SNL Church Lady says, "isn't that special!" ....I say of course it is, and the Circle Theorem applied to this problem using radius and tangent angle 90⁰ is a very special case of the 7 main Circle Theorems. I absolutely love it. 🙂
I did the same as you, but didn't use OP = x + 27 for the Pythagorean theorem which then requires the quadratic formula; I used OP as a whole to get 39, and then subtracted 21 and 6 to get 12.
My brain did this in more steps. I took the OP distance for my C in the Pythagorean Theorem. 15^2 + 36^2 = C^2. This comes to C = Sqrt(1,521) = 39. Then the basic math of 39 - 21 - 6 = 12. Just another way to solve the same problem. Love these!
I like how you broke down the geometric problem, but it would be SO MUCH SIMPLER to solve for OP and then subtract the two radii. The expasion of (x+27)**2 seems needlessly complicated. Thank you for the video.
With you as far as constructing triangle OEP. Then I realized that OEP is an integer Pythagorean triple, 15-36-39; So OP = 39 and x = 39 -- 27) = 12. Elapsed time about 90 seconds. Cheers. 🤠
If PB is extended beyond B and the parallel to OP is drawn through A until it meets the extension of PB in R, the parallelogram OPRA is obtained. PR=AO=21 so BR=21-6=15. In the right triangle ABR we have BR=3*5, AB=3*12 so AR closes the Pythagorean triple (5,12,13) with 3*13. OP = AR --> 21+x+6=3*13 --> 3*7+x+3*2=3*13 --> x= 3(13-7-2)=3*4=12.
Generalized: _x = √((R − r)² + a²) − R − r_ where _R_ is the radius of the large circle, _r_ is the radius of the small circle and _a_ is the tangent AB.
As OC and PD are radii of their semicircles, they equal 21 and 6 respectively, and OP = 21+x+6. Let E be the point on OA where PE is perpendicular to OA. As AB is tangent to both semicircles, PE is parallel to AB, and EPBA is a rectangle. Triangle ∆PEO: a² + b² = c² (21-6)² + 36² = (x+27)² x² + 54x + 729 = 225 + 1296 = 1521 x² + 54x - 792 = 0 x² + 66x - 12x - 792 = 0 x(x+66) - 12(x+66) = 0 (x-12)(x+66) = 0 x - 12 = 0 | x + 66 = 0 x = 12 ✓ | x = -66 ❌ x > 0 x = 12
In this example, there is no need to use the cumbersome way of quadratic equation: On the 15^2+36^2=(x+27)^2 just leave the term on the right side as is: 225+1296=(x+27)^2 => 1521=(x+27)^2 Take the square root now: sqrt(1521)=sqrt((x+27)^2). Since OP is a positive value, we only need to take care about the positive value: 39=x+27 => 39-27=x -> x=12. Alternatively, we also can use 15^2+36^2=OP^2 and 21+x+6=OP So it will be obvious to solve the first equation for OP first which will result in OP=39. Then just replace OP on the second equation: 21+x+6=39 which is trivial to solve.
let the distance be 21+6+x = root 2 of (36^2+15^2)..... root 2 of (1296+225)= root 2 og 1521 = 39....easy bc 40^2 = 1600 and 1 step back is 1600 -40-39 = 1521... now 39-21-6 = 12 all fast and in head.. as long as you know your square numbers
@@ashieshsharmah1326 Draw radii of two circles to get a trapezium, and a line parallel to the external tangent joining the center of the small semicircle, we get one rectangle dimension 36x6 and a right angled triangle 15x36x39(in ratio 5:12:13),and draw two lines from the two end points of perpendicular to the external tangent,.......
Another way: Extend the line BP down with 21 length. Connect the point to O to form a rectangular. Then, you will form a right angle triangle below with the same of yours. The rest of calculations is the same as yours 😊
Absolutely wonderful ❤
Glad you think so!
Thanks for your feedback! Cheers!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
OP^2= 36^2 + 15^2
OP = 39
x=39-21-6=12
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
yes i did that exactly the same way before watching tthe explanation that is surprisingly more complicated :]
I also solved it the same way before watching the video.
