Move the square down to touch the x axis. The left triangle has hypotenuse 2. The right triangle has hypotenuse 6. Then 2*cos(t)=a and 6*sin(t)=a where t is the smaller angle between the bottom vertex of the square and the x axis and a is the square side length. Therefore (a/6)^2 + (a/2)^2 = 1. A = a^2 = 18/5
8:00 a more quicker way! youre trying to find out the length of other sides but it's not needed.if you look carefully the yellow traingle(except the smaller one) (at the timestamp i mentioned) is a right angle triangle so (3x)² + x² = (2+2+2)² 10x²=36 and x²=3.6.
I solved it differently. My square has an area of (3x)². The small triangle between 5 and 7 has the sides x and 3x, thus the hypotenuse is 2 = sqrt(10)x => x = sqrt(2/5). The total area of the square is then 9x² = 18/5.
Here’s a vector solution for anyone interested A is the position vector for the top of the square. b is one side of the square as a vector, c is the other. A + 5b = (13 0) A + 2b + c = (7 0) 3b - c = (6 0) and b dot c = 0 is then solvable (let b = (x y)) 3x - y = 6 3y + x = 0 b = (9/5 -3/5) c= (-3/5 -9/5) |b x c| = 3.6
I approached it along the lines of the last of the methods shown, but reasoned a little differently. Consider the entire right triangle (i.e., with hypothenuse 10). Now the extension of the tilted bottom of the square cuts the hypothenuse into two segments, of length 6 (i.e., 10-4) and 4 (i.e., 7-3). This means (due to parallel line segments) that the red segment is 2/3 times x, or that the left leg of the triangle has length 5x/3. Similarly, the extension of the tilted right side of the square cuts the hypotenuse into two segments, of length 8 (i.e., 13-5) and 2 (i.e., 5-3). This means the longer leg of the triangle is cut into segments of length x and 4x, or that the longer leg has length 5x. Thus, the right triangle has legs in the ratio of three to one. Since the length of the hypothenuse is 10, the short leg must be SQRT(10) and the longer leg 3* SQRT(10). Hence x = 3*SQRT(10)/5 and x^2 = 90/25 = 3.6
What a long winded way to get to the answer! I did it using 3 similar triangles, as follows: Let the side of the square be S. Let the red dotted line be X. Let the yellow dotted line be Y. Then, using similar triangles, (S+X)/X = 5/2 &. (S+Y)/Y = 5/4 Solving for X and Y gives, X= ⅔S and Y =4S Using Pythagoras' in the largest triangle gives, (X + 5)² + (Y + S)² = 10² Substituting X and Y from above gives (⅔S +S)² + (4S + S)². = 100 (5S/3)² + (5S)² = 100 25S²/9 + 25S² = 100 250S² = 900 S² = 900/250 = 18/5 Therefore, area of square = 18/5 square units.
I think the area is 3.6. If x is the side length of the square. Make a point P where a line from 7 intersects the yellow line at 90 degrees you have a right triangle 7,P,13. The triangle formed by the baseline and the purple line and the green line is also a right triangle that has the same proportions as the bigger one since the green and the yellow line are parallel. Since the 5 to 7 is length 2 and 7 to 13 is length 6 it means the sides of the smaller triangle is 1/3 of the bigger. That means the purple line is 1/3 of the shortest side of the larger triangle, which is length x. If you shift the green line 2 steps to the left it will create an identical triangle 3,5 and the point where the shifted green line intersects the red line. Meaning the green line also has length x. Therefore the new rectangle formed by the initial new line is a square identical to the blue square. And since the big triangle has sides 3 times longer, and the green line is x long then the yellow line of the bigger triangle is 3x and the whole lenth of the yellow line is 4x. Which means x^2 + (3x)^2 = 6^2 x^2 + 9 x^2 = 36 10x^2 = 36 x^2 = 3.6
The four congruent triangles form each vertex of the square 2^2 = x^2 + y^2 4^2 = (2x)^2 + (2y)^2 8^2 = (4x)^2 + (4y)^2 10^2 = (5x)^2 + (5y)^2 If we take the closest point to the x-axis if we travel along one orthogonal it projects an orthogonal lines on the x-axis of -2. If we move along the other orthogonal, the orthogonal projection to the x axis is 6. This soul indicate that from that closest point the triangle so that point is z/6 to 5 and z/2 to 7 2^2 = z^2 /36 + 9z^2/36 = 10z^2/36 2^2*6^2 = 10z^2 12/SQRT(10) = z 6/SQRT(10) = z 36/10 or 3.6 = Area
Well one method can be as line through 13 is y = m(x-13) and through 7 is y = m(x-7) now use distance btw ll lines formula for side x in terms as 6m/sq root(1+m^2) now lines from 3&5 and distance between them as x = 2/sq root(1+m^) now equate two x same we get m as -1/3 and hence x ^2 as 3.6
It is simple. Purple line is equal to x/3 cause of similarity of triangles. Green line is equal to x cause of similarity of triangles. x^2+(x/3)^2=4. x^2= 18/5
Fast solution: By inspection, x/(cos theta) = 2, x/(sin theta) = 6. Divide the equations for tan theta = 1/3. Substitute for x = 2 cos(arctan 1/3) => area = 3.6.
Solution without additional constructions: x/sin(θ)=2 and x/cos(θ)=6 => x=2sin(θ)=6cos(θ) => Area A=4*sin(θ)^2 and A=36*cos(θ)^2. Multiply the first equation by 9 and sum them => 10A=36(sin(θ)^2+cos(θ)^2) => 10A=36 => A=3.6
In general the area of the square is A=1/(1/d1^2 + 1/d2^2), where d1 and d2 are the differences between the line crossings (in the given example d1=5-3 and d2=13-7). The general result follows from the point-line distance formula and the fact that the product of the slopes of two orthogonal lines equal -1.
