You don't need all those constructions! As soon as you determine the congruency of the 2 triangles, you know both are 2 - 3 - √13 triangles. Side of the square is √13, so its area is 13.
thank you for the comment, the video wasn't necessarily about finding the area of the square but more to prove the Pythagorean theorem in a roundabout, serendipitous manner. Think of it as an exercise to introduce the Pythagorean theorem to students learning it for the first time.
This could be easily done by showing the triangles congruent at first and then using Pythagoras theorem to find out the side of the square and then the area
or... you could have half a brain, take one look at the drawing and realize that the two triangles are equal and know that the area of the square is 2 squared plus 3 squared.
Brains aren't created with the knowledge of the Pythagorean Theorem already impregnated in them, it had to be learnt by the brain at some point. The point of the video was more to prove that 2 squared plus 3 squared is the area of the square.
@@mathu6514 the title of your video reflects to finding the area rather than proving it. If it would have to be done in an Olympiad exam, then this approach shouldn't be applied.
I reflected each triangle along their hypotenuse and added two more on the other sides of the square. You end up with a 5*5 square, area 25. If you subtract 4 of the triangles, 4*1/2*2*3 = 12 you get 13.
Let the side of the square be a. a*sin(alpha)=3. ….(1) a*sin(belta)=2 or a*cos(alpha)=2. …..(2) Dividing (1) by (2) tan(alpha)=3/2 from which sec(alpha)=sqrt(1+(3/2)^2)=sqrt(13)/2 hence cos(alpha)=2/sqrt(13) From (2) a*2/sqrt(13)=2 then a=sqrt(13) Area=a*a=13
wow. why not just realize that the side of the square equals sqrt(2^2 + 3^2). (Pythagorean!) Square that to get the area 13 Btw, the exercise in the vid was to get the area using the most primitive method possible, which doesnt include trig function, or Pythagorean either, for that matter
You don't need all those constructions! As soon as you determine the congruency of the 2 triangles, you know both are 2 - 3 - √13 triangles. Side of the square is √13, so its area is 13.
thank you for the comment, the video wasn't necessarily about finding the area of the square but more to prove the Pythagorean theorem in a roundabout, serendipitous manner. Think of it as an exercise to introduce the Pythagorean theorem to students learning it for the first time.
@@mathu6514Then it's a good one.
But the thought process of doing it in a other way was also good.
This could be easily done by showing the triangles congruent at first and then using Pythagoras theorem to find out the side of the square and then the area
or... you could have half a brain, take one look at the drawing and realize that the two triangles are equal and know that the area of the square is 2 squared plus 3 squared.
yea even i was thinking the same
Brains aren't created with the knowledge of the Pythagorean Theorem already impregnated in them, it had to be learnt by the brain at some point. The point of the video was more to prove that 2 squared plus 3 squared is the area of the square.
Do not be mean to the nice math man.
@@mathu6514 the title of your video reflects to finding the area rather than proving it. If it would have to be done in an Olympiad exam, then this approach shouldn't be applied.
I reflected each triangle along their hypotenuse and added two more on the other sides of the square. You end up with a 5*5 square, area 25. If you subtract 4 of the triangles, 4*1/2*2*3 = 12 you get 13.
that's a nice way of looking at it, like a flower blooming.
c² = (a + b)² - 4(ab/2) = a² + 2ab + b² -2ab = a² + b² → a² + b² = c²
Let the side of the square be a.
a*sin(alpha)=3. ….(1)
a*sin(belta)=2 or a*cos(alpha)=2. …..(2)
Dividing (1) by (2)
tan(alpha)=3/2 from which sec(alpha)=sqrt(1+(3/2)^2)=sqrt(13)/2 hence
cos(alpha)=2/sqrt(13)
From (2) a*2/sqrt(13)=2 then a=sqrt(13)
Area=a*a=13
wow. why not just realize that the side of the square equals sqrt(2^2 + 3^2). (Pythagorean!)
Square that to get the area
13
Btw, the exercise in the vid was to get the area using the most primitive method possible, which doesnt include trig function, or Pythagorean either, for that matter
Very good 👍
very obvious