The last question is not well-defined. The answer to that is "you can make the area arbitrarily small". Just put two points arbitrary close to one vertex and the third point to the opposite side. Resulting triangle will be arbitrarily close to a line, so its area will be as small as you want. It's all "legit" too - because you avoid the "cheating" by turning the triangle into a line exactly, but it goes to show that there is no answer if we constrain ourselves only to a "proper" triangle.
@@loadstone5149So you chose your points on lines 3 and 4 to be arbitrarily close to the 3,4 vertex of the original triangle. Then you chose your points on the 5 line to be arbitrarily close to the 3,5 vertex. Now you have a triangle that is arbitrarily close to matching the 3 line.
The minimum triangle (question at the end) is the one, where two of the points are almost at the tip of the smaller angle and the the third point almost at the right angle. The area is miniscule, basically zero, as the triangle is almost a line by that point.
Similar solution: the point on the hypotenuse is close to 1 corner, the point on the vertical is close to the bottom, and the point on the bottom is anywhere. Since you can get as close as you want to the corners, the area approaches 0.
Another way of doing it is by finding the roots of the parabola. Doing so, you get that the area is smallest at x=0 or x=4. Assuming you want a non-zero answer, then there isn't an exact answer, because for any non-zero value you plug in to the area equation (x-4)*y, there will always be a smaller number that you could plug in.
@@qqqalo this has nothing to do with the parabola, we're talking about the puzzle he gave at the end, which is completely different to whatever problem you're trying to solve
Double the triangle, make a rectangle. Construct a perpendicular intersection within the interior of the rectangle perpendicular to the sides of the rectangle. There will be exactly one complete rectangle where the hypotenuse does not intersect the rectangle, until the interior corner of the rectangle hits the midpoint of the hypotenuse, at which point the single rectangle duplicates. This maximizes size of the rectangle, which is 1/4 the area of the larger rectangle defined by the shorter sides of the right triangle.
Minimal triangle would be A-->0. Pick a vertex. Point at the hypotenuse and on the adjacent side approach that vertex, point at the opposite side approaches the adjacent side (or hypotenuse, doesn't matter). The triangle collapses to a line, and with infinitessimal height right before it does, ie, right before all 3 points become collinear, the area of that triangle approaches 0.
I made x the ratio of the left piece (where you put x) and the full length (4); therefore the size of the rectangle is (3x) * (4 - 4x), which simplifies to 12 * (x - x^2). Setting the derivative to 0 gets: 0 = 12 * (1 - 2x) 0 = 1 - 2x x = 0.5 And the area is: 12 * (x - x^2) = 12 * (0.5 - 0.25) = 12 * 0.25 = 3
I solved it for a rectangular base being shared with the '5' side. The answer is still 3 ! Do to so, I minimised the 3 small triangles formed by inserting a diagonal rectangle. If you shuffle those 3 triangles around, you can recover the bprp initial assumption .
For any height h>0 and any base b>0 (in this example 3 and 4 respectively) the equation for the point on the diagonal is y = (h/b) * x and the area as a function of x is (b-x) *y = h*x - (h*x*x) / b. the derivative for the area as a function of x is h - (2*h*x) / b = h(1- (2*x)/b). setting this equal to zero, either h = 0, which is impossible, or 1 = (2*x) / b b/2 =x. replacing x in the earlier equation :y = (h/b) * (b/2) y = h/2. In other words, the max area for a rectangle inscribed in a right triangle is always when the base and height of the rectangle are half of the base and height of the triangle( if the base and height are parallel to its sides, idk if putting one of the sides on the hypotenuse might give a bigger area).
Enjoyable problem. Not too hard but still requires some thought. I solved it exactly as our host, except I had forgotten the equation for the max/min of a parabola; probably because I learned it 40 years ago and hadn't used it since. However I had not forgotten how to take a derivative of a simple quadratic.
For the original question I used the fact that the smaller triangles left after taking away the rectangle are similar to the original. If you define a scaling factor s ( 0 < s < 1) where is the proportion of the base of the triangle that the rectangle occupies then the area of the rectangle is 4s * (3-3s) - 12s - 12s². To get the maximum, take the root of derivative i.e. 12 - 24 s = 0. Thus, s = ½. and A = 3.
For largest area of triangle, dA/dx = 0, or 6x/4=3 or x=2. This way I don't have to remember equation of parabola or its peak point. (To make sure we get the highest area, not the lowest, one can take second derivative of A, if needed).
Assume the vertex on the right vertical side is sliding up, the vertex on the horizontal slide is moving right. So, as the hight of the rectangle increases, the width decreases. With this assumption, and mapping the both extremes to 0 and 1, we can get the area with the formula 3x*4(1-x). simplifying it we get 12x-12x^2. Take the derivative (12-24x), compute the local maximum(x=0,5), plug it into the original formula 12*0,5-12*0,5^2 and you get 3.
Man this was amazing, I solved this using differential calculus but this approach was superlative, and outstandingly genius, gave me a dopamine rush fr.
given the formula for area, A and that the maxinun is where dA/dX = 0. So differentiate the area formula and set to 0. A = 3x -(3/4)x^2 d/dx = 3 -(6/4)x set = 0 0 = 3-(6/4)x or -3(-4/6)=x 4/2 = x so x = 2
For the question at the end of the video, I selected 3 points (a,0) where a is between 0 and 4 (4,b) where b is between 0 and 3 and (c, ¾c) where c is between 0 and 4 The area they make is (12c-3ac+4bc+4ab) all over 8 I need to learn how to get the minimum, though.😅
Same answer, different setup. I imagined the vertex on the hypotenuse as a bead on a string that I can slide left or right. That reduces the problem to one variable, since the bead's position determines all other vertices and lengths. I set the length of string left of the bead to "x", which makes the length of string right of the bead "5-x". Since the inner triangles are also 3-4-5 ratio, the rectangle is (3/5)(x) tall, and (4/5)(5-x) wide. Multiplied to get area, and found maximum from there.
An easy reflexion can help us fond the solution without any calculus If we take a isocele right triangle, by symmetry, the largest rectangle is the one with half the size of the side of the triangle Then if you stretch the triangle horizontaly from a certain factor, the aera of any rectangle will increase in a linear manner from the same factor, this operation doesn’t change the local maximum
I had a smaller 3:4:5 triangle above the square with vertical h. This gave me the area if the square A = 4h - h^2(4/3). Seeing this was a (inverted) parabola, I differentiated the RHS and said at a max/min 4 - (8/3)h = 0. So h = 1.5 and is halfway the length of the vertical. The horizontal is (4/3) h = 6/3 =2 and the area is 1.5 x 2 =3 So apart from labelling everything differently I got the same sides and the same area for the rectangle. I’ll have to look for your calculus solution!