@@numbers93Sorry I can’t underline anything, this is horrible to type lmao A is the position vector for the top of the square. b is one side of the square as a vector, c is the other. A + 5b = (13 0) A + 2b + c = (7 0) 3b - c = (6 0) and b dot c = 0 is then solvable (let b = (x y)) 3x - y = 6 3y + x = 0 b = (9/5 -3/5) c= (-3/5 -9/5) |b x c| = 3.6
Without changing the puzzle, you can eliminate the smaller triangle with hypotenuse from 5 to 7. Imagine it scaling down to nothing, while keeping the 5 point fixed. You're left with the same blue square, it's lower right point now touching the x-axis. It's now embedded in triangle with hypotenuse from 3 to 11. Now let s be the square's side and y the (now scaled down) yellow dashed line. By similar triangles "x/2 = y/6". By pythagoras, x^2 + y^2 = 6^2". Solving these two equations gets us the same answer.
Segment(3-5)=segment(5-7)=2 Segment(7-13)=6=3*segment(3-5),(5-7) --> you can line 5 squares with the second's low right vertex in x=7 and the fifth one's high right vertex in x=13 --> Segment(7-13) is the hypotenuse of a 3LxL right triangle --> (3L)^2+L^2=36 --> L^2=3,6
Label side of square x, we want area A: A=xx Label the colored segments red purple green and yellow r, p, g, and y respectively. We've got 5 congruent triangles of differing sizes: p/2=r/4=x/6=(p+x)/8=(r+x)/10, (*8,-p -->) 3p=2r-p=4x/3-p=x=.8(r+x)-p, (*10,-r -->) 5p-r=1.5r=5x/3-r=1.25(p+x)-r=x g/2=(g+x)/4=(y-g)/6=y/8=(y+x)/10, (*4,-g -->) g=x=.5y-g=(2y-5g)/3=.4(y+x)-g, (*10,-y -->) 5g-y=2.5(g+x)-y=(2y-5g)/3=y/4=x Note the likely most useful: x=3p=1.5r=g=y/4, p=x/3, r=2x/3, g=x, y=4x Pythagorean theorem: (y+x)^2+(r+x)^2=100 (4x+x)^2+(2x/3+x)^2=100 25xx+25xx/9=100 9*25xx+25xx=900 250xx=900 xx=900/250= A=18/5 We can also see from the above that the exercise gave us more information than minimally needed to solve it. Notably where the sides of the square were extended to intersect the x axis, we don't need all four points, any three would suffice, and two of the three would restrict the square to within column between pair of parallel lines, and third point would fix the angle of that column and the square's position within the column, and with that, the size of the square's side, and thus its area is then set. And can also check the consistency of the proportionality of the triangles to assure us we haven't been given a set of 4 points that are impossible or have no solution. In fact given any of those 3 points, the 4th needs to be where it is as shown in the exercise for it to be possible/correct.
*Some hints to my solution...* • Notice that the distance between the intersections of the red and purple lines, being extended versions of two of the square's sides, with the X-axis is 2. • Also notice how that aforementioned value of 2 never changes, no matter where the square is located. • Based on that, how many copies of the square would fit between the intersections of the green and yellow lines with the X-axis? • #PythagoreanTheorem
A simplification; draw a horizontal that touches the lower-right corner of the square and complete the two obvious similar right triangles. From the x-axis, the ratio of their bases (2 and 6) is 3, so the larger triangle has sides x,3x and hypotenuse 6. Proceed.
We can use projection lemma: Given a line segment L in 2d, call L1=Proj(L,x) and L2=Proj(L,y) (Proj is the projection, x,y are an abitrary couple of perpendicular directions of projection), then a=L1/L, b=L2/L are fixed ratios of transformation which satisfy a^2+b^2=1 so we have L^(-2)=L1^(-2)+L2^(-2). Using this lemma, we can solve the problem without introducing any new variable. Also, using similar arguments, we can solve 3d version of this problem: Given a cube lying completely above a plane (P), call S1,S2,S3 are areas of parellelograms that projected from each face of the cube, call S the area of a face of the cube, then we have S^(-2)=S1^(-2)+S2^(-2)+S3^(-2), from there we can actually compute the volume of the cube!
Slide the square down until it just touches the X axis. Or think of drawing a horizontal line that passes through the lowest corner. As long as the square's angle remains the same, the intercept points will still be spaced 2 apart on the left and 6 apart on the right (thanks to the parallel lines). There will only be two triangles now, and they'll be similar. The left one's top side and (horizontal) hypotenuse are x and 2 respectively. The corresponding sides of the triangle on the right will be sqrt(6^2 - x^2) and 6 (via Pythagoras). Corresponding sides of similar triangles have equal ratios: x/2 = sqrt(6^2 - x^2)/6 thus 3x = sqrt(6^2 - x^2) (via multiplying both sides by 6) thus 9x^2 = 6^2 - x^2 (via squaring both sides) thus 10x^2 = 36 (via basic algebraic simplification) thus x^2 = 36/10 = 3.6 (via dividing both sides by 10).
Let the purple dotted line have length y and the green one have length x. Let the square have side length s. The smallest triangle has the same angles as the triangles with hypotenuses 4 and 8, so they are similar and we can use ratio's to get a system we can solve. We have 2/4 = x/(x+s), 2/8 = y/(y+s), and 2^2 = x^2 + y^2 from the small triangle. From the 1st two equations we get that x = s and (1/3)s = y. Substitute for x and y in the third equation to get 4 = s^2 + (1/9)s^2 which simplifies to s^2 = 3.6, giving the area of the square.
Start by positioning the lowest corner at 0. The projection lines go to 6 and -2. cos theta = x/6 cos (theta + pi/2) = -x/2 = sin theta By the Pythagorean identity, sine squared + cosine squared must equal 1, so: (x/6)^2 + (-x/2)^2 = 1 (x^2)/36 + (x^2)/4 = 1 (x^2)/36 + (9x^2)/36 = 1 10x^2 = 36 x^2 = 3.6
The points 3-13 are the hypotenuse of a 6-8-10 (3-4-5) right triangle. The points 5-13 are a hypoteneuse of a right triangle with the same angles as a 3-4-5 triangle, so the lengths will all be the same proportions. From here you can get the length of the longer leg of that smaller triangle and subtract that from 8 to get X.
Once you realise that the smallest triangle has sides x, 3x and 2, and that you can neatly tile 6 of those triangles inside the square, its easily solved...