@@kokodin5895 yes...he really likes to be professor- aknowledged i mean..
take an easy problem and make it hard.
Right, clean and simple. I dont unnderstand why the complicating
SNL Church Lady says, "isn't that special!" ....I say of course it is, and the Circle Theorem applied to this problem using radius and tangent angle 90⁰ is a very special case of the 7 main Circle Theorems. I absolutely love it. 🙂
U can use Pythagoras for the triangle of 36 and 15 sides to have 39 units for the 3rd side and consequently x=39-(6+21)=12
I did the same as you, but didn't use OP = x + 27 for the Pythagorean theorem which then requires the quadratic formula; I used OP as a whole to get 39, and then subtracted 21 and 6 to get 12.
Πολύ σωστά. Ακριβώς έτσι και αποφεύγουμε την εξίσωση 2ου βαθμού.
In the RT angle triangle OEP, OE= 15, EP=36.
Using Pythagorean theorem
OE^2+EP^2=OP^2
225+1296=OP^2
OP=√1521= 39
Hence X = 39-21-6= 12
Absolutely and it won't give you a X square to dwell on!
Nothing like squaring things just to then have to root them later... the 6 and 21 will still be 6 and 21
There is a 5, 12, 13 Pythagorean triple hidden in there. 😊
Good spot. I saw the 15, 36, 39 but didn't connect with 5, 12, 13 even though they are all integers.
My brain did this in more steps. I took the OP distance for my C in the Pythagorean Theorem. 15^2 + 36^2 = C^2. This comes to C = Sqrt(1,521) = 39. Then the basic math of 39 - 21 - 6 = 12. Just another way to solve the same problem. Love these!
I like how you broke down the geometric problem, but it would be SO MUCH SIMPLER to solve for OP and then subtract the two radii. The expasion of (x+27)**2 seems needlessly complicated. Thank you for the video.
Is that a scaled 5-12-13 triangle I see? I do like a whole number answer.
Sir outstanding video❤
Glad you think so!
Thanks for your continued love and support!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
Triangle OPE is of the 5,12,13 type.😊
شكرا لكم
يمكن استعمال X نقطة تقاطع (AB) و (OP) والمثلث القاءم الزاوية AOX.....
Thanks for video.Good luck sir!!!!!!!!!!!!!
Using Pytagorean theorem over the APB triangle:
y=AP
So, y²=15²+36²
y²=225+1296
y²=1521
y=√1521
y=39
y=21+6+x
39=21+6+x
x=12
Coffee time problem, but only because I recognised that you used this approach in a similar question ...
Glad I was paying attention 🤓👍🏻
there is a shorter calculation... 15^2+36^2=op^ therefore op=39, then x=39 - (21+6) which is 12... anyway nice one thx
With you as far as constructing triangle OEP. Then I realized that OEP is an integer Pythagorean triple, 15-36-39; So OP = 39 and x = 39 -- 27) = 12. Elapsed time about 90 seconds.
Cheers. 🤠
If PB is extended beyond B and the parallel to OP is drawn through A until it meets the extension of PB in R, the parallelogram OPRA is obtained. PR=AO=21 so BR=21-6=15. In the right triangle ABR we have BR=3*5, AB=3*12 so AR closes the Pythagorean triple (5,12,13) with 3*13. OP = AR --> 21+x+6=3*13 --> 3*7+x+3*2=3*13 --> x= 3(13-7-2)=3*4=12.
Good point. When you see the 5-12-13 right triangle you can figure out the solution in your head.
Generalized: _x = √((R − r)² + a²) − R − r_ where _R_ is the radius of the large circle, _r_ is the radius of the small circle and _a_ is the tangent AB.
As OC and PD are radii of their semicircles, they equal 21 and 6 respectively, and OP = 21+x+6.
Let E be the point on OA where PE is perpendicular to OA. As AB is tangent to both semicircles, PE is parallel to AB, and EPBA is a rectangle.