I solved it as an optimisation problem. After finding the area as A(x)= -3/4 xˆ2 + 3x, I took the first derivative and set it to zero to find the optimal x which makes the largest area. A'(x)= -3/2x + 3 = 0 ==> x = 2
most of the optimization tasks in geometry end up in nice values like 1/2, 1/3 or 1/4 anyways i think there was some theorem that states if a+b=c, then a*b reaches its peak value when a=b, you could also see when a,b = c/2, apparently it's because of the AM-GM inequality
To me, the answer to the challenge question at the end is infinitesimally small, so I would say the limit of the area is zero. The points would approach 2 of the three verticies but never actually be at them.
Here’s an alternative way to express the height of the rectangle in terms of x: The small triangle in the bottom left is similar to the larger 3:4:5 triangle. Using the same labels as in the video, we have y/x=3/4, which gives y=3x/4. However, it might feel more intuitive (at least to me) to label the base of the small triangle as 4x instead of x. This way, it clearly suggests that the height of the small triangle (and therefore the rectangle) is 3x.
Solution: a = horizontal side of the rectangle, b = vertical side of the rectangle. 2nd radiation theorem: a/(3-b) = 4/3 ⟹ a = 4/3*(3-b) Area of the rectangle = A = a*b = 4/3*(3-b)*b = 4b-4/3*b² = -4/3*b²+4b = -4/3*(b²-3b) = -4/3*(b²-3b+1.5²-1.5²) = -4/3*[(b²-3b+1.5²)-1.5²] = -4/3*[(b-1.5)²-1.5²] = -4/3*[(b-1.5)²-2.25] = -4/3*(b-1.5)²+2.25*4/3 = -4/3*(b-1.5)²+9/4*4/3 = -4/3*(b-1.5)²+3 This is a quadratic parabola that opens downwards with the vertex at S = (1.5;3). This means that the maximum area of the rectangle is at b = 1.5 and is then 3 area units. Then: a = 4/3*(3-1.5) = 4/3*3/2 = 2
Question at end: given the constraints, I would put a vertex on the short leg just at the right angle, another on the hypotenuse just at the most acute angle, and the third vertex literally anywhere on the long leg. This forms a degenerate triangle with area 0. If we nudge the first two points ever so slightly up, the triangle is a very flat one with barely any area, and by moving these points closer to the vertices I mentioned, we can make this triangle;s area arbitrarily small, all the way to 0.
If we divide the segment 5 into two segments, we get a+b=5. If we now multiply a*b, we get the area of a rectangle. The largest area of a rectangle will be when the rectangle is a square, that is, when a=b. For example, a=b=2.5 because 2.5+2.5=5. We get 2.5*2.5=6.25. And now let's see what the area will be when ab. For example, a=3, b=2. We get that 3*2=6 < 6.25. And the next example a=1, b=4 then 1*4=4 < 6.25 This relationship proves to us that the largest area will be when side 5 is divided equally into two, which means that side 3 and side 4 must also be divided by two, so the largest area will be (3/2)*(4/2)=3
I tried to solve for the maximum area in my head as a way to get myself to fall asleep last night, and I think I actually got x = 2 by the end. This is my version of "counting sheep".
I wasn't sure (before I watched the video) that the rectangle was to share the right angle with the triangle, so after solving for that arrangement (area of the rectangle is maximally 1/2 the area of the triangle), I decided to make sure there wasn't a rectangle with one line on the hypotenuse which would be larger, but after some careful setup for an arbitrary non-obtuse triangle (vertices at (-a,0), (0,b), and (c,0)), it (of course) turns out that the maximal such rectangle is also exactly half the area of the triangle, with the upper vertices at the center points of edges ab and bc.
Alternative method (high school math competition 101): Let's say the rectangle has height a and width b, the triangle's hippo side has a slope of k, then we have 4 * k = 3 and a + b * k = 3 using arithmetic inequality: a * kb
Actually there's a bit simpler way to solve this problem. The question is equivalent to that the min of the sum of the area of the 2 triangles. Therefore let's say the ratio of one of the small triangle to the big one is x, and the other is 1-x. So the question is to calculte the min of x^2 + (1-x)^2 => 1/2 (x - 1/x)^2 + 1/2, so the answer is 1/2. And we know when the sum of the 2 small triangle equals to 1/2 of the big triangle, we got the max area of the square.
I did it by differentiating the A with respect to x, equating to zero to find the value of x giving maximum...then replacing that value in the equation for the area. By the way, the second derivative is negative, so we have a maximum area.
About the last question, can’t you just put two of your points on the 90 degree angle, and the 3rd point on the 5 long line? That would be an area of 0 because it’s a line, and if you want an actual triangle, move one of the points ever so little, and get an area infinitely close to 0. Right?
@@Crazy-mx9dc The problem is that you can get the area as close to zero as you like, by putting 2 of the points as close to a corner as you like. Any value you can name that's greater than zero, I can tweak it to be closer to zero. So, either zero *is* the minimum, or there is no minimum.
De modo geral suas questões não são fáceis, mas antes de ver a sua resolução eu tento resolver. Resolvi essa por derivada. Acertei. A = 3. (In general, your questions are not easy, but before seeing their resolution, I try to resolve them. I solved this by derivative. I got it right. A = 3.)
Watched the first minute, solved it by myself, skipped to the end, got a completely wrong answer of 3/8(9-sqrt(17)) or about 1.8288. I found that the final area would be (4-x)(3-y) and y=3x/4, then substituted 3x/4 for y in the first equation. I decided to graph y=(4-x)(3-3x/4) (for whatever reason) and y=3x/4 and see where they intersect to find the answer. I should've realized I was on the wrong path when they intersected at 2 points, one being completely outside of the triangle.
if you put 2 points on the right angle (equal to each other, but "touching" different sides) and the other point on the 5 side anywhere, wouldn't the area be 0? or would this question be better answered by figuring out the shortest distance from each corner to the opposite sides?
the largest inscribed rectangle is formed from a point half way up the diagonal. It’s area is half the area of the triangle and that is true for a right angled triangle of any shape or size
You are assuming the the desired rectangle is orientated as you drew it , but that is not necessary. In fact, if the "bottom" of our rectangle is the slanted line segment through (2,0) and (4, 3/2) and the "top" is along the hypotenuse, that rectangle has area 3 (the same as this video's area)..... To fully answer the original question, one must show that there is not another orientation for the desired rectangle....