What if we just consider the right triangle with hypotenuse running from x=5 to x=7 (small triangle), and the right triangle that runs from x=3 to x=13 (large triangle). a = blue side, b = green side, s=side of square Because the ratio of their hypotenuses is large:small=5:1, then the length of sides of the large triangle is 5 times the length of sides of small triangle. 5a = red+s The trapezoid on the left of the small triangle consists of 3 small triangles. The red is twice of blue. red = 5a-s = 2a which implies a = (1/3)s By Pythagorean, a = sqrt(4-s^2) Then it's just (1/3)s = sqrt(4-s^2).
Er, at 1:24 Presh U said to me that side linked w 3, 5 that side (which u also said is x) is = 2 not ~ 2 but specifically you said it = 2. So x = 2. All sides of a square are equal thus 2 sqrd = 4, so the area is 4. Yet you run through this (skipping steps by the way) and get 3.5?
Due to how parallel lines intersect, you could move the square left-down until the corner hits 5, eliminating the 5-7 triangle completely and leaving the triangles with hypothenuse 2 and 6.
Можно подвинуть пару правых прямых так, чтобы точка 7 перешла в 5, а 13 в 11. Получился прямоугольный треугольник (с неизвестными катетами), в который вписан квадрат, и вершина квадрата на гипотенузе делит её на отрезки 2 и 6. Надо найти площадь (ну, или сторону) квадрата. Это очень простая задача, где-то класса для 7. Так как вершина на гипотенузе лежит на биссектрисе прямого угла, то катеты относятся, как 1:3. Отсюда легко найти и катеты, и сторону квадрата, точнее, площадь 3,6
My observation: A. the square has sides of length x, B. line on axis from 3 to 5, is the hypotenuse of a right triangle such that x is the longer leg. C. The line segment from 7 to 13 is similar to triangle in [B], BUT scaled up by 3, therefore longer leg is 3x, shorter leg is [ direct observation ] parallel to square & therefore length x .... So right triangle has all 3 lengths ... Same Math as you to get answer.
I have another approach, using the area of similar triangles. Let's say the area of the triangle with the hypotenuse between 5 and 7 is A1, the area of the triangle with the hypotenuse between 3 and 7 is A2, the area of the triangle with the hypotenuse between 5 and 13 is A3, and the area of the square is S. Since the triangle with the hypotenuse between 5 and 7 and the triangle with the hypotenuse between 3 and 7 are similar, so using the area of similar triangles: A1 / A2 = (2 / 4)^2 => A2 = 4A1 Since the triangle with the hypotenuse between 5 and 7 and the triangle with the hypotenuse between 5 and 13 are similar, so using the area of similar triangles: A1 / A3 = (2 / 8)^2 => A3 = 16A1 The total area of the triangle with the hypotenuse between 3 and 13 is: A2 + A3 - A1 + S = 4 A1 + 16 A1 - A1 + S = 19A1 + S Since the triangle with the hypotenuse between 3 and 7 and the triangle with the hypotenuse between 3 and 13 are similar, so using the area of similar triangles: 4A1 / (19A1 + S) = (4 / 10)^2 => 4A1 / (19A1 + S) = 16 / 100 => 400 A1 = 304 A1 + 16 S => 96 A1 = 16S => S = 6A1 => S^2 = 36A1^2 Let's say the leg extended to 7 and the leg extended to 5 for the triangle with the hypotenuse between 5 and 7 have lengths b and a respectively. Since the triangle with the hypotenuse between 5 and 7 and the triangle with the hypotenuse between 3 and 7 are similar, the side length of the square is also b. This means that: S = b^2 Using the area of a triangle: A1 = a * b / 2 => A1^2 = a^2 * b^2 / 4 => A1^2 = a^2 * S / 4 => S^2 = 36 * a^2 * s / 4 => S = 9a^2 => a^2 = S / 9 Using the Pythagorus' Theorem on the triangle with the hypotenuse between 5 and 7: a^2 + b^2 = 4 => a^2 + S = 4 =>. S / 9 + S = 4 => 10 / 9 * S = 4 => 10 S = 36 => S = 3.6
Using similar triangles, you can determine that the area of the square is 6 times the area of the little right triangle having points (5,0) and (7,0) on its hypotenuse. Also using similar triangles and relating sides of the right triangle with sides of the square, you find the longest side of that small right triangle is 3 times the length of its shortest side. Call the short side b and the longer side h. By pythagorean theorem we know h^2+b^2=9b^2+b^2=4. This implies b is sqrt(2/5). h is 3 times that. So the area of the little triangle is (1/2)b(3b)=(3/2)b^2=3/5. The area of the square is six times that so 18/5.
Let s = the side length of the square. Let the x-axis intersection points be labeled C, E, G, and M in order left to right, and let the square's vertices be labeled PQRS clockwise from top left. This problem gives us 4 similar triangles: ERG (green and purple lines), CSG (red and green lines plus bottom side of the square), EQM (yellow and purple lines plus right side of the square), and CPM (yellow and red lines plus top/left sides of the square). Let x be the length of ER (purple line). As CG = 2EG, then CS (red line) = 2x and as SR = s, SG = 2s. This means that RG (green line) = s. ER/EG = CP/CM x/2 = (2x+s)/10 2(2x+s) = 10x 4x + 2s = 10x 2s = 6x s = 3x ER² + RG² = EG² x² + s² = 2² x² + (3x)² = 4 x² + 9x² = 4 10x² = 4 x² = 4/10 = 2/5 x = √(2/5) s = 3√(2/5) Area = s² = (3√(2/5))² = 9(2/5) = 18/5 sq units
The dotted line that passes through 5 is parallel to one side of the large right triangle, so it divides the other two sides in the same proportion. Ditto the dotted line that passes through 7. Finally, apply Pythagoras to the three sides of the large right triangle, whose hypotenuse is given. Three equations, three unknowns. Simple.
