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its not only visual, trying to solve that without remembering all the things about 30-60 triangles i proved that those are congruent triangles with basic trigonometry and so the problem is solved
If you bisect the shaded area between the diagonal and the 120 degree angle, you get 3 identical triangles in the lower right half of the rectangle. The shaded area is 2/3 of that, or 1/3 of the entire rectangle.
@@paulvansommeren You know the interior angle of the shaded triangle is 120 degrees from the alternate angle ('Z') theorum. That means it's a 30 30 120 triangle, which is isosceles; the bisector of the opposite angle goes through the midpoint of the hypotenuse. The smaller unshaded triangle and one half of the bisected shaded area are now congruent. It's a nice way of looking at it without ever having to assign values to sides.
@@paulvansommerenWhen you bisect that angle, you create two triangles. The corner angle is 30, the bisected angle is 60, so the remaining angle must be 90. That right triangle shares a hypotenuse with the small white triangle, and they're both 30-60-90. Therefore they must be identical.
"Hey guys, this looks like a fun one." It's been my experience that when you approach something with a positive attitude you can make that something fun, and that's what I see with your videos. I appreciate that.
The shaded area is twice the small triangle, because If you split it in half, the small triangle shares a side and an angle. So 3 small triangles fit in half the rectangle. Six in the full rectangle. 2 are shaded. 2/6. 1/3.
Feels like the scenic route was taken on this one. Just turn the rectangle into 4 30/60/90 triangles, find the hypotenuse of the upper left triangle, divide by 2 and call it a day.
You can also shift the top point of the triangle into the top left corner of the rectangle, then when we're told that the baseline of the triangle is 2/3's of the long side of the rectangle we can see that the triangle occupies half of 2/3 of the box, aka. 1/3 of the box. (The rule of shifting any corner of the triangle is that it must be shifted in a line parallel to the opposite side of the triangle. The area of the triangle is held constant since A=1/2*b*h doesn't care about the angles, and the baseline and height is the same)
@@keith6706draw a line that bisects the red from the 120 degree angle inside the red and you’ll get two triangles. If you work out all the angles of the triangles, i.e. the 2 red and 1 white, all the triangles must be equal. So red must be 2x white
I did something very similar using trig. We know from the top triangle that the length of the left side of the rectangle must be some multiple of sin30 and the top must be some multiple of cos30. For simplicity, let’s say that the diagonal along the rectangle is 1 so that they are just sin30 and cos30 respectively. This means that the right side of the rectangle must also be sin30. We know from the bottom triangle that the length of right side must be some multiple of cos30. But we already know this length is sin30. So we can compute the hypotenuse of the bottom triangle to be sin30/cos30. Then the bottom side of the bottom triangle must be sin^2(30)/cos30. Now we can compute the area of the top triangle to be √3/8 and the area of the bottom triangle to be √3/24. We can also compute the area of the entire rectangle to be √3/4. Then we can compute the area of the middle rectangle to be √3/12. Finally (√3/12)/(√3/4) = 1/3 so the middle triangle is roughly 33% of the rectangle.
When you found that the small triangle's hipothenuse was 2x, you could've noted that the shaded triangle had two equal sides due to the 30° angles, so the rest of the rectangle's side would've been 2x as well. From then on the areas would've been x.h and 2x.h, and h's value would've been irrelevant.
The larger of the two white triangles is clearly exactly half the rectangle’s area (its hypothenuse being the rectangle’s diagonal). The smaller white triangle is similar to the larger one, as both are 30-60-90 triangles. If you set the legs of the smaller to be 1 and SQRT(3), then the corresponding legs of the larger would be SQRT(3) and 3. From there, you can see that the smaller is one-third the area of the larger, which means it’s one-sixth the area of the rectangle. That leaves the red triangle with the remaining one-third of the rectangle’s area.
Nice, I did it this way: the hypotenuse is dividing the rectangle into 1/2, so I took the smaller right triangle and tried how many times it would fit into the triangle, so I know that two pairs would fit 3 times, so the small triangle is 1/6, now 1/2-1/6=4/12, 4/12=1/3.
So in your opinion the 3 areas are the same? The white and the other white is also 1/3? How can that be when there is a diagonal which halves a rectangle by definition? The white in top is exactly the same as the red + lower white.
