Find the Distance CD in this Semicircle | Easy Step-by-Step Tutorial

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PreMath

PreMath

Күн бұрын

Пікірлер: 90
@hellosirjamshed9876
@hellosirjamshed9876 3 жыл бұрын
Good Work
@PreMath
@PreMath 3 жыл бұрын
Thanks for the visit sir. Stay blessed😀
@vvijayalakshmi2560
@vvijayalakshmi2560 Жыл бұрын
Bb
@JLvatron
@JLvatron 3 жыл бұрын
Amazing question. However, since drawings aren’t always to scale, you should say at the beginning that points D & E are tangent to the circle. Without that information, the solution just assumes it.
@ghhdcdvv5069
@ghhdcdvv5069 Жыл бұрын
تمرين جيد جميل. رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا. تحياتنا لكم من غزة فلسطين .
@waiphyoemg1668
@waiphyoemg1668 3 жыл бұрын
I use Pythagorean theorem. Let CD=x. Draw BC. Angle ACB is right angle because this angle subtends on the diameter. In rt triangle ACB, BC^2 = AB^2 - AC^2=9^2 - (5+x)^2 = 81 - 25 - 10x -x^2 =56 -10x - x^2 In rt triangle DCB, BC^2 = DB^2 - CD^2 = 6^2 - x^2 = 36 - x^2 therefore, 56 - 10x - x^2 =36 - x^2 So, 10x=20 then x=2 CD=2
@philipkudrna5643
@philipkudrna5643 3 жыл бұрын
Nice problem! I had figured out that the diameter of the semi-circle must be nine. Knowing Al-Akashi’s Law of Cosines would have very much helped me in calculating angle alpha, which I had arrived at correct 38,94 degrees, but only with help of a calculator using tangent and the height of the inner smaller triangle. Very cumbersome and far less elegant. From there I also arrived at x=2 using Thales, but also only with a calculator. I like that it is possible in a more elegant way without the help of a calculator! Unfortunately I hadn‘t realized that you could also have solved it using Pythagoras and the Triangles ACB and DCB easily without a calculator!
@procash1968
@procash1968 3 жыл бұрын
Superb Sir. Wish someone like you was my Maths teacher at school. If so, then I may have done my graduation & post graduation in Maths & become a maths teacher myself
@MrPaulc222
@MrPaulc222 2 ай бұрын
2-tangents give D=9, so R = 9/2 Thales gives ABC and therefore, BCD, as right triangles. CD is x. CB is y. (x+5)^2 + y^2 = 9^2 (1) y^2 + x^2 = 6^2 (2) x^2 + 10x + 25 + y^2 = 81 (1) y^2 + x^2 = 36 (2) x^2 = 81 - y^2 - 10x - 25 (1) x^2 = 36 - y^2 (2) 56 - y^2 - 10x = 36 - y^2 Simplify: 20 = 10x x = 2 Wow! Astonishingly, I have used a simpler, cleaner method than your good self. As this is such a rarity, I might frame it :) .
@thirumalaithathachary3111
@thirumalaithathachary3111 3 жыл бұрын
We can do it without involving trigonometry. Let CD=x. Join points C and B.. ' Triangle ACB and triangle DCB are right triangles. Therefore CB^2=36-x^2 ( Pythogorean Theorem ). Also, AC^2+BC^2=AB^2. Diameter of the circle=6+3=9. ( obvious ). Therefore (5+x)^2 + (36-x^2) = 81. Solve this equation to get x=2
@bkj556
@bkj556 3 жыл бұрын
This the level which I wanted. Thanks....
@Sultan-nx9jn
@Sultan-nx9jn Жыл бұрын
It can be solved without trigonometry as well. If we take a length of CD as x, then CB will be equal to 9^2-(5+x)^2 or 6^2 - x^2. Therefore, 9^2-(5+x)^2=6^2-x^2 => 81-36=5*(5+2*x) => 5+2*x=9 => x=2 or CD=2.