Let x be the angle between bottom left side and horizontal axis. Then, side = (5-3)cosx = 2cosx and side = (13-7)sinx = 6sinx So 2cosx = 6sinx As sin²x + cos²x = 1 → sin²x + 9sin²x = 1 → sin²x = 0.1 Area = side² = 36sin²x = 36×0.1 = 3.6
Let purple segment = x. Then, by similar triangles, red segment = 2x, and entire red side of the big triangle is 5x, so square is 3x per side. Also by similar triangles, extended green edge is 6x. and green segment is half of that, or 3x. Now, by Pythagoras, we have x^2 + (3x)^2 = 2^2. 10x^2 = 4, and x^2 = 2/5. But the area of the square is (3x)^2, or 9x^2, so 18/5.
I solved this in a much more complicated manner: I extended a line (parallel to the x axis) from the leftmost point of the square to the right side of the square, and noted that this line has length 2 (since it must be the same length as the distance from 3 to 5). Then considering the angle between this line and the bottom of the square to be θ, we have cos(θ) = x/2 (adjacent over hypotenuse). I then extended a line from the bottom of the square off to the left, and also drew another x-axis-parallel line leftward from the rightmost corner of the square, and found that the length from the rightmost corner to where those two new lines meet must have length 6 (again, same distance as from 7 to 13). The angle between the length-6 line and the rightmost side of the square is π/2-θ radians (since the sum of this angle and our earlier θ angle is π/2 radians, or 90 degrees), so we get cos(π/2-θ) = x/6. Now that I had two equations and two variables, I could plug them into an equation solver and get the value for x, which I squared to get the final answer.
Fourth solution: 1. make a vertical line at point 7, the distance on this line between the parallel lines must be 2. 2. Now we have a right triangle with sides 2, 6 and an hypotenuse. 3. The height relative to this hypotenuse is the size of the side of the square, x. 4. The inverse square of the height relative to the hypotenuse is equal to the sum of the inverse square of the sides of the right triangle. 1/x^2 = 1/2^2 + 1/6^2 1/x^2 = (9+1)/36 1/x^2 = 10/36 = 5/18 x^2 = 18/5 = 3.6
The 5-7 right triangle is repeated in the 3-5 section plus another 2 triangles together making a rectangle.. if we complete the triangle to make it a rectangle .. we can see that it isn’t a square yet.. but each 1 triangle high rectangle is 1/3rd of the way towards being a square … Red dotted line being 2x/3 & blue dotted line being x/3 Long adjacent side is the X The hypotenuse is the length between the 5-7. We go across, to the other side of the diagram like you did in “solution 2” and all these lengths are tripled up , because of similar shapes with same angles.. And so.. - Pythagorean formula starts: • A²+B² =C² • 3x² + x² = 6² ↪️• 4x² = 36 • 4x² /4 = 36 /4 ↪️• X² = 9 • √ X² = √ 9 ↪️• X = 3 X = 3 Area of 🟦 = X² X² = 3² ↪️• Area of 🟦 = X² = 9 🔸 Answer is = 9 🔸
1:36 What's the point of this insight? I don't think it's immediately intuitive why the triangle's sides correspond to the axis segments, and you don't seem to even need that fact in your solutions.
I used geometry to conclude the triangles were similar: The sides of a square are parallel so the traversals are also parallel which means the triangles so formed are similar. Thus, the sides of the second quadrilateral must also be a square of side x.
I find this super complicated. Here is my solution. Let x be the side of the square and y the little segment over 5. From similar triangles : (x+y)/(13-5) = y /(7-5) so x = 3 y. From Thales on the bottom triangle 3-7, the line over 5 cuts the line at the square bottom in the middle. We can now use Pythagorean theorem on the little triangle 5-7: 4 = y^2 + x^2 = 10/9 x^2 so x^2 =3.6
- take top corner (x1, y1), construct equation for yellow and red lines in terms of x1 nad y1; apply their perpindicularity - get x1 or y1, plug one into line equation to get another - take left corner (x2,y2), do the same around red and green - find distance between these corners, square it in the meantime realise purple line is irrelevant as condition of shape being a square suffice. For more general rectangle case must have taken third corner
Just make 4 point slope formula for the line for the side of squares and use general variable for slope. Then get the distance of the sides with distance formula for 2 different sides, set them equal to find what the slope would have to be because it’s a square (1/3) then just find area
I did it Called the slope of red and purple lines m, and the slope of green and yellow lines -1/m (because the product of slopes of 2 perpendicular lines is -1) Found the intersection of purple line and green line in terms of m and called it point M Found the intersection of red line and yellow line in terms of m and called it point N Found the intersection of red line and green line in terms of m and called it point Q Found the distance QM and QN in terms of m using distance between 2 points formula Made QM and QN equal because the sides of a square are equal to each other Found out that m=3 Which means the side of the square is (3sqrt10)/5 So the area of the square is 3.6
..another approach.. using intersection points ..once u realised the tangent is 1/3 one can construct the lines intersecting the x axis and a pependicular line (purple in the video) line1(13 x axis intersect) y1 = -(1/3)x +13/3 line 2(7 x axis intersect) y2 = -(1/3)x + 7/3 Perpendicular line intersecting x axis at 5 y3 = 3x - 15 intersection pooints between line 1 and 3, and line 2 and 3 give the points equation 1: -(1/3)x +13/3 = 3x - 15 -> x=5.8 equation 2: -(1/3)x + 7/3 = 3x - 15 -> x=5.2 put value of x into line3 equation ang get the y points (it intersects both lines) 1: 5.8, 2.4 2: 5.2, 0.6 a=Δx=0.6, b=Δy=1.8 [a²+b²=c², c in this case is the side of the square] 1.8² + 0.6² = 3.6 sq units ...tada... ps, corrected an error -1/3x vs -(1/3)x ...
If the difference between the first two values is x and the difference between the last two values is y then the area of the square is xxyy/(xx+yy) or 3.6 in this case.
Solving this problem is very easy. I think the real challenge happens when you are only given the word description of the problem and you have to convert words into a diagram. You have to be careful and make sure your drawing is completely accurate. If given the diagram, this is a piece of cake.