For the second question at the end, the maximum and minimum areas of an inside triangle cannot be defined. If you try to create a maximum-sized inside triangle by placing the 3 points at the 3 vertices of the outside triangle, each of those 3 points will be on the 2 lines of the outside triangle that intersect at that vertex point and therefore there would be 2 points on each of the 3 lines of the outside triangle. The problem specifies to "pick a point" on each of the 3 lines of the outside triangle, but each point placed at a vertex of the outside triangle would result in an outside triangle line with 2 points on it. Therefore, an inside triangle point cannot be placed at an outside triangle vertex, an inside triangle equal in size to the outside triangle cannot be achieved, and a maximum inside triangle area cannot be defined. A minimum inside triangle area can not be defined because you cannot define the minimum distances between any 2 points of the inside triangle and a common vertex of the outside triangle.
Euclid 101: The maximum area of a rectangle that can be inscribed in a right triangle is half the area of the triangle. The rectangle's base is also half the base of the triangle. So the maximum area of the rectangle is simply h x b / 4.
no matter the side lengths of a triangle, the rectangle with the largest area that could fit in the triangle has the half of the area of the triangle. The height and base length of the rectangle is the half of the triangle's. that could be proven by using some simple geometry
Imagine folding left triangle to the right and top one down. These triangles cover entire rectangle, and one of them sticks out - unless the triangles are exactly halves of the rectangle, thus minimizing their total area (and maximizing area of the rectangle therefore).
let the base of the triangle be "b" and the height of the triangle be "h" if we put the x and y axes like bprp did, and use the slope formula, we would get the line equation as y=hx/b now lets take a look at the rectangle's area, the area of the rectangle (A) = (b-x)y = (b-x)(hx/b) distributing hx/b to the two terms would give us A = -hx^2/b + hx if we use the same formula of bprp then we would get the x value at the maximum area i.e., x=-b/2a = -h/2(-h/b) = (-h/2)(b/-h) cancelling the "-h"s we would get x=b/2 now, to find y we must go back to the line equation y=hx/b substituting x=b/2, the equation becomes y=(h/b)(b/2) cancelling the "b"s we would get y=h/2 now, we should go back to the area of the rectangle and substitute x and y in there i.e., A = (b-x)y = (b-b/2)(h/2) simplifying, we get A=(b/2)(h/2) A=bh/4 we can write that as A=(1/2)(bh/2) A=(1/2)(Area of triangle) Therefore, Area of rectangle = half of Area of triangle So, answering your question, yes, the maximum of the rectangles area is always half of the original triangle :D
@@IMayBeAnAlien We can get to "yes" much easier than that. If you scale the triangle along the X axis, the area of every enclosed rectangle is scaled by the same factor, so the largest rectangle remains the one that's half the width and height of the triangle. Then you can scale along the Y axis, it's it's still true.
lets see length times height is 4 times 3 = 12 , the area of the triangle is half that or 6 units if you go along half the length of 4 (or the adjacent) or 2 units to the hypotenuse and back to the opposite side you will get half way along opposite of 3 units. multiple 2 (or half the length) times 1.5 (half the height) equals 2*1.5 or equals 3 square units. 3 square units is the answer to this youtube . My answer came to me a lot faster. than watching the bprp math basics youtube.
But I knew from early maths classes that the max rectangle (in a right-angle triangle) was always (x/2).(y/2) which here gives 3. Would I have lost points by declaring that?
How do you know the biggest rectangle has the edges along x- and y-axes? We might find a bigger rectangle running along the hypotenuse, extending towards the right angle of the triangle.
@@yawninglion Analytic geometry told me that. 😉 Another way to approach the problem was shown in my other comment: three triangles beside that rectangle could be moved in such way so they cover it entirely and (if rectangle vertices aren't midpoints) there will be some extra triangular area outside the rectangle or overlap inside it. Consequently, these triangles have minimum total area when there's no overlap or extra; thus rectangle's maximum area is half of the big triangle's.
@@yawninglion Any point on the leg uniquely defines two rectangles: one with vertex at the right angle, another with side on hypotenuse. Those rectangles obviously have equal area (because they can be transformed into each other by shearing, which preserves area).
Solution: When we put the triangle into a coordinate system, with the right angle and the zero point, the hypotenuse is either y = -3/4 * x + 3 or y = -4/3 * x + 4. This way, the target triangle is x * y or x * f(x), which is x * (-3/4 * x + 3) or x * (-4/3 * x + 4). Let's continue with the first: g(x) = x * (-3/4 * x + 3) g(x) = -3/4 * x² + 3x We are looking for the maximum value, so we need g'(x) = 0 and g''(x) < 0 g'(x) = -3/2 * x + 3 g''(x) = -3/2 → constant, therefore always a maximum g'(x) = 0: 0 = -3/2 * x + 3 |+3/2 * x 3/2 * x = 3 |*2/3 x = 2 So the maximum of g(x) = x * (-3/4 * x + 3) is at x = 2 The corresponding maximum area value is: g(2) = 2 * (-3/4 * 2 + 3) g(2) = -3/4 * 4 + 3 * 2 g(2) = -3 + 6 g(2) = 3 Therefore the maximum rectangle is 3 areal units
Very tedious method ....compkete the square of 3×4 of triangle arms. 1/4 of square area is max in the triangle hence mwx area is 3..draw the picture to clear it... Now minimum triangle area in 1/2 of this quardent area that is 3/2
Coud you explain the unit of área you have used ?? " unit squared" as you said ?? What unit system is that ?? I was interested to apply this, in a chart "Power-rpm" (axis 'y' and 'x', KW vertical axis, rpm horizontal axis), but the unit of area non-standardized you used is disconcerning !!! It's imposible to square a KW unit.
Well, it's the same unit the original triangle had, but squared. if the triangle was 3cm - 4cm - 5cm, then, the area of the rectangle is 3cm². In this case, there wasn't any unit specified, 3 units - 4 units - 5 units right triangle. So unit squared for the answer.
@msolec2000 And why don't he put cm or cm² ?? from the begining. "Cm" is and standard unit in the international metric system. You can't go to the market to buy 3 units squared of something. And in my case, in my chart x-y Kw-rpm, I can't square any unit, are differents applications !!