Shaded area is isosceles therefore the 2 short sides are equal. Drop a line from the vertex of that isosceles to the base to make 2 congruent right triangles. The unshaded right triangle is also congruent as it shares a hypotenuse. So there are 6 congruent right triangles in the rectangle and the shaded area contains 2. (1/3)
I got there by a different route, but I liked your way too :) I had 1/2 + 1/3rd of 1/2 = 4/6 or 2/3, so the remaining was 1/3. However, like your method, mine also needed some prior knowledge.
Please correct me if I am wrong; wouldn't the fraction of the corner that it takes up always be equal to the fraction of the area of the rectangle it is in? Based on the fact that the most any triangle can take up of a rectangle (where it must touch the bottom opposite corner) is half and that angle would be half of 90?
I thought only 2 of them are equal so i got the result: (l²+b²)2l² where l is length and b is breadth of the rectangle. Plugging in the sides he got (3x and x√3) you still get 1/3 :D
I was bored and paused the video to do it myself before I saw your solution. If the angles are all equal, they're all 30 degrees. You can also figure out that the angle the shaded triangle makes with the bottom side of the rectangle is also 30 degrees based on some geometric theorem (I think it's alternate interior angles or something). Using that, you can determine that the shaded triangle is an isoceles triangle, with the base running along the diagonal. From there, you need to determine the side lengths. If the short side of the rectangle is a length x, then the base of the shaded triangle is 2x. To determine the height, you need to figure out that the height makes yet another 30 60 90 with half the base and one of the sides as the hypotenuse. Using ratios, you find the height is x/sqrt(3). Multiplying the area of the shaded triangle you get x^2 over sqrt(3). The area of the rectangle is sqrt(3) times x^2, so now you can find the proportion by dividing the shaded area by the rectangle's area. x^2 over sqrt(3) times 1 over (sqrt(3) times x^2) equals 1/3.
I think this problem has a smallest way so can you . I think it is 90/3 3 is for the equal angles hence 30/90. Also the first angle which is 30 is at the half side of the rectangle so 1/2 then the left angle is 60 so all angles are equal so 30 and 30 so 1/3
I don't see why you would want or need to. As the right angle is trisected, you get that these angles are 30 degrees for free. Once you know that, the rest is simple. Sure, we could do the extra steps - noting that an equilateral triangle has all angles congruent, and triangles in a plane have interior angles totaling 180, so each angle is 60, then drawing an altitude for the equilateral triangle, noting that the bisected angle must be 30, noting the bisected base is half the hypotenuse, and using Pythagoras to find the long leg, and then yeah, you don't need to know the 30-60-90 triangle. But why not just remember it and save yourself a lot of re-proving the basics? Knowing the 30 60 90 triangle in geometry and trigonometry is like knowing how to clip your toenails in real life.
@@Qermaq Thanks for the explanation. See, the thing is that I totally understand the reason behind the angles having their respective values, But i thought of asking my question bcuz often in my school we have to prove a bunch of things like this in geometry. Most of the times we need to do it without any angle measurements and just with the formulas. That's why I thought it would be good ask Andy, even tho it would be easy to work it out if I try.
Nah the angles are same if small triangles 60 degree root 3 big triangles 30 degree becomes root 3 so 60 degree become 3 3-1 2 twice size of small triangle 1/2.2/3= 1/3
Can you calculete whats The probality of being in a game named Valorant and The 2 team picked same agents. There are 24 agents and in one team 2 people can not choose The same agent
Damn I used the fact that the biggest triangle is half of the rectangle and then used the similar triangle thing to calculate that the small triangle is 3 times as small as the big one so together they’re 1/2 + 1/6 = 2/3 so the shaded triangle is 1/3.
This problem only has a solution for this specific case of rectangle. If you allow for a slightly wider or narrower rectangle, then there is not a clear solution
@@Grizzly01-vr4pn The three angles would be equal, as that was the only condition on the problem. This rectangle was chosen specifically such that its long side was sqrt(3) * its short side. If the side lengths did not have this condition, then none of the lines would be a diagonal of the rectangle.
@@sweetcornwhiskey if you change the relative dimensions of the rectangle, the 3 angles would no longer be equal. It would no longer be the same question.