Triangle ∆PEO:
a² + b² = c²
(21-6)² + 36² = (x+27)²
x² + 54x + 729 = 225 + 1296 = 1521
x² + 54x - 792 = 0
x² + 66x - 12x - 792 = 0
x(x+66) - 12(x+66) = 0
(x-12)(x+66) = 0
x - 12 = 0 | x + 66 = 0
x = 12 ✓ | x = -66 ❌ x > 0
x = 12
I drew BE as a parallel to OP. E is on AO. Then used Pythagorean theorem to find BE=39 which is equal to OP thus 39-21-6=12.
In this example, there is no need to use the cumbersome way of quadratic equation: On the 15^2+36^2=(x+27)^2 just leave the term on the right side as is: 225+1296=(x+27)^2 => 1521=(x+27)^2 Take the square root now: sqrt(1521)=sqrt((x+27)^2). Since OP is a positive value, we only need to take care about the positive value: 39=x+27 => 39-27=x -> x=12. Alternatively, we also can use 15^2+36^2=OP^2 and 21+x+6=OP So it will be obvious to solve the first equation for OP first which will result in OP=39. Then just replace OP on the second equation: 21+x+6=39 which is trivial to solve.
nice one yr..superb
Thanks for your continued love and support!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
Extend BP by 15 to R and join RO ( this is 36). OPsq = 36sq + 15sq = 1521. OP = 39. So X = 39 - (21 + 6) = 12.
36^2 + 15^2 = 1296 + 225 = 1521, so OP also equals sqrt(1521) = 39.
39 - 21 - 6 = 12.
x = 12.
a = 15 = 3 * 5
b = 36 = 3 * 12
Therefore, we have a 5:12:13 right triangle and
c = OP = 3 * 13 = 39
x + 27 = 39
x = 12
I solved it. Thanks for such questions 😊
Yay! I solved it.
Please start some questions about cubic equation , please it is my request :(
Draw OX parallel to AB. XB = 21, therefore(sic) XP=21 and XP=15.... [X+27] = sqrt[1521]
The -66 answer is correct as well. It places the smaller semicircle on the lefthand side of the larger semicircle with a gap of 12 between them.
Correcto; físicamente la ecuación nos entrega dos respuestas valederas.
∆ OPE; PE = 36, OE = 15 so OP= 39, X= 39-21-6=12
that was an easy one
let the distance be 21+6+x = root 2 of (36^2+15^2).....
root 2 of (1296+225)= root 2 og 1521 = 39....easy bc 40^2 = 1600 and 1 step back is 1600 -40-39 = 1521...
now 39-21-6 = 12
all fast and in head.. as long as you know your square numbers
Tripel pithagoras 5,12,13
X = 39 - 27 = 12
Nice!
Some steps could have been dropped.
1521 = 39² = (x+25)²
x = ±39 - 25
x = 12
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Using tecnique of similar triangle, x=(36-21x12/13-6x12/13)x13/12=(36-27x12/13)x13/12=39-27=12.🙂
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Can you please explain little bit .
@@ashieshsharmah1326 Draw radii of two circles to get a trapezium, and a line parallel to the external tangent joining the center of the small semicircle, we get one rectangle dimension 36x6 and a right angled triangle 15x36x39(in ratio 5:12:13),and draw two lines from the two end points of perpendicular to the external tangent,.......
[]《 *_Absolutely Beautiful Presentation and sharing_* 》 []
[]《 *_Greetings and Blessings from Hamida Qayyum_* 》 [] LIKE 240
Another way:
Extend the line BP down with 21 length. Connect the point to O to form a rectangular. Then, you will form a right angle triangle below with the same of yours.
The rest of calculations is the same as yours 😊
Lo hizo demasiado largo.
Se puede hacer en dos pasos.
1-calcular OP con el teorema de pitágoras.
2-Al resultado anterior restarle 21 y 6.
Eso es todo
The Pythagorean Theorem did pay off
12
Second one to addlike.😚
Thank you! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
x= 66😮😮😮
x= (36^2+15^2)^(1/2)- 21- 6= 12😊
Pythagoras theorem
Easy, x=12
12 ezzz