@@marioalb9726 Yeah, but this is geometry. When you want to adapt that to your situation, you use the units you need. In other words, the important thing here is the calculation, not the unit. :)
You would probably call the "units" used here "unitless." There is no measurement system attached. If you take a course in college level linear algebra you'll be introduced to what are called basis vectors. These are the vectors you use to define your coordinate axes, and you divide them by their own magnitudes so that they are one unit long. That's the unit being referenced here. If we have the expression: 1 + 1 = 2, in math, that's one unit plus one unit equals two units. Marbles, cows, space launches, stuck zippers, kilometers traveled, we don't really care for the purposes of doing the math what the external labels are, it's still one unit plus one unit, and mathematically it will always behave the same way. There's not a problem with squaring kW. You just do the math as normal. It's the same principle as squaring velocity when calculating kinetic energy. Just keep track of your fundamental S.I. units and cancel where you can to make keeping track easier. Depending on what you choose to multiply in S.I. the end result may not have a defined physical meaning, but that's an error in choosing what equation to set up, the actual math and S.I. units will handle it just fine.
@@ericschori5519 For me, it is a serious mistake to talk about surface area and not include a surface unit in the final result. When these same teachers talk about angles, they always include the unit "sexagesimal degrees", when they talk about temperature, they always include the unit (centigrade or fahrenheit degrees), they never say "angular unit" Which represents a contradiction, Sometimes they say the unit, sometimes they don't, it's incoherent. You can't make an engineering specification, and put "square units, cubic units" Just as you can't go to the hardware store and buy 4 square units of sheet metal And it is totally impractical to square a KW, as you wrote. This is just a reflection of how divorced the university is (mathematicians), from practical applications and the world of work. And also from the international standardization rules
Sorry... you answered the wrong question. You made the extra assumption that one of the corners of the rectangle is in the right angle of the triangle.... and that is just an assumption. If you lay one of the sides of the rectangle on the hypotenuse of the triangle... you get the same size. I have not worked out yet what the maximum size is if the rectangle is ANYWHERE inside the triangle.... WIP.
@@torlumnitor8230 The area of the triangle is 1/2 * base * height. Substituting the values for this triangle results in: 1/2 * 3 * 4, which is also 3 * 2, or 6. The area of the square is 3, so the area of the triangle is twice that area.
Tried solving it before the video and got 3 as the biggest possible area. Let’s see if that’s right. Edit: I was right 😁 I used a different method, but I got the same solution.
Was gonna make exactly this post. My method: any rectangle is going to leave two congruent 3-4-5 triangles. Therefore we can restate the problem as minimizing the total area of these two triangles. There is one situation where the two triangles are identical (1.5-2-2.5 triangles) which each have an area of 1.5 units. We can see that if we change the rectangle we make one triangle larger and the other one smaller. Since the area scales quadratically with the side-lengths, the one that grows gains more area than the one that shrinks loses. So, it is clear that if the two triangles are equal, their summed area is minimal and it is equal to 3. The inscribed rectangle is 2x1.5= 3 units.
My intuition says that the biggest possible rectangle for most right triangles will have one side aligned along the hypothenuse. So, yes you spotted the question for rectangles that snuggle into the right angle, so you did what you planned to do However I am not convinced that this is actually the very biggest rectangle that could fit inside that triangle.
Real life application: under the eaves of a roofline, I want to install the largest possible air duct: what’s the largest duct I could possibly use? I’ve done this exact exercise numerous times graphically (because math hurts my head) using drafting software. But real world: the answer is your outer diameter and be sure to reduce your results by .25-.5” or you’ll never get it installed! (Ask me how I know!) 🫣
The last question is not well-defined. The answer to that is "you can make the area arbitrarily small". Just put two points arbitrary close to one vertex and the third point to the opposite side. Resulting triangle will be arbitrarily close to a line, so its area will be as small as you want. It's all "legit" too - because you avoid the "cheating" by turning the triangle into a line exactly, but it goes to show that there is no answer if we constrain ourselves only to a "proper" triangle.
Each vertex must be on a different side
@@loadstone5149So you chose your points on lines 3 and 4 to be arbitrarily close to the 3,4 vertex of the original triangle. Then you chose your points on the 5 line to be arbitrarily close to the 3,5 vertex. Now you have a triangle that is arbitrarily close to matching the 3 line.
This is the answer I came up with.
3/2×9/8=27/16 sq
for maxima and minima Dy/Dx can b made 0.
The minimum triangle (question at the end) is the one, where two of the points are almost at the tip of the smaller angle and the the third point almost at the right angle. The area is miniscule, basically zero, as the triangle is almost a line by that point.
Similar solution: the point on the hypotenuse is close to 1 corner, the point on the vertical is close to the bottom, and the point on the bottom is anywhere. Since you can get as close as you want to the corners, the area approaches 0.
Another way of doing it is by finding the roots of the parabola. Doing so, you get that the area is smallest at x=0 or x=4.
Assuming you want a non-zero answer, then there isn't an exact answer, because for any non-zero value you plug in to the area equation (x-4)*y, there will always be a smaller number that you could plug in.
One could say infinitesimally small.
@@qqqalo this has nothing to do with the parabola, we're talking about the puzzle he gave at the end, which is completely different to whatever problem you're trying to solve
Double the triangle, make a rectangle. Construct a perpendicular intersection within the interior of the rectangle perpendicular to the sides of the rectangle. There will be exactly one complete rectangle where the hypotenuse does not intersect the rectangle, until the interior corner of the rectangle hits the midpoint of the hypotenuse, at which point the single rectangle duplicates. This maximizes size of the rectangle, which is 1/4 the area of the larger rectangle defined by the shorter sides of the right triangle.
Why does it maximise the size of the rectangle
Minimal triangle would be A-->0.
Pick a vertex. Point at the hypotenuse and on the adjacent side approach that vertex, point at the opposite side approaches the adjacent side (or hypotenuse, doesn't matter). The triangle collapses to a line, and with infinitessimal height right before it does, ie, right before all 3 points become collinear, the area of that triangle approaches 0.
I made x the ratio of the left piece (where you put x) and the full length (4); therefore the size of the rectangle is (3x) * (4 - 4x), which simplifies to 12 * (x - x^2).
Setting the derivative to 0 gets:
0 = 12 * (1 - 2x)
0 = 1 - 2x
x = 0.5
And the area is:
12 * (x - x^2) =
12 * (0.5 - 0.25) =
12 * 0.25 =
3
I solved it for a rectangular base being shared with the '5' side. The answer is still 3 !
