Area of Blue Triangle

I hope you guys like this one!

Let me know if you see an easier way to solve it!

Any requests for other things I should work on?

Thank you!

[oblivion]

instead of using the law of cosines, side c that is being solved for can be found more quickly with the pythagorean theorem. its touching the bottom of the small square and the top left of the medium square so its length is equal to the square root of 2^2 + 6^2

[EarlofCheese]

I’m proud of myself! I did it a little differently (not using law of cosines) and got the right answer all in my head. I’ve been enjoying your videos lately 😁

[jeffhearn]

Andy, congrats on the 100K subscribers…..”How exciting!”
Love the vids, keep up the great work.

[yurio]

I love your videos. For fun or studies they are the best ni doubt, just like you. Keep being amazing.

[manofmartin]

Im not sure what i like more, the solving of the equation or how exciting it is!

[PetDogScout]

100k subscribers for my favorite youtuber!!! Congratulations Andy. How exciting.

[Marzy]

This is the first video I’ve ever attempted one of these questions and I happened to almost get it correct

[ScubaBob-zmwo]

I think I can give a creative solution
Say the blue triangle with corners ABC, Boeing AB the side of the small square (you solved as x), and AC the other catheto (lets name it y), and BC the hypotenuse (lets name it z) equal to the side of the large square
The area of the blue triangle would be 1/2(xy)
Lets name the rightmost common point between the medium and small sqare D, and the common point between the medium and large square E
You'll now have 2 congruent right triangles, ABC and DBE (AB=BD=x, BC=BE=z, therefore AC=DE=y)
Now, you'll have another right triangle ADE, with AD=AE=4, where we can also conclude that DE=2x
So, the final blue triangle area is going to be 2x×x/2=x²=8

[guyonabike]

Solved with geometry:
Use a^2 + b^2 = c^2 to calculate the smallest square side length as sqrt8.
Noting that we have 3 squares in the diagram, the diagonal across the medium square can be completed up to its top left corner, and be of total length 2sqrt8.
We now have an identical triangle to the blue one, with short side length sqrt8 and perpendicular medium side length 2sqrt8.
Area = 1/2 x base x height
A = 1/2 x 2sqrt8 x sqrt8
1/2 x 2 cancels, so A = sqrt8 squared
A = 8 cm^2

[Ruija]

I just used the Pythagorean theorem a bunch. Everything about sines and cosines and such has escaped my mind since school.
The diagonal of the medium square is 4*sqrt(2), half of that is the side length of the small square (whose diagonal is 4). So the short side of the blue triangle is 2*sqrt(2).
With parts of these squares, the side of the large square forms the hypotenuse of a right triangle with sides 2 and 6. The large square's side length is 2*sqrt(10).
Then you get the missing side length of the blue triangle via Pythagoras again, that's sqrt(32). Half base times height [½ * sqrt(32) * 2*sqrt(2)] and you get an area of 8 cm^2.

[zg]

A generalized solution: Let a=side of the smallest square; the area of the blue triangle is always a^2.

Proof: Let ABC the blue right triangle, A=90°, AC=a. Let D the upper right corner of the medium square. We can easily prove that B, A, D are collinear, thus BCD is an isoceles triangle (BC=CD=CE, E is the upper left corner of the medium square), and AB=AD= 2a.

[GrandeZizinho]

Really like your channel. Thanks for the content

[stixoimatizontas]

Question: why does he check if the smallest square is actually a square, when all of the sides are equal? Did I miss something?
Edit: I just realised why he called it a rhombus. 4 equal sides doesn't necessarily mean square.

[danpliska]

congrats on the 100k subscribers. it has been a long journey. keep making great videos.

[thiagomotta]

i would have never thought that one of my favorite yt channels would be a guy doing maths

[pantumaka]

i did it by just figuring out the side of the rhombus, then making a right triangle with the shortest side of the blue triangle and half of the side of the big square to figure out that half, multiplied it by two, and did pythagorean on the blue triangle. in my opinion the way i solved it is a little better since it doesn't make use of the law of cosines and thus angles as a way of calculating sides but yours is still nice

[rupom_]

Got everything first try except the theta angle in the laws of cosine part
Although i understood it after your explanation

[SubhanSaeen]

Which software did you used to make these exciting videos.😊

[OptimusPhillip]

Huh, I had modeled the triple-ticked line segment as a vector, basing its horizontal and vertical components off the dimensions of the squares it touches the corners of. Got the same solution, though.

[a]

Side of medium square is b=4.
1. Side on small square is a=b sin(45)=2sqrt(2)
2. Mirror small square along right side of triangle.
3. Prove, that new triangle is equal to original one.
It's a right angled triangle with hypothenuse, equal to side of big square. And side equal to small square.
4. Since one side is a and other is 2a, S=2a²/2= a²=8

Completely no reason for equations.