Here’s my solution, Extend the square so that it becomes a rectangle which has its vertice on the point 13 . Form there you can use the three triangles , one with base 57 other with 35 another with 7 13 to solve the problem
If you move the square to the x axis you get 2 similar triangles. By comparing the hypotenuse of the smaller (2) and the bigger (6), you know that the samller is 1/3 of the bigger. With a hypotenuse of 2, and one cathetus of x and the other of 1/3 x (because of the similar triangles), you get the equasion : (1/3 x)^2 + x^2 = 2^2 x = 6/(10^0.5) A=x^2 therefore A=3.6
Just by trying to figure it out in my head, I knew it was going to involve similar triangles, but without writing it down and labeling everything, I kind of got lost in the algebraic manipulation. I was too lazy to grab a piece of paper and a pen(cil), so I just gave up just to see the conclusion. Another method of solving this would be to use systems of linear equations however, we only had enough information for each line to know 1 point and not 2. We know where they cross the x-axis or when y is 0, solutions to the lines, but we didn't have any more information to construct any of the general forms of the linear equations. We don't know their slopes nor their y-intercepts so trying to solve these for these lines and using their slopes to find other points on the lines we couldn't easily determine the actual vertices or corner points of the square preventing us from using the distance formula for finding the length of one of its sides. It is another method of solving this, but not enough information was given to use this method. Another possible way to solve it would be to use the inscribed angles of a circle, but again, we still don't have enough information for that technique. So, using the properties of self-similar triangles appears to be the only valid approach towards solving it. My initial attempt of using self-similar triangles was the correct approach.
I did it by writing equations of line green, yellow, purple and red, in terms of slope m of line green. Then equated the distance between line green and yellow to be equal to the distance between line red and purple since, they are sides of a square & got the value of m = ± 1/3 then I put m = ± 1/3 in one of the above equations to get Area = 3.6
I have a fun question for you. Q. I will provide you a side fragment of a circular dish and you have to find the radius of that original dish. Please show in several ways
Let X be the side length of the square Let Y be the length of the purple line We know that the red-yellow-axis triangle is similar to the red-green-axis triangle and the purple-green-axis traingle For Purple-Green-Axis: X²+Y²=2² For Red-Green-Axis, we double it since the hypotnuse is doubled (2X)²+(2Y)²=4² For Red-Yellow-Axis, we multiply by 5 since the hypotnuse goes from 2 to 10 (5X)²+(5Y)²=10² We can now look at the Red side of Red-Yellow-Axis Its length is 5Y But it also has segments of X along the square and 2Y on the red dashes Thus X=3Y Subbing in we get X²+(X/3)²=2² 10/9 • X² =4 X² = 36/10 or 18/5
The easiest problems to solve are those that have several approaches because you just need to visual one of several. This is one of those problems. I solved it differently than Presh and got the right answer. My solution was to get the red and purple lines in terms of the side length of the square and use pythagorean theorem. Really a blend of what he did. The hardest problems to solve are those that only have one approach that leads to the correct answer and what makes the problem hard, the approach is not obvious.
This puzzle seemed to be easy, and I quickly found the solution. But then I wondered: how can I initially prove that this figure is even possible? I mean, I have a feeling that there are too many inputs to be correlated: the given figure is a square and all its faces lie on lines intersecting the x-axis at given points. Could it be that some of the inputs are redundant? What if the left coordinate is 2.9 instead of 3? Will we get the solution which won't make sense or will just the θ be different so that the figure remains square? More generally, is it true that for any tuple of four differing x coordinates the right value of θ forming the square exists (and if yes, how can it be found)? This problem can be simplified in terms of the input data, for example, we can move the x-axis up so that the lower angle of the square lies on the x-axis and then reduce the number of parameters to just two lengths of segments on x-axis. Then we can ask ourselves: what is the condition making this problem consistent? First of all, we can compose some function of, lets say, θ describing the ratio of the rectangle (formed with four lines) faces, find its extremum and prove that there is indeed some value of θ (expressed in any useful terms) with which the formed rectangle is a square so we can make further conclusions about this particular problem consistency (as well as for the general case with different lengths). It's just an assumption of a possible way, I didn't do it myself, but I expected such discussion to be in the video...
My thoughts exactly! How can we be sure that for every set of numbers a square can be formed? Turns out that there's always a θ that can make the obtained rectangle a square. If the y-axis is shifted to the right to coincide with the first point on the x-axis, then the problem is reduced to 3 parameters, say a, b and c, where a
I have a much faster solution! If the tilt angle is â and side s: (5-3)cos â=s (13-7)sin â=s Then cos â=s/2 sin â=s/6 cos²+sin²=1 Then s²/4+s²/(4*9)=1 10s²/36=1 s²=36/10 😊
I tried assuming the area of the big triangle 6x8 because of the 10 hypothenuse , then try to subtract the area of 24 with the small triangle + trapezoid to get the area of the square , but my answer is 3,2 something Probably make a mistake in my calculation but nice video
🤔Just complete the full rectangle (area=30), and subtract the lower strip (area=12). The remainder is 5 squares (area=18), so each square has area=18/5.
Square? Yani kare , o zaman 4 kenarı da birbirine eşit. Bu durumda 2×2= 4 Eğer mavi alanın alanını deseydiniz, o zaman kare olarak düşünmeyiz. Square deyince ben " kare" anlıyorum. Acaba yanılıyor muyum?
Just one minor adjustment @8:05. Instead what you could do, is use the Pythagorean theorem to show that (x/3)^2 + x^2 = 4. It’s just a tad bit easier I think
I have snatched the pebble from MindYourDecision's hand! Here's a simpler solution, sort of a solution by observation. The x axis is a hypotenuse. Call the scaling factor to the long leg of the biggest right triangle x. That makes the hypotenuse 10 and the longest leg 10x. The ratio of the whole x axis segment to the smallest hypotenuse is 10:2, making the smallest right triangle's long leg 2x. Add a perpendicular at that intersection on the x axis 6 units from the right. That divides the topmost, longest leg into segments of 6x and 4x. We know that littlest triangle has a long leg of 2x, so the distance along the longest leg between the perpendicular we added and the inset square is also 2x. That long leg is 10x. Subtract 6x for the part above the segment of length 6, and subtract 2x for the length just determined, and the side of the square is 2x. Now we can look at the rightmost inner triangle and see 6x^2 + 2x^2 = 36, or 40x^2 = 36. x^2=36/40 and x = approximately 0.95. The sides of the square are 2x, so (2x)^2 is 3.6. Woo-hoo! Proud of myself. Got that before I hit play!
