I solved it in a different way. Draw a height of this triangle which belongs to the hypotenuse. Because of isosceles right triangle, it goes through midpoint of hypotenuse and also the vertex with right angle. We can find that lenght easily. sqrt2 + 2*sqrt2 + 1. The lenght of the hypotenuse will be twice larger. The area will be equal (1+3*sqrt2)^2 and that is 19 + 6 sqrt2. Thanks for the video. Nice problem 😃
@@Zollaho yes i did it mentally too.Although my memory is not too good so i have to keep rechecking. So not 90 seconds. I think its a useful discipline. It can stop you going down a rabbit hole. In this case, 1) measure (x) the dropped perp. across the middle of the triangle corner to centre, centre to centre, centre to edge= root2 +2root2 +1 = x. 2) check that area is x^2 3) 3root2+1 squared = Square each of the terms 9×2 +1 =19 Twice the cross product= 6root2
We can also calculate the area using the height (from right angle) h = √2 +2√2+1 = 1+3√2 in this triangle height is also median, so hypotenuse = 2h the triangle is straight and isosceles -> side = hypotenuse/√2 = h√2 and area = h^2 area = 19 + 6√2
Well actually you can find 'b' even with similar triangles! I did the same exact thing but except using trig I used similar triangles ratio of the sides , notice if you join the centres of the circles then you will get a right triangle with angles 45° 45° and 90° exactly same as the main triangle and the lengths will be 4,4,4√2 using ratio of hyp to any leg you will get 4√2/4= 4√2+2b/5+b And then could be solved for b = √2+1 and then just calculate the area which is quite straight forward ,
I just cut the triangle exactly in the half ( with the rect of right corner and hipotenusa middle point) Then I got a isosceles right triangle with de half of the originals triangle area The side of the triangle can be easily calculated: Diagonal of the square formed from the 4 circles centers = 2√2 Distance between right corner and center of the closet circle = √2 Distance of tangent point an center of circle in the middle of hipotenusa = 1 Total = 3√2 + 1 This squared is= 19 + 6√2 Divide by 2 and multiplied by 2(because it's the half of the originals triangle area) Final answer: 6√2 + 19
@@joseaugustogomes9598 your solution was great also you should have written "Hypotenuse" instead of "hipotenusa" and "rectangle" also "note:" instead of "notice" other than those minor English mistakes you're good to go!
your solution is far fetched. The length of the left bottom corner of the triangle to the middle of the hypothenuse is 3*sqrt(2) + 1. This is one of the right hand sides of a isosceles right hand triangle which is half of the full triangle so the area of the full triangle is (3*sqrt(2)+1)².
Sometimes he shows not the simplest solution, but a solution he finds more interesting. Many ways to solve this, and if you use a proper method and get the right answer it's fine.
Man, I love your videos. Keep up the great work.
I appreciate that!
@@SyberMath Thanks..i thought you would just ise a calculator for the tangent of pi/8..why not..to show how else to do it? Thanks.for sharing..
Please check MY SOLUTION: ruclips.net/video/cZ3oPgXcpyY/видео.html
to see if you like it or not.
i love to see this channel grow. you obviously deserve it fam
Thank you! I enjoy solving problems and people seem to enjoy seeing it
I solved it in a different way. Draw a height of this triangle which belongs to the hypotenuse. Because of isosceles right triangle, it goes through midpoint of hypotenuse and also the vertex with right angle. We can find that lenght easily. sqrt2 + 2*sqrt2 + 1. The lenght of the hypotenuse will be twice larger. The area will be equal (1+3*sqrt2)^2 and that is 19 + 6 sqrt2. Thanks for the video. Nice problem 😃
did the same
Did the same mentally in 90 secs !
Nice! Thank you!
Yep that was my approach as well.
@@Zollaho yes i did it mentally too.Although my memory is not too good so i have to keep rechecking. So not 90 seconds.
I think its a useful discipline. It can stop you going down a rabbit hole.
In this case,
1) measure (x) the dropped perp. across the middle of the triangle corner to centre, centre to centre, centre to edge= root2 +2root2 +1 = x.
2) check that area is x^2
3) 3root2+1 squared =
Square each of the terms
9×2 +1 =19
Twice the cross product=
6root2
Finding the base of triangle was awesome
Thank you!
We can also calculate the area using the height (from right angle)
h = √2
+2√2+1
= 1+3√2
in this triangle height is also median, so hypotenuse = 2h
the triangle is straight and isosceles -> side = hypotenuse/√2 = h√2
and area = h^2
area = 19 + 6√2
Good!
Please check MY SOLUTION: ruclips.net/video/cZ3oPgXcpyY/видео.html
to see if you like it or not.
Well actually you can find 'b' even with similar triangles! I did the same exact thing but except using trig I used similar triangles ratio of the sides , notice if you join the centres of the circles then you will get a right triangle with angles 45° 45° and 90° exactly same as the main triangle and the lengths will be 4,4,4√2 using ratio of hyp to any leg you will get
4√2/4= 4√2+2b/5+b
And then could be solved for b = √2+1 and then just calculate the area which is quite straight forward ,
Nice!
Please check MY SOLUTION: ruclips.net/video/cZ3oPgXcpyY/видео.html
to see if you like it or not.
Very good question sir
I see your math videos regularly
I am lecturer of mathematics in India
Thank you
All the best!
I solved it orally within 2 mins
Great video!
Thank you!
If we can fit 6 circles in that way in a right triangle it is automatically a isosceles one
No need to mention
Why did you need the tan?!
You calculated the length without it
I used tan(π/8) to calculate the base, hence the area.
@@SyberMath you calculated the tan by calculating the actual base of the same small yrllow triangle
I just cut the triangle exactly in the half ( with the rect of right corner and hipotenusa middle point)
Then I got a isosceles right triangle with de half of the originals triangle area
The side of the triangle can be easily calculated:
Diagonal of the square formed from the 4 circles centers = 2√2
Distance between right corner and center of the closet circle = √2
Distance of tangent point an center of circle in the middle of hipotenusa = 1
Total = 3√2 + 1
This squared is=
19 + 6√2
Divide by 2 and multiplied by 2(because it's the half of the originals triangle area)
Final answer: 6√2 + 19
Notice: I don't speak English very well, sorry I wrote something wrong or confused.
Np. Looks good to me!
@@joseaugustogomes9598 your solution was great also you should have written "Hypotenuse" instead of "hipotenusa" and "rectangle" also "note:" instead of "notice" other than those minor English mistakes you're good to go!
Please check MY SOLUTION: ruclips.net/video/cZ3oPgXcpyY/видео.html
to see if you like it or not.
reminds me of presh talwalker
Wow
I know you probably meant well, but please don't insult him like that. He is so much better than Presh.
Thanks for the compliment! I really appreciate that! 🥰
@@chrissekely that is not an insult
"that reminds me of a person" is not an insult. how did i insult him.
your solution is far fetched. The length of the left bottom corner of the triangle to the middle of the hypothenuse is 3*sqrt(2) + 1. This is one of the right hand sides of a isosceles right hand triangle which is half of the full triangle so the area of the full triangle is (3*sqrt(2)+1)².
Sometimes he shows not the simplest solution, but a solution he finds more interesting. Many ways to solve this, and if you use a proper method and get the right answer it's fine.
But trigonometry is fun! 😁
Güzel içerik :)
Sevindim!
Nycccc vedio
Thanks1
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