@mustafizrahman2822
@mustafizrahman2822 3 жыл бұрын
What a tremendous question it is! But I have failed to solve it. Thanks a lot for posting. I have saved it.
@PreMath
@PreMath 3 жыл бұрын
No worries. Keep it up dear 😃
@HassanLakiss
@HassanLakiss 3 жыл бұрын
Thank you for a nice question and as usual a good explanation
@wafikhwijeh4006
@wafikhwijeh4006 3 жыл бұрын
It's exciting problem and very interesting to me when i was watching your kindy explaning... Then i ask you sir to present most of your solution problems if that is possible with most best Regardes fir you sir.
@johnbrennan3372
@johnbrennan3372 3 жыл бұрын
Very good method------ always interesting.
@PreMath
@PreMath 3 жыл бұрын
Thank you John! Cheers!👍 Keep it up dear 😃
@lindafromcalifornia1155
@lindafromcalifornia1155 3 жыл бұрын
That was so cool to watch you work that out.
@ovalteen4404
@ovalteen4404 3 жыл бұрын
I figured out AC by placing point o at the center of the semicircle and drawing an isosceles triangle with both radii, splitting it into 2 right triangles, using the cosine to find the adjacent bisector of AC, and doubling it. So 4.5/(7/9)*2=7. Of course I've never been known to do things the easy way.
@predator1702
@predator1702 3 жыл бұрын
Thank you teacher. 👏👏👍
@PreMath
@PreMath 3 жыл бұрын
You are very welcome Thank you 👍 Keep it up dear 😃
@ferroggon
@ferroggon 3 жыл бұрын
I applied the Pythagorean theorem to the DCB and ACB triangles.
@ohiovic1236
@ohiovic1236 3 жыл бұрын
same
@markp7262
@markp7262 3 жыл бұрын
I also did that. As long as you knew Thales, it is a much simpler solve without needing to use Law of Cosines. 9^2 - (x+5)^2 = 6^2 - x^2, solve for x.
@ohiovic1236
@ohiovic1236 3 жыл бұрын
@@markp7262 Agree .. Bravoo
@Mathematician6124
@Mathematician6124 3 жыл бұрын
I drew a perpendicular from D to AB at point P to derive the cos theta. Let AP be equal to x and PB be equal to (9-x).This gave 25-x2=36-81+18x-x2, finally x =35/9 and cos theta equals AP /AD =7/9.
@وجديالنصرالله
@وجديالنصرالله 3 жыл бұрын
اسهل مما ظننت..شكرا
@harikatragadda
@harikatragadda 3 жыл бұрын
ACB and DCB are both right triangles with a common side CB. (5+CD)²-9²=6²-CD² CD=2
@gandharvagrover8396
@gandharvagrover8396 3 жыл бұрын
How is DCB right triangle? AD doesn't include the diameter for 90° angle between radius and tangent. Or you see it other way... Just asking...
@tosuchino6465
@tosuchino6465 3 жыл бұрын
@@gandharvagrover8396 In a semicircle, the angle between the lines drawn from the edges of the diameter to ANY point on its arc is always 90 degrees.
@rolandhutapea6390
@rolandhutapea6390 2 жыл бұрын
Yes, I did the same simple way...no need using cosine law
@xyz9250
@xyz9250 2 жыл бұрын
Great idea. But it should be 9^2 - (5+CD)^2 ,
@mahalakshmiganapathy6455
@mahalakshmiganapathy6455 3 жыл бұрын
Simply superb
@theoyanto
@theoyanto Жыл бұрын
Great example, didn't get, had to watch the video, and didn't think to use cosine law, and you did say in your note that you would be using it, so my fault, should pay more attention, thanks again 👍🏻
@242math
@242math 3 жыл бұрын
love this question, very well explained, thanks for sharing
@luigipirandello5919
@luigipirandello5919 3 жыл бұрын
Wonderful, Sir.
