Fascinating Geometry Challenge: Find the Shaded Area - Semicircle inside a Right Triangle
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- Опубликовано: 16 сен 2024
- In this super-fun and challenging geometry problem, we've got a semicircle with an unknown radius partially located inside a right triangle with unknown side lengths. The truth is, the only lengths we know in this problem are two line segments on two of the sides of the triangle. And they look like they are totally unconnected to what we are trying to find. And we are asked to find the area of the turquoise-colored shaded region. This geometry problem has a medium level of difficulty, so it is not super-difficult but also certainly not too easy to solve. That makes it good for developing your skills in finding the area of the shaded region type of questions. If you want to try solving this on your own, go ahead and try. Here is a clue for you. You will also practice your skills in finding the areas of circles, sectors, and triangles using the radius, pi, base, and height. If you cannot solve it on your own, no need to worry because I am here to show you how to do it in five minutes. I hope you will think my video is cool, and feel like you have actually learned something helpful and nice by watching every second of this video!
Draw a line connecting the left end of the diameter to the left end of the chord. Since the angle is on the circumference and connects to each end of the diameter, the angle is 90°, so we have a 30-60-90 special right triangle. That means the line we just drew is the short leg, which is half the length of the hypotenuse and 1/√3 the length of the long leg. The hypotenuse is the diameter of the semicircle, or 2r, so the segment is equal to r, and the chord is thus equal to √3r.
Let the vertical segment on the left equal x. As x is also the short leg of a 30-60-90 special right triangle, the long leg (2r+1) equals √3x and the hypotenuse (√3r+2√3) equals 2x.
√3x = 2r + 1
x = (2r+1)/√3
2x = √3r + 2√3
x = (√3r+2√3)/2
(2r+1)/√3 = (√3r+2√3)/2
2(2r+1) = √3(√3r+2√3)
4r + 2 = 3r + 6
r = 4
Draw a radius from the center of the semicircle to the upper left end of the chord. As the triangle thus formed is isosceles (having two sides of length r), the internal angle at that intersection is also 30°, so the angle at the center of the circle is 180-2(30) = 120°.
The turquoise area is a circular segment, and its area is equal to the area of the sector it covers minus the area of the overlapping isosceles triangle.
A = (120/360)πr² - r²sin(120°)/2
A = π(4²)/3 - 4²(√3/2)/2
A = 16π/3 - 4√3 ≈ 9.827 sq units
Oh how did I forget to reply this for so long. Thank you very much for this alternative solution quigon.
A solution with similarity between two right trianle of 30,60,90 degree angles, the first one is that shown in the figure, the second is that inside the circle drawn with the intersection of the hypotenuse of the external triangle and the diameter. Setting the sides of the external triangle as:
x, 2x, x√ 3 we can write:
2x : (x√ 3 - 1) = x√ 3 : (2x - 2√ 3)
x = 3√ 3
radius = (x√ 3 - 1)/2 = (3√ 3*√ 3 - 1)/2 = (9-1)/2 = 4
and what follows is similar...
That's also a certainly valid and equally interesying method leading to the same solution. Thanks for taking the time.
Including the little professorreally put a smile on my face!
Thanks for the comment bro! I am showing all these great comments to my son!
Notación: Empezando por el vértice superior y en sentido antihorario el triángulo rectángulo es ABC; "D" es la intersección de AC con el semicírculo de radio "r" y centro "O", y "E" su proyección ortogonal sobre BC. El extremo izquierdo del diámetro horizontal del semicírculo es el punto "F". Llamamos "G" a la proyección ortogonal de D sobre AB.
Ángulo ADG=ACB=30º→ Ángulo central correspondiente =DOF =2*30º=60º→ DE=r√3 /2 → Si AD=2√3→ AG=√ 3→ GD=3 → EC=BC-BE =BC-GD =(1+2r)-3 =2r-2.
Los triángulos rectángulos AGD y DEC son semejantes → AG/GD=DE/EC→ r=4
Área del segmento circular sombreado = (Área del sector circular de radio=r y ángulo central=180º-60º=120º) - (Área triángulo OCD) = (4²π/3) - 4*(4√3/2)/2 =16π/3 - 4√3 =4[(4π/3)-√3] =9,8269.
Gracias por el vídeo. Un saludo cordial.
Gracias muchas for taking the time to provide this detailed solution Señor. It's great and encouraging to know that people from all parts of the world are watching!
Really cool excellent problem thank you for sharing,so happy your son is engaged at such an early age in developing a special talent☺️
Thanks for watching and the cool comment! I am showing all of these awesome comments to my son snd he loves them!
This is the best math channel i've ever struck. Keep it up man 💪💪💪
Thanks sir! I appreciate that!
Nice, A better path to solution
Feel free to share your solution with us whenever you feel like it.
cos30°= (2r+1)/x
hyp=(2r+1)/√3/2
hyp=(4r+2)/√3
chord=(4r+2)/√3- 2√3
(4r+2-6)/√3
(4r-4)/√3= 2rsin60°
4r-4= √3*2*r √3/2
4r-4=3r
r=4
Area=16π/3- ½(16)sin120°
Area=16π/3-8 √3/2
Area=16π/3-4√3
Thank you for taking the time to provide this explanation sir.
Never knew that geometry can be so fun
How did forget to reply this concise and awesome comment! Thank you for this!
Great problem. I think r =4 and the final answer is 9.83.
but after solving they got the value of r as 4 .... i didnt get your comment . a bit of explanation ?
r=4
@@ashis2000plusxD Edited
@@JSSTyger oh ok nice , I did think it might have been a mistake we all make mistakes but people like you fixed it :D... nice , I hope the very best for your future and have a great day ahead :)
Wow! That is a good geometric problem to solve. Also, I like how your son helps you in solving. Keep up the great work 👏🏻🙂👍🏻🥜
Thanks for the support dude. I love this comment! And if you enjoy this stuff, make sure you stick around for our future videos.
The professor was genius👍🏻😅
Thanks for this comment dude 😊 I'll show it to my son.
Which computer programs are you using to produce your videos?
Super easy for a JEE advance aspirant
You should try the latest one I published.
Shaded area 9.827 area units.
but can you find the diarhea
Sorry. I can only find the area.
you don't need to apply any formulas for this question the way you solved it is the way AI would
You talk to fast
Feedback is always useful to me. So people who read this comment, if you have time, why not find a video in my channel that you think has the best playback speed and tell me in the comments that it is the pace I need to replicate in my future videos. Knowing what people think really helps me!