Do to so, I minimised the 3 small triangles formed by inserting a diagonal rectangle. If you shuffle those 3 triangles around, you can recover the bprp initial assumption .
For any height h>0 and any base b>0 (in this example 3 and 4 respectively) the equation for the point on the diagonal is y = (h/b) * x and the area as a function of x is (b-x) *y = h*x - (h*x*x) / b. the derivative for the area as a function of x is h - (2*h*x) / b = h(1- (2*x)/b). setting this equal to zero, either h = 0, which is impossible, or 1 = (2*x) / b b/2 =x. replacing x in the earlier equation :y = (h/b) * (b/2) y = h/2. In other words, the max area for a rectangle inscribed in a right triangle is always when the base and height of the rectangle are half of the base and height of the triangle( if the base and height are parallel to its sides, idk if putting one of the sides on the hypotenuse might give a bigger area).
Enjoyable problem. Not too hard but still requires some thought. I solved it exactly as our host, except I had forgotten the equation for the max/min of a parabola; probably because I learned it 40 years ago and hadn't used it since. However I had not forgotten how to take a derivative of a simple quadratic.
For the original question I used the fact that the smaller triangles left after taking away the rectangle are similar to the original. If you define a scaling factor s ( 0 < s < 1) where is the proportion of the base of the triangle that the rectangle occupies then the area of the rectangle is 4s * (3-3s) - 12s - 12s². To get the maximum, take the root of derivative i.e. 12 - 24 s = 0. Thus, s = ½. and A = 3.
For largest area of triangle, dA/dx = 0, or 6x/4=3 or x=2.
This way I don't have to remember equation of parabola or its peak point.
(To make sure we get the highest area, not the lowest, one can take second derivative of A, if needed).
Assume the vertex on the right vertical side is sliding up, the vertex on the horizontal slide is moving right. So, as the hight of the rectangle increases, the width decreases. With this assumption, and mapping the both extremes to 0 and 1, we can get the area with the formula 3x*4(1-x). simplifying it we get 12x-12x^2. Take the derivative (12-24x), compute the local maximum(x=0,5), plug it into the original formula 12*0,5-12*0,5^2 and you get 3.
Man this was amazing, I solved this using differential calculus but this approach was superlative, and outstandingly genius, gave me a dopamine rush fr.
All you need is effing geometry to solve it in less than half the time this guy did!
given the formula for area, A
and that the maxinun is where dA/dX = 0. So differentiate the area formula and set to 0.
A = 3x -(3/4)x^2
d/dx = 3 -(6/4)x
set = 0
0 = 3-(6/4)x
or
-3(-4/6)=x
4/2 = x
so
x = 2
Exactly how I did it.
3 sq units
equation for hypotenuse is: y = (3/4)xo and xo is: xo = 4.- x and 0
A=xy then sub for y, A=3x-6/4x^2 find the first derivative and then the second derivative to prove its max
For the question at the end of the video, I selected 3 points
(a,0) where a is between 0 and 4
(4,b) where b is between 0 and 3
and (c, ¾c) where c is between 0 and 4
The area they make is (12c-3ac+4bc+4ab) all over 8
I need to learn how to get the minimum, though.😅
Same answer, different setup.
I imagined the vertex on the hypotenuse as a bead on a string that I can slide left or right. That reduces the problem to one variable, since the bead's position determines all other vertices and lengths. I set the length of string left of the bead to "x", which makes the length of string right of the bead "5-x".
Since the inner triangles are also 3-4-5 ratio, the rectangle is (3/5)(x) tall, and (4/5)(5-x) wide. Multiplied to get area, and found maximum from there.
An easy reflexion can help us fond the solution without any calculus
If we take a isocele right triangle, by symmetry, the largest rectangle is the one with half the size of the side of the triangle
Then if you stretch the triangle horizontaly from a certain factor, the aera of any rectangle will increase in a linear manner from the same factor, this operation doesn’t change the local maximum
I had a smaller 3:4:5 triangle above the square with vertical h. This gave me the area if the square A = 4h - h^2(4/3). Seeing this was a (inverted) parabola, I differentiated the RHS and said at a max/min 4 - (8/3)h = 0. So h = 1.5 and is halfway the length of the vertical. The horizontal is (4/3) h = 6/3 =2 and the area is 1.5 x 2 =3 So apart from labelling everything differently I got the same sides and the same area for the rectangle. I’ll have to look for your calculus solution!
I solved it as an optimisation problem. After finding the area as A(x)= -3/4 xˆ2 + 3x, I took the first derivative and set it to zero to find the optimal x which makes the largest area.
A'(x)= -3/2x + 3 = 0 ==> x = 2
that's exactly equivalent to working out the discriminant. A = ax^2 + bx +c => A' = 2ax + b => 0 => 2ax + b => x = -b / 2a
Same thing here. Straight forward.
That's how I would have done it.
Same here. Just did it and checked the video to verify that the result is correct. Took me 90 seconds.
@@valuemastery Yes, It was pretty easy. I did in my head ... no pen and paper.
most of the optimization tasks in geometry end up in nice values like 1/2, 1/3 or 1/4 anyways
i think there was some theorem that states if a+b=c, then a*b reaches its peak value when a=b, you could also see when a,b = c/2, apparently it's because of the AM-GM inequality
Or you could differentiate the equation for A with respect to x, set it to zero to find the x that maximizes A, and then calculate A.
This is exactly what he does when he writes Xmax= -b/2a 😉
That is exactly how I did it.
The solution is simple
Area rec = y(4 - x)
2(¾x) = 3 ⇒ x = 2
⇒ y = ¾(2) = 1.5
∴ Area rec = 1.5(4 - 2) = 3 u²
You could use max min. Or, you could just take a 45, 45, 90 triangle, with a square inscribed in it, and stretch it by 1/3.
To me, the answer to the challenge question at the end is infinitesimally small, so I would say the limit of the area is zero. The points would approach 2 of the three verticies but never actually be at them.
Here’s an alternative way to express the height of the rectangle in terms of x:
The small triangle in the bottom left is similar to the larger 3:4:5 triangle. Using the same labels as in the video, we have y/x=3/4, which gives y=3x/4. However, it might feel more intuitive (at least to me) to label the base of the small triangle as 4x instead of x. This way, it clearly suggests that the height of the small triangle (and therefore the rectangle) is 3x.