@xyz9250
@xyz9250 2 жыл бұрын
I didn’t apply the law of cosine, but dropped a line from D perpendicular to AB, call it DG, as it’s shared side of two right angle triangles, AD^2 - AG^2 = DB^2 - BG^2, as BG= 9 - AG, can get AG= 35/9, as AC/AB = AG/AD , (5+ DC) / 9 = ( 35/9)/ 5, and DC= 2
@pedroloures3310
@pedroloures3310 3 жыл бұрын
Nice problem! I tried using chords theorems but it didn't work very well. After some time I also came up with this solution. Thanks for the video.
@مجديمحمود-ذ5ذ
@مجديمحمود-ذ5ذ 2 жыл бұрын
فيثاغورث
@kunalchakraborty9735
@kunalchakraborty9735 2 жыл бұрын
I have solved this by pythagorous theorem 9^2 - ( 5 + CD) ^ 2 = 6 ^ 2 - CD^2 By solving this equation CD can be found. @cd=2
@Waldlaeufer70
@Waldlaeufer70 2 жыл бұрын
1) AE = AF = 3 2) BD = BF = 6 3) AB = AF + BF = 3 + 6 = 9 4) Pythagoras: AC² + BC² = AB² => AC = 5 + CD => (5 + CD)² + BC² = 9² => 25 + 10CD + CD² + BC² = 81 => 10CD + CD² + BC² = 56 5) Pythagoras: CD² + BC² = BD² => CD² + BC² = 6² => CD² + BC² = 36 6) 4)-5) => 10CD = 20 => CD = 2
@مجديمحمود-ذ5ذ
@مجديمحمود-ذ5ذ 2 жыл бұрын
هو ده الحل الاسهل
@Waldlaeufer70
@Waldlaeufer70 2 жыл бұрын
@@مجديمحمود-ذ5ذ بالنسبة لي ، كانت الأقرب.
@s1ng23m4n
@s1ng23m4n 3 жыл бұрын
Looks hard but solved in 2 mins. Nice problem :)
@janvanroekel6787
@janvanroekel6787 3 жыл бұрын
AB=3+6=9. Angle ACB =90 degrees. Combine Pythagoras in triangle ABC and triangle BCD. Resulting in DC=2.
@dipinds6446
@dipinds6446 3 жыл бұрын
Nice👍
@PreMath
@PreMath 3 жыл бұрын
Thank you Dipin! Cheers! 👍 Keep it up dear 😃
@tahasami597
@tahasami597 3 жыл бұрын
Thank for premath
@billylowgroundvytaszukas5797
@billylowgroundvytaszukas5797 2 жыл бұрын
beautiful
@kiranbarnwal8503
@kiranbarnwal8503 3 жыл бұрын
CD=2 by laws of sine and cosine Nice question
@PreMath
@PreMath 3 жыл бұрын
Excellent Kiran 👍 Keep it up dear 😃
@phanimaheswara7492
@phanimaheswara7492 3 жыл бұрын
Thank you sir
@PreMath
@PreMath 3 жыл бұрын
So nice of you Phani Thank you 👍 Keep it up dear 😃
@pranavamali05
@pranavamali05 3 жыл бұрын
Awesome
@mauriciosahady
@mauriciosahady Жыл бұрын
Teorema de Pitágoras duas vezes. That's enough! Triangulos ACB and ADB
@recepgocmen6848
@recepgocmen6848 3 жыл бұрын
very good
@rodolforathakrishnan8128
@rodolforathakrishnan8128 3 жыл бұрын
cute and cool! 💎💪🏽😉
@PreMath
@PreMath 3 жыл бұрын
Thank you Rodolfo 👍 Keep it up dear 😃
@kamaljain5228
@kamaljain5228 3 жыл бұрын
Since you did not really use \alpha, but just cos(\alpha), it is possible to solve the problem without trigonometry.
@ExpressStaveNotation
@ExpressStaveNotation 3 жыл бұрын
I wasn't content to just find CD = 2. So I worked out the radius of the green circle. (6r2/5)
@Mathematician6124
@Mathematician6124 3 жыл бұрын
You can make it harder. Find the length of the chord or find out the radius. Would you like to do this? It is a good sum. Please respond, would you like to make it harder?
