+4*14 X=9

[@mohamedatteiya5077]

Hi, like ur videos. There is a much easier solution. Take a parallel line to AC from point O to intercept with CB in point G. Now OG will equal to half of AC. Also GB = half CB = 13. So GP=5. X^2= OG^2+ GP^2 = 4*14 + 25 =81 then X=9

[@Copernicusfreud]

Yay! I solved the problem .

[@johnwindisch1956]

Yea! I did it!

[@PreMath]

Bravo ❤️

[@leonardochrockattrodrigues7613]

Hello Professor, I propose an alternative solution for the problem: Draw, from O, a straight line parallel to AC, with the intersection point being close to BC as O'. The triangle OO'P will therefore be a rectangle in O'. Furthermore, O' is the midpoint of side AC, concluding that OO' is the midbase and is worth AB/2 = 2. sqrt(14). Now applying Pythagoras' theorem to the triangle OO'P: OP² = ÓP² + OO'² x² = [2.sqrt(14)]² + 5² where we obtain x = 9. Thank you very much for your channel

[@PreMath]

❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@Hayet-jb2sd]

Tres bien

[@Istaphobic]

Feel like I cheated a bit. Figured that the diameter was 30 using Pythagoras' like everyone else. Then I just used the cosine rule. Let ∠ABC = θ. ⇒ cos θ = 26/30 = 13/15. Now, using the cosine rule: x² = 15² + 8² - 2.15.8.cos θ ⇒ x² = 225 + 64 - 240.(13/15) ⇒ x² = 289 - 208 ⇒ x² = 81 ∴ x = 9.

[@andirijal9033]

Cosinus law in triangle OPB

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@chelliahRaveedrarajah]

Sir I have a doubt. Could you please help me to do this question? a²+b²+c²=1 a³+b³+c³=1 a+b+c=?.

[@egillandersson1780]

Thank you ! I get it by another way : find the radius by Pythagore (as you did) but then use cosine law on the triangle OBP

[@PreMath]

You are very welcome! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@abdulkadirbuyuksoy2076]

3teorem used woow

[@PreMath]

Thanks 😀

[@joeschmo622]

Too complicated. I did the solution in my head. Make a rectangle with short side 4rad14 and long side 26 (18+8). Now you got a line bisecting that rectangle whose base is 10 (26=8+10+8) and height 4rad14. Use pythagoras where 2x is the hypotenuse, come up with 224+100=324 = 81*4, sqrt that to 9*2. That gives length 2x, so length x=9.

[@AmirgabYT2185]

9

[@prossvay8744]

X=9

[@PreMath]

❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@comdo777]

asnwer=2/13 is it

[@講場愛丁堡]

Actually no need to find the radius

[@wackojacko3962]

Nine! 🙂

[@PreMath]

Thank you! Cheers! 😀 You are awesome. Keep it up 👍

[@Teamstudy4595]

1st like

[@PreMath]

Thanks for liking, dear ❤️

[@Teamstudy4595]

1st view

[@PreMath]

Super star 🌹

[@محمدالنجار-و2ف]

I' ve solved it more easily than that solution.

[@marioalb9726]

Pytagorean theorem: (2R)²=CB² + AC² 4.R² = (18+8)² + (4√14)² 4R² = 676 + 224 R= √225 = 15 cm Cosine rule: x² = a² + b² - 2 ab cos α x² = R² + 8² - 2.R.8.(26/2R) x² = R² + 64 - 208 x² = R² - 144 x = 9 cm ( Solved √ )

[@whoh3222]

9

[@DB-lg5sq]

X=9

[@PreMath]

Thanks 😀

[@yakupbuyankara5903]

X=9

[@devondevon4366]

x=9

[@PreMath]

Thanks ❤️