Solution:
a = horizontal side of the rectangle,
b = vertical side of the rectangle.
2nd radiation theorem:
a/(3-b) = 4/3 ⟹ a = 4/3*(3-b)
Area of the rectangle = A = a*b = 4/3*(3-b)*b = 4b-4/3*b² = -4/3*b²+4b
= -4/3*(b²-3b) = -4/3*(b²-3b+1.5²-1.5²) = -4/3*[(b²-3b+1.5²)-1.5²]
= -4/3*[(b-1.5)²-1.5²] = -4/3*[(b-1.5)²-2.25] = -4/3*(b-1.5)²+2.25*4/3
= -4/3*(b-1.5)²+9/4*4/3 = -4/3*(b-1.5)²+3
This is a quadratic parabola that opens downwards with the vertex at S = (1.5;3). This means that the maximum area of the rectangle is at b = 1.5 and is then 3 area units. Then: a = 4/3*(3-1.5) = 4/3*3/2 = 2
Question at end: given the constraints, I would put a vertex on the short leg just at the right angle, another on the hypotenuse just at the most acute angle, and the third vertex literally anywhere on the long leg. This forms a degenerate triangle with area 0. If we nudge the first two points ever so slightly up, the triangle is a very flat one with barely any area, and by moving these points closer to the vertices I mentioned, we can make this triangle;s area arbitrarily small, all the way to 0.
If we divide the segment 5 into two segments, we get a+b=5. If we now multiply a*b, we get the area of a rectangle. The largest area of a rectangle will be when the rectangle is a square, that is, when a=b. For example, a=b=2.5 because 2.5+2.5=5. We get 2.5*2.5=6.25. And now let's see what the area will be when ab. For example, a=3, b=2. We get that 3*2=6 < 6.25. And the next example a=1, b=4 then 1*4=4 < 6.25 This relationship proves to us that the largest area will be when side 5 is divided equally into two, which means that side 3 and side 4 must also be divided by two, so the largest area will be (3/2)*(4/2)=3
I tried to solve for the maximum area in my head as a way to get myself to fall asleep last night, and I think I actually got x = 2 by the end.
This is my version of "counting sheep".
I wasn't sure (before I watched the video) that the rectangle was to share the right angle with the triangle, so after solving for that arrangement (area of the rectangle is maximally 1/2 the area of the triangle), I decided to make sure there wasn't a rectangle with one line on the hypotenuse which would be larger, but after some careful setup for an arbitrary non-obtuse triangle (vertices at (-a,0), (0,b), and (c,0)), it (of course) turns out that the maximal such rectangle is also exactly half the area of the triangle, with the upper vertices at the center points of edges ab and bc.
Alternative method (high school math competition 101):
Let's say the rectangle has height a and width b, the triangle's hippo side has a slope of k, then we have 4 * k = 3 and a + b * k = 3
using arithmetic inequality: a * kb
It’s obvious that x = 1/2l is the solution (otherwise there would be two solutions to the problem found by exchanging xl-x)
By generalizing any right triangle, the area of the rectangle will always equal ab/4.
Actually there's a bit simpler way to solve this problem.
The question is equivalent to that the min of the sum of the area of the 2 triangles. Therefore let's say the ratio of one of the small triangle to the big one is x, and the other is 1-x. So the question is to calculte the min of x^2 + (1-x)^2 => 1/2 (x - 1/x)^2 + 1/2, so the answer is 1/2. And we know when the sum of the 2 small triangle equals to 1/2 of the big triangle, we got the max area of the square.
I did it by differentiating the A with respect to x, equating to zero to find the value of x giving maximum...then replacing that value in the equation for the area. By the way, the second derivative is negative, so we have a maximum area.
About the last question, can’t you just put two of your points on the 90 degree angle, and the 3rd point on the 5 long line?
That would be an area of 0 because it’s a line, and if you want an actual triangle, move one of the points ever so little, and get an area infinitely close to 0. Right?
that might be considered cheating because the limit of the area is zero, and not the actual area? please correct me if i'm wrong 😅
@@zoinkerspoinker4326maybe if we get this to be top comment, we’ll get an answer
@@zoinkerspoinker4326 I’m just a 15 year old student, I’ve got no idea if that’s right or wrong 😅
@@Crazy-mx9dc The problem is that you can get the area as close to zero as you like, by putting 2 of the points as close to a corner as you like. Any value you can name that's greater than zero, I can tweak it to be closer to zero. So, either zero *is* the minimum, or there is no minimum.
@@jursamaj that’s how far I understood it, but still thanks
I like what you do.
De modo geral suas questões não são fáceis, mas antes de ver a sua resolução eu tento resolver. Resolvi essa por derivada. Acertei. A = 3. (In general, your questions are not easy, but before seeing their resolution, I try to resolve them. I solved this by derivative. I got it right. A = 3.)
Watched the first minute, solved it by myself, skipped to the end, got a completely wrong answer of 3/8(9-sqrt(17)) or about 1.8288. I found that the final area would be (4-x)(3-y) and y=3x/4, then substituted 3x/4 for y in the first equation. I decided to graph y=(4-x)(3-3x/4) (for whatever reason) and y=3x/4 and see where they intersect to find the answer. I should've realized I was on the wrong path when they intersected at 2 points, one being completely outside of the triangle.
You have to define "large" first. Area? Perimiter? Height?
Based on what I got for my final answer of the question in the end. I don’t feel like typing it in, it’s way too long.
My answer is
1536/625 or 2.4576
❤ other method
taking vertices as (0,0) , (4,0) and (0,3)
(x/4)+(y/3) = 1
by AM -GM inequality
x/4+y/3 > = 2 sqrt(xy/12)
1 > = xy/3
Max area is 3
Can you explain to me where (x/4)+(y/3)=1 came from?
@assassin01620 intercept form of straight line in co - ordinate geometry
@raghvendrasingh1289 wow. How have I never heard of intercept form before? Thanks
@@assassin01620 👍
1 1 √2 triangle is 1/4, scale it by 12, you get 3
if you put 2 points on the right angle (equal to each other, but "touching" different sides) and the other point on the 5 side anywhere, wouldn't the area be 0? or would this question be better answered by figuring out the shortest distance from each corner to the opposite sides?
There is no minimum area of triangle. Because it can't be 0 but it can be 0,00......001.
The area is equal to 3*4*r(1-r), where r is in [0,1]. It is maximized at r=.5
the largest inscribed rectangle is formed from a point half way up the diagonal.