@cherie1940
@cherie1940 Жыл бұрын
CD=x, BC=y x*2+y*2=6*2 (x+5)*2+y*2=9*2 25+10x+x*2+y*2=9*2 25+10x=81-36=45 10x=20 x=2 CD=2
@JC-xq2ec
@JC-xq2ec 3 жыл бұрын
I solved it just using the Pyth. Theorem.
@tosuchino6465
@tosuchino6465 3 жыл бұрын
Yes. So did I. No need for trig to solve this proboem.
@niltonsilveira4199
@niltonsilveira4199 2 жыл бұрын
CB=y: CD^2 + y^2=36 and (5+CD)^2+y^2=81 ====> CD=2
@qobilruzmatov48
@qobilruzmatov48 3 жыл бұрын
Nice good
@PreMath
@PreMath 3 жыл бұрын
Thank you 👍 Keep it up dear 😃
@VolkGreg
@VolkGreg 3 жыл бұрын
AB = AF + FB = AE + DB = 3 + 6 = 9 cos a = (AB^2 + AD^2 - DB^2) / (2 AB AD) = (9^2 + 5^2 - 6^2) / (2*9*5) = 70/90 = 7/9 AC = AD + DC = AB cos a DC = AB cos a - AD = 9 (7/9) - 5 = 2
@theophonchana5025
@theophonchana5025 3 жыл бұрын
Angle ACB = 90°
@theophonchana5025
@theophonchana5025 3 жыл бұрын
cos (alpha) = (5 + CD) ÷ 9
@adgf1x
@adgf1x 2 жыл бұрын
CD=7-5=2 unit
@micke_mango
@micke_mango 2 жыл бұрын
I'm sorry, but I think this is sloppy math! At the very least, we're presented far to little information at the beginning of the problem. * The only tangent point we know of is F. D and E is never mentioned to be tangent points so the lines that presumably (?) connect to these points aren't necessarily tangents. Which means that we cannot use the two-tangent theorem without adding assumptions. * We also don't know if AC is a straight line, i. e. there could be an angle in D, between AD and DC (≠ 180°). That means we cannot use Thales' theorem without proving (or assuming) that point D is a point on the straight line AC. Your intentions with this problem are understandable, but unfortunately not mathematically obvious - the necessary geometric details are understated. A good practice is to describe how the geometry details are constructed in the beginning of the problem - that will pinpoint necessary assumptions
@مجديمحمود-ذ5ذ
@مجديمحمود-ذ5ذ 2 жыл бұрын
بمكن الحل بتطبيق فيثاغورث اسهل جدا جدا
@theophonchana5025
@theophonchana5025 3 жыл бұрын
AD = 5 AE = 3 BD = 6 CD = ?
@theophonchana5025
@theophonchana5025 3 жыл бұрын
AB = 9
@Nishant-us2il
@Nishant-us2il 3 жыл бұрын
Watch at 1.5 x , it's slow other wise
@theophonchana5025
@theophonchana5025 3 жыл бұрын
CD + 5 = 7 CD = 2
@PreMath
@PreMath 3 жыл бұрын
Keep it up Theo dear 😃
@theophonchana5025
@theophonchana5025 3 жыл бұрын
Diameter = 9
@theophonchana5025
@theophonchana5025 3 жыл бұрын
BD = BF BF = 6
@theophonchana5025
@theophonchana5025 3 жыл бұрын
AE = AF AF = 3
@sumitsagar1993
@sumitsagar1993 3 жыл бұрын
2
@TomKaren94
@TomKaren94 2 жыл бұрын
Good one. But, again, too many steps from 5+CD=7 to CD=2.Too detailed explanation. I jump ahead through these videos out of frustration.
@Teamstudy4595
@Teamstudy4595 3 жыл бұрын
2nd comment
@PreMath
@PreMath 3 жыл бұрын
Thank you Jayant 👍 Keep it up dear 😃
@Teamstudy4595
@Teamstudy4595 3 жыл бұрын
Really tricky but. After seeing. Comments I realized x =2 (@ns)
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