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I was way off... I wanted to come up with analytical equations for all lines, given the parallel and perpendicular properties of a slope but it didn;t give me enough to go on
I suck at math and I am a med student (Ofc I didn't took med stream because I sucked at math) But IDK I miss this subject so this channel is my daily dose of it..... YK physics is good but not good as math.
Move the square down to touch the x axis. The left triangle has hypotenuse 2. The right triangle has hypotenuse 6. Then 2*cos(t)=a and 6*sin(t)=a where t is the smaller angle between the bottom vertex of the square and the x axis and a is the square side length. Therefore (a/6)^2 + (a/2)^2 = 1. A = a^2 = 18/5
I did the same approach but went through arctan and sin. your answer is more elegant.
how could i got 11.5??!!???
I know it. It's the side length squared. Do I win?
Bros so smart
Yeah man,... You've won...
It says
" The " Square
Not " a " Square
Tit for Tat 😂
yeah man you did it you did it!!
bro discovered the answer that stumbled most people who didnt know math
Similar triangles is the most economical route to approach the solution
my brain cannot process the word 'economical'
8:00 a more quicker way!
youre trying to find out the length of other sides but it's not needed.if you look carefully the yellow traingle(except the smaller one) (at the timestamp i mentioned) is a right angle triangle so (3x)² + x² = (2+2+2)² 10x²=36 and x²=3.6.
I solved it differently.
My square has an area of (3x)². The small triangle between 5 and 7 has the sides x and 3x, thus the hypotenuse is 2 = sqrt(10)x => x = sqrt(2/5).
The total area of the square is then 9x² = 18/5.
That was my approach as well.
Same here.
Here’s a vector solution for anyone interested
A is the position vector for the top of the square. b is one side of the square as a vector, c is the other.
A + 5b = (13 0)
A + 2b + c = (7 0)
3b - c = (6 0) and b dot c = 0 is then solvable (let b = (x y))
3x - y = 6
3y + x = 0
b = (9/5 -3/5)
c= (-3/5 -9/5)
|b x c| = 3.6
I approached it along the lines of the last of the methods shown, but reasoned a little differently. Consider the entire right triangle (i.e., with hypothenuse 10). Now the extension of the tilted bottom of the square cuts the hypothenuse into two segments, of length 6 (i.e., 10-4) and 4 (i.e., 7-3). This means (due to parallel line segments) that the red segment is 2/3 times x, or that the left leg of the triangle has length 5x/3. Similarly, the extension of the tilted right side of the square cuts the hypotenuse into two segments, of length 8 (i.e., 13-5) and 2 (i.e., 5-3). This means the longer leg of the triangle is cut into segments of length x and 4x, or that the longer leg has length 5x. Thus, the right triangle has legs in the ratio of three to one. Since the length of the hypothenuse is 10, the short leg must be SQRT(10) and the longer leg 3* SQRT(10). Hence x = 3*SQRT(10)/5 and x^2 = 90/25 = 3.6
What a long winded way to get to the answer!
I did it using 3 similar triangles, as follows:
Let the side of the square be S.
Let the red dotted line be X.
Let the yellow dotted line be Y.
Then, using similar triangles,
(S+X)/X = 5/2 &. (S+Y)/Y = 5/4
Solving for X and Y gives,
X= ⅔S and Y =4S
Using Pythagoras' in the largest triangle gives,
(X + 5)² + (Y + S)² = 10²
Substituting X and Y from above gives
(⅔S +S)² + (4S + S)². = 100
(5S/3)² + (5S)² = 100
25S²/9 + 25S² = 100
250S² = 900
S² = 900/250 = 18/5
Therefore, area of square = 18/5 square units.
I think the area is 3.6.
If x is the side length of the square.
Make a point P where a line from 7 intersects the yellow line at 90 degrees you have a right triangle 7,P,13.
The triangle formed by the baseline and the purple line and the green line is also a right triangle that has the same proportions as the bigger one since the green and the yellow line are parallel. Since the 5 to 7 is length 2 and 7 to 13 is length 6 it means the sides of the smaller triangle is 1/3 of the bigger.
That means the purple line is 1/3 of the shortest side of the larger triangle, which is length x.
If you shift the green line 2 steps to the left it will create an identical triangle 3,5 and the point where the shifted green line intersects the red line. Meaning the green line also has length x. Therefore the new rectangle formed by the initial new line is a square identical to the blue square. And since the big triangle has sides 3 times longer, and the green line is x long then the yellow line of the bigger triangle is 3x and the whole lenth of the yellow line is 4x.
Which means x^2 + (3x)^2 = 6^2
x^2 + 9 x^2 = 36
10x^2 = 36
x^2 = 3.6
Very good, I used also similar triangles
The four congruent triangles form each vertex of the square
2^2 = x^2 + y^2
4^2 = (2x)^2 + (2y)^2
8^2 = (4x)^2 + (4y)^2
10^2 = (5x)^2 + (5y)^2
If we take the closest point to the x-axis if we travel along one orthogonal it projects an orthogonal lines on the x-axis of -2. If we move along the other orthogonal, the orthogonal projection to the x axis is 6. This soul indicate that from that closest point the triangle so that point is z/6 to 5 and z/2 to 7
2^2 = z^2 /36 + 9z^2/36 = 10z^2/36
2^2*6^2 = 10z^2
12/SQRT(10) = z
6/SQRT(10) = z
36/10 or 3.6 = Area
Well one method can be as line through 13 is y = m(x-13) and through 7 is y = m(x-7) now use distance btw ll lines formula for side x in terms as 6m/sq root(1+m^2) now lines from 3&5 and distance between them as x = 2/sq root(1+m^) now equate two x same we get m as -1/3 and hence x ^2 as 3.6
i did the same
It is simple. Purple line is equal to x/3 cause of similarity of triangles. Green line is equal to x cause of similarity of triangles. x^2+(x/3)^2=4. x^2= 18/5
Fast solution: By inspection, x/(cos theta) = 2, x/(sin theta) = 6. Divide the equations for tan theta = 1/3. Substitute for x = 2 cos(arctan 1/3) => area = 3.6.