It’s area is half the area of the triangle and that is true for a right angled triangle of any shape or size
□1.5X2.
Light work
Max area is 3 and occurs when b = 2 and h = 3/2
Simplify sqrt 464
@ 2sqrt(116)
4sqrt(29)
I love the Christmas tree on the right of the whiteboard. Can you post that somewhere, please, or link to the source of it belongs to someone else
You are assuming the the desired rectangle is orientated as you drew it , but that is not necessary. In fact, if the "bottom" of our rectangle is the slanted line segment through (2,0) and (4, 3/2) and the "top" is along the hypotenuse, that rectangle has area 3 (the same as this video's area)..... To fully answer the original question, one must show that there is not another orientation for the desired rectangle....
For the second question at the end, the maximum and minimum areas of an inside triangle cannot be defined.
If you try to create a maximum-sized inside triangle by placing the 3 points at the 3 vertices of the outside triangle, each of those 3 points will be on the 2 lines of the outside triangle that intersect at that vertex point and therefore there would be 2 points on each of the 3 lines of the outside triangle.
The problem specifies to "pick a point" on each of the 3 lines of the outside triangle, but each point placed at a vertex of the outside triangle would result in an outside triangle line with 2 points on it.
Therefore, an inside triangle point cannot be placed at an outside triangle vertex, an inside triangle equal in size to the outside triangle cannot be achieved, and a maximum inside triangle area cannot be defined.
A minimum inside triangle area can not be defined because you cannot define the minimum distances between any 2 points of the inside triangle and a common vertex of the outside triangle.
What happens if "WE" decide to place the rectangle adjacent to the HYPOTENUSE? Is the answer still 3?
And the area of the inscribed circle is only slightly larger at pi...
Euclid 101: The maximum area of a rectangle that can be inscribed in a right triangle is half the area of the triangle. The rectangle's base is also half the base of the triangle. So the maximum area of the rectangle is simply h x b / 4.
Why is Y 3/4x?
no matter the side lengths of a triangle, the rectangle with the largest area that could fit in the triangle has the half of the area of the triangle. The height and base length of the rectangle is the half of the triangle's. that could be proven by using some simple geometry
Imagine folding left triangle to the right and top one down. These triangles cover entire rectangle, and one of them sticks out - unless the triangles are exactly halves of the rectangle, thus minimizing their total area (and maximizing area of the rectangle therefore).
holy shit its wx 78
Could this be generalized, and will the general maximum area of an inscribed square always be half the area of the original triangle?
let the base of the triangle be "b" and the height of the triangle be "h"
if we put the x and y axes like bprp did, and use the slope formula, we would get the line equation as y=hx/b
now lets take a look at the rectangle's area,
the area of the rectangle (A) = (b-x)y = (b-x)(hx/b)
distributing hx/b to the two terms would give us
A = -hx^2/b + hx
if we use the same formula of bprp then we would get the x value at the maximum area
i.e., x=-b/2a = -h/2(-h/b) = (-h/2)(b/-h)
cancelling the "-h"s we would get
x=b/2
now, to find y we must go back to the line equation y=hx/b
substituting x=b/2, the equation becomes y=(h/b)(b/2)
cancelling the "b"s we would get
y=h/2
now, we should go back to the area of the rectangle and substitute x and y in there
i.e., A = (b-x)y = (b-b/2)(h/2)
simplifying, we get
A=(b/2)(h/2)
A=bh/4
we can write that as
A=(1/2)(bh/2)
A=(1/2)(Area of triangle)
Therefore, Area of rectangle = half of Area of triangle
So, answering your question, yes, the maximum of the rectangles area is always half of the original triangle :D
@@IMayBeAnAlien We can get to "yes" much easier than that. If you scale the triangle along the X axis, the area of every enclosed rectangle is scaled by the same factor, so the largest rectangle remains the one that's half the width and height of the triangle. Then you can scale along the Y axis, it's it's still true.
Yes the general case is true (the triangles outside the rectangle are similar, so the maximum must be when dimensions are equal)
but what if you lay the rectange on the hypotenuse?
The "so-called" derivative!
And that maximum area is half the area of the 3-4-5 triangle.
Did you mean the minimum size of a similar triangle, or a triangle of any arbitrary proportions?
The answer of the 2nd question is 0. Lets' align the 3 points of the triangle on the same line, then the area of this triangle is 0.
Does the rectangle have to match the sides of the triangle? Is that provable?
lets see length times height is 4 times 3 = 12 , the area of the triangle is half that or 6 units
if you go along half the length of 4 (or the adjacent) or 2 units to the hypotenuse and back to the opposite side you will get half way along opposite of 3 units.
multiple 2 (or half the length) times 1.5 (half the height) equals 2*1.5 or equals 3 square units.
3 square units is the answer to this youtube .
My answer came to me a lot faster. than watching the bprp math basics youtube.
smallest triangle is infinitely close to zero.
Now do it for the largest possible circle.
But I knew from early maths classes that the max rectangle (in a right-angle triangle) was always (x/2).(y/2) which here gives 3. Would I have lost points by declaring that?
i think you would have had to show your working.
How do you know the biggest rectangle has the edges along x- and y-axes? We might find a bigger rectangle running along the hypotenuse, extending towards the right angle of the triangle.
Actually no, rectangle along hypotenuse have exactly the same maximum area (two vertices are midpoints of the triangle's legs again). 😉
@-wx-78- You'd still need to argue why that's not bigger than the one presented in the video.
@@yawninglion Analytic geometry told me that. 😉
Another way to approach the problem was shown in my other comment: three triangles beside that rectangle could be moved in such way so they cover it entirely and (if rectangle vertices aren't midpoints) there will be some extra triangular area outside the rectangle or overlap inside it. Consequently, these triangles have minimum total area when there's no overlap or extra; thus rectangle's maximum area is half of the big triangle's.
Well, I know the largest rectangle is indeed half of the triangle - but the argument in the video is incomplete
@@yawninglion Any point on the leg uniquely defines two rectangles: one with vertex at the right angle, another with side on hypotenuse. Those rectangles obviously have equal area (because they can be transformed into each other by shearing, which preserves area).
WILL THE AREA ALWAYS BE THE ORIGINAL Y VALUE SQUARED
OR WAS THAT PURELY COINCIDENCE
Solution:
When we put the triangle into a coordinate system, with the right angle and the zero point, the hypotenuse is either y = -3/4 * x + 3 or y = -4/3 * x + 4.