Solution without additional constructions: x/sin(θ)=2 and x/cos(θ)=6 => x=2sin(θ)=6cos(θ) => Area A=4*sin(θ)^2 and A=36*cos(θ)^2. Multiply the first equation by 9 and sum them => 10A=36(sin(θ)^2+cos(θ)^2) => 10A=36 => A=3.6
Great! Congratulation 🎉
@@hans-rudigerdr.zimmermann5672 Thanks, professor!
I like that you go through the many different ways to do it. I solved it with the last method but knew there was other ways.
Frankly, this is precisely why I love math so much. There's never only a single way to solve a problem. Only solutions and verifications of solutions.
In general the area of the square is
A=1/(1/d1^2 + 1/d2^2),
where d1 and d2 are the differences between the line crossings (in the given example d1=5-3 and d2=13-7).
The general result follows from the point-line distance formula and the fact that the product of the slopes of two orthogonal lines equal -1.
I have the feeling that it also has a solution using the rule: "The dot product of perpendicular vectors is zero." But I'm too lazy to do it :)
not really
@@numbers93Sorry I can’t underline anything, this is horrible to type lmao
A is the position vector for the top of the square. b is one side of the square as a vector, c is the other.
A + 5b = (13 0)
A + 2b + c = (7 0)
3b - c = (6 0) and b dot c = 0 is then solvable (let b = (x y))
3x - y = 6
3y + x = 0
b = (9/5 -3/5)
c= (-3/5 -9/5)
|b x c| = 3.6
@MegaUpstairs I got you bro :)
Without changing the puzzle, you can eliminate the smaller triangle with hypotenuse from 5 to 7. Imagine it scaling down to nothing, while keeping the 5 point fixed. You're left with the same blue square, it's lower right point now touching the x-axis. It's now embedded in triangle with hypotenuse from 3 to 11. Now let s be the square's side and y the (now scaled down) yellow dashed line. By similar triangles "x/2 = y/6". By pythagoras, x^2 + y^2 = 6^2". Solving these two equations gets us the same answer.
Segment(3-5)=segment(5-7)=2
Segment(7-13)=6=3*segment(3-5),(5-7)
--> you can line 5 squares with the second's low right vertex in x=7 and the fifth one's high right vertex in x=13
--> Segment(7-13) is the hypotenuse of a 3LxL right triangle
--> (3L)^2+L^2=36 --> L^2=3,6
Label side of square x, we want area A:
A=xx
Label the colored segments red purple green and yellow r, p, g, and y
respectively.
We've got 5 congruent triangles of differing sizes:
p/2=r/4=x/6=(p+x)/8=(r+x)/10,
(*8,-p -->) 3p=2r-p=4x/3-p=x=.8(r+x)-p,
(*10,-r -->) 5p-r=1.5r=5x/3-r=1.25(p+x)-r=x
g/2=(g+x)/4=(y-g)/6=y/8=(y+x)/10,
(*4,-g -->) g=x=.5y-g=(2y-5g)/3=.4(y+x)-g,
(*10,-y -->) 5g-y=2.5(g+x)-y=(2y-5g)/3=y/4=x
Note the likely most useful:
x=3p=1.5r=g=y/4, p=x/3, r=2x/3, g=x, y=4x
Pythagorean theorem:
(y+x)^2+(r+x)^2=100
(4x+x)^2+(2x/3+x)^2=100
25xx+25xx/9=100
9*25xx+25xx=900
250xx=900
xx=900/250=
A=18/5
We can also see from the above that the exercise gave us more information than minimally needed to solve it.
Notably where the sides of the square were extended to intersect the x axis, we don't need all four points, any three would suffice,
and two of the three would restrict the square to within column between pair of parallel lines, and third point would fix the angle of that column and the square's position within the column,
and with that, the size of the square's side, and thus its area is then set.
And can also check the consistency of the proportionality of the triangles to assure us we haven't been given a set of 4 points that are impossible or have no solution. In fact given any of those 3 points, the 4th needs to be where it is as shown in the exercise for it to be possible/correct.
*Some hints to my solution...*
• Notice that the distance between the intersections of the red and purple lines, being extended versions of two of the square's sides, with the X-axis is 2.
• Also notice how that aforementioned value of 2 never changes, no matter where the square is located.
• Based on that, how many copies of the square would fit between the intersections of the green and yellow lines with the X-axis?
• #PythagoreanTheorem
What if when you rotate the square -30° then whats the intersection will be on X-Axis, so its not a solution to square 2 and get 4 😂
A simplification; draw a horizontal that touches the lower-right corner of the square and complete the two obvious similar right triangles. From the x-axis, the ratio of their bases (2 and 6) is 3, so the larger triangle has sides x,3x and hypotenuse 6. Proceed.
We can use projection lemma: Given a line segment L in 2d, call L1=Proj(L,x) and L2=Proj(L,y) (Proj is the projection, x,y are an abitrary couple of perpendicular directions of projection), then a=L1/L, b=L2/L are fixed ratios of transformation which satisfy a^2+b^2=1 so we have L^(-2)=L1^(-2)+L2^(-2). Using this lemma, we can solve the problem without introducing any new variable. Also, using similar arguments, we can solve 3d version of this problem:
Given a cube lying completely above a plane (P), call S1,S2,S3 are areas of parellelograms that projected from each face of the cube, call S the area of a face of the cube, then we have S^(-2)=S1^(-2)+S2^(-2)+S3^(-2), from there we can actually compute the volume of the cube!
Slide the square down until it just touches the X axis. Or think of drawing a horizontal line that passes through the lowest corner. As long as the square's angle remains the same, the intercept points will still be spaced 2 apart on the left and 6 apart on the right (thanks to the parallel lines). There will only be two triangles now, and they'll be similar. The left one's top side and (horizontal) hypotenuse are x and 2 respectively. The corresponding sides of the triangle on the right will be sqrt(6^2 - x^2) and 6 (via Pythagoras). Corresponding sides of similar triangles have equal ratios:
x/2 = sqrt(6^2 - x^2)/6
thus 3x = sqrt(6^2 - x^2) (via multiplying both sides by 6)
thus 9x^2 = 6^2 - x^2 (via squaring both sides)
thus 10x^2 = 36 (via basic algebraic simplification)
thus x^2 = 36/10 = 3.6 (via dividing both sides by 10).
Great illustration of the use of Pythagoras' Theorum
Could we not use pythagorean theorum to solve for x at 7:48?
Let the purple dotted line have length y and the green one have length x. Let the square have side length s. The smallest triangle has the same angles as the triangles with hypotenuses 4 and 8, so they are similar and we can use ratio's to get a system we can solve. We have 2/4 = x/(x+s), 2/8 = y/(y+s), and 2^2 = x^2 + y^2 from the small triangle. From the 1st two equations we get that x = s and (1/3)s = y. Substitute for x and y in the third equation to get 4 = s^2 + (1/9)s^2 which simplifies to s^2 = 3.6, giving the area of the square.
Start by positioning the lowest corner at 0. The projection lines go to 6 and -2.
cos theta = x/6
cos (theta + pi/2) = -x/2 = sin theta
By the Pythagorean identity, sine squared + cosine squared must equal 1, so:
(x/6)^2 + (-x/2)^2 = 1
(x^2)/36 + (x^2)/4 = 1
(x^2)/36 + (9x^2)/36 = 1
10x^2 = 36
x^2 = 3.6
The points 3-13 are the hypotenuse of a 6-8-10 (3-4-5) right triangle. The points 5-13 are a hypoteneuse of a right triangle with the same angles as a 3-4-5 triangle, so the lengths will all be the same proportions. From here you can get the length of the longer leg of that smaller triangle and subtract that from 8 to get X.
Once you realise that the smallest triangle has sides x, 3x and 2, and that you can neatly tile 6 of those triangles inside the square, its easily solved...
X, 1/3X*, and 2?
@@davidnewell3232depending on what x is defined as, some solved for 9x^2 by defining a different x
@@nol2521 I assumed the labeled diagram was the appropriate reference.
What if we just consider the right triangle with hypotenuse running from x=5 to x=7 (small triangle), and the right triangle that runs from x=3 to x=13 (large triangle).
a = blue side, b = green side, s=side of square
Because the ratio of their hypotenuses is large:small=5:1, then the length of sides of the large triangle is 5 times the length of sides of small triangle.
5a = red+s
The trapezoid on the left of the small triangle consists of 3 small triangles. The red is twice of blue.
red = 5a-s = 2a which implies a = (1/3)s
By Pythagorean, a = sqrt(4-s^2)
Then it's just (1/3)s = sqrt(4-s^2).
Er, at 1:24 Presh U said to me that side linked w 3, 5 that side (which u also said is x) is = 2 not ~ 2 but specifically you said it = 2. So x = 2. All sides of a square are equal thus 2 sqrd = 4, so the area is 4. Yet you run through this (skipping steps by the way) and get 3.5?
Due to how parallel lines intersect, you could move the square left-down until the corner hits 5, eliminating the 5-7 triangle completely and leaving the triangles with hypothenuse 2 and 6.
Можно подвинуть пару правых прямых так, чтобы точка 7 перешла в 5, а 13 в 11. Получился прямоугольный треугольник (с неизвестными катетами), в который вписан квадрат, и вершина квадрата на гипотенузе делит её на отрезки 2 и 6. Надо найти площадь (ну, или сторону) квадрата. Это очень простая задача, где-то класса для 7. Так как вершина на гипотенузе лежит на биссектрисе прямого угла, то катеты относятся, как 1:3. Отсюда легко найти и катеты, и сторону квадрата, точнее, площадь 3,6
My observation: A. the square has sides of length x, B. line on axis from 3 to 5, is the hypotenuse of a right triangle such that x is the longer leg. C. The line segment from 7 to 13 is similar to triangle in [B], BUT scaled up by 3, therefore longer leg is 3x, shorter leg is [ direct observation ] parallel to square & therefore length x .... So right triangle has all 3 lengths ... Same Math as you to get answer.
I have another approach, using the area of similar triangles.
Let's say the area of the triangle with the hypotenuse between 5 and 7 is A1, the area of the triangle with the hypotenuse between 3 and 7 is A2, the area of the triangle with the hypotenuse between 5 and 13 is A3, and the area of the square is S.
Since the triangle with the hypotenuse between 5 and 7 and the triangle with the hypotenuse between 3 and 7 are similar, so using the area of similar triangles:
A1 / A2 = (2 / 4)^2
=> A2 = 4A1
Since the triangle with the hypotenuse between 5 and 7 and the triangle with the hypotenuse between 5 and 13 are similar, so using the area of similar triangles:
A1 / A3 = (2 / 8)^2
=> A3 = 16A1
The total area of the triangle with the hypotenuse between 3 and 13 is:
A2 + A3 - A1 + S = 4 A1 + 16 A1 - A1 + S = 19A1 + S
Since the triangle with the hypotenuse between 3 and 7 and the triangle with the hypotenuse between 3 and 13 are similar, so using the area of similar triangles:
4A1 / (19A1 + S) = (4 / 10)^2
=> 4A1 / (19A1 + S) = 16 / 100
=> 400 A1 = 304 A1 + 16 S
=> 96 A1 = 16S
=> S = 6A1
=> S^2 = 36A1^2
Let's say the leg extended to 7 and the leg extended to 5 for the triangle with the hypotenuse between 5 and 7 have lengths b and a respectively.
Since the triangle with the hypotenuse between 5 and 7 and the triangle with the hypotenuse between 3 and 7 are similar, the side length of the square is also b.
This means that:
S = b^2
Using the area of a triangle:
A1 = a * b / 2
=> A1^2 = a^2 * b^2 / 4
=> A1^2 = a^2 * S / 4
=> S^2 = 36 * a^2 * s / 4
=> S = 9a^2
=> a^2 = S / 9
Using the Pythagorus' Theorem on the triangle with the hypotenuse between 5 and 7:
a^2 + b^2 = 4
=> a^2 + S = 4
=>. S / 9 + S = 4
=> 10