This way, the target triangle is x * y or x * f(x), which is x * (-3/4 * x + 3) or x * (-4/3 * x + 4).
Let's continue with the first:
g(x) = x * (-3/4 * x + 3)
g(x) = -3/4 * x² + 3x
We are looking for the maximum value, so we need g'(x) = 0 and g''(x) < 0
g'(x) = -3/2 * x + 3
g''(x) = -3/2 → constant, therefore always a maximum
g'(x) = 0:
0 = -3/2 * x + 3 |+3/2 * x
3/2 * x = 3 |*2/3
x = 2
So the maximum of g(x) = x * (-3/4 * x + 3) is at x = 2
The corresponding maximum area value is:
g(2) = 2 * (-3/4 * 2 + 3)
g(2) = -3/4 * 4 + 3 * 2
g(2) = -3 + 6
g(2) = 3
Therefore the maximum rectangle is 3 areal units
i had an intuition that it would be nine based on the geometry, is it correct to say that all of those triangles are 3,4,5 ?
I suspect you are right. I was guessing all the triangles were scaled versions of each other
Yes, all the triangles are similar. Their 3 sides are all parallel between triangles.
Is the biggest rectangle inscribed in a triangle always going to have half the area of the triangle?
Very tedious method ....compkete the square of 3×4 of triangle arms. 1/4 of square area is max in the triangle hence mwx area is 3..draw the picture to clear it...
Now minimum triangle area in 1/2 of this quardent area that is 3/2
Three units square is 9 square units
Last one: 0.
👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏
I'm just gonna guess at 144/49....
Why is the other one blue 😢😢😢😢
Coud you explain the unit of área you have used ??
" unit squared" as you said ??
What unit system is that ??
I was interested to apply this, in a chart "Power-rpm" (axis 'y' and 'x', KW vertical axis, rpm horizontal axis), but the unit of area non-standardized you used is disconcerning !!! It's imposible to square a KW unit.
Well, it's the same unit the original triangle had, but squared. if the triangle was 3cm - 4cm - 5cm, then, the area of the rectangle is 3cm². In this case, there wasn't any unit specified, 3 units - 4 units - 5 units right triangle. So unit squared for the answer.
@msolec2000
And why don't he put cm or cm² ?? from the begining.
"Cm" is and standard unit in the international metric system.
You can't go to the market to buy 3 units squared of something.
And in my case, in my chart x-y Kw-rpm, I can't square any unit, are differents applications !!
@@marioalb9726 Yeah, but this is geometry. When you want to adapt that to your situation, you use the units you need. In other words, the important thing here is the calculation, not the unit. :)
You would probably call the "units" used here "unitless." There is no measurement system attached. If you take a course in college level linear algebra you'll be introduced to what are called basis vectors. These are the vectors you use to define your coordinate axes, and you divide them by their own magnitudes so that they are one unit long. That's the unit being referenced here. If we have the expression: 1 + 1 = 2, in math, that's one unit plus one unit equals two units. Marbles, cows, space launches, stuck zippers, kilometers traveled, we don't really care for the purposes of doing the math what the external labels are, it's still one unit plus one unit, and mathematically it will always behave the same way.
There's not a problem with squaring kW. You just do the math as normal. It's the same principle as squaring velocity when calculating kinetic energy. Just keep track of your fundamental S.I. units and cancel where you can to make keeping track easier. Depending on what you choose to multiply in S.I. the end result may not have a defined physical meaning, but that's an error in choosing what equation to set up, the actual math and S.I. units will handle it just fine.
@@ericschori5519
For me, it is a serious mistake to talk about surface area and not include a surface unit in the final result.
When these same teachers talk about angles, they always include the unit "sexagesimal degrees", when they talk about temperature, they always include the unit (centigrade or fahrenheit degrees), they never say "angular unit"
Which represents a contradiction, Sometimes they say the unit, sometimes they don't, it's incoherent.
You can't make an engineering specification, and put "square units, cubic units"
Just as you can't go to the hardware store and buy 4 square units of sheet metal
And it is totally impractical to square a KW, as you wrote.
This is just a reflection of how divorced the university is (mathematicians), from practical applications and the world of work. And also from the international standardization rules
gdzie mas gdzie mas tymi rekami cie zab nie chcem ciem juz znac
Something tells me you're using analytic geometry.
Sorry... you answered the wrong question.
You made the extra assumption that one of the corners of the rectangle is in the right angle of the triangle.... and that is just an assumption.
If you lay one of the sides of the rectangle on the hypotenuse of the triangle... you get the same size.
I have not worked out yet what the maximum size is if the rectangle is ANYWHERE inside the triangle.... WIP.
Which is surprisingly being half of the big triangle area
How did you come to that conclusion?
@@torlumnitor8230I have contacts in high places
@@torlumnitor8230 The area of the triangle is 1/2 * base * height. Substituting the values for this triangle results in:
1/2 * 3 * 4, which is also 3 * 2, or 6.
The area of the square is 3, so the area of the triangle is twice that area.
Tried solving it before the video and got 3 as the biggest possible area. Let’s see if that’s right.
Edit: I was right 😁 I used a different method, but I got the same solution.
Was gonna make exactly this post. My method: any rectangle is going to leave two congruent 3-4-5 triangles. Therefore we can restate the problem as minimizing the total area of these two triangles. There is one situation where the two triangles are identical (1.5-2-2.5 triangles) which each have an area of 1.5 units. We can see that if we change the rectangle we make one triangle larger and the other one smaller. Since the area scales quadratically with the side-lengths, the one that grows gains more area than the one that shrinks loses. So, it is clear that if the two triangles are equal, their summed area is minimal and it is equal to 3. The inscribed rectangle is 2x1.5= 3 units.
He teaches us the max, in the shortest time😂!
Waaaaao math hahahaha do you naw abcd thish math
My intuition says that the biggest possible rectangle for most right triangles will have one side aligned along the hypothenuse.
So, yes you spotted the question for rectangles that snuggle into the right angle, so you did what you planned to do
However I am not convinced that this is actually the very biggest rectangle that could fit inside that triangle.
👍👍👍
Real life application: under the eaves of a roofline, I want to install the largest possible air duct: what’s the largest duct I could possibly use? I’ve done this exact exercise numerous times graphically (because math hurts my head) using drafting software. But real world: the answer is your outer diameter and be sure to reduce your results by .25-.5” or you’ll never get it installed! (Ask me how I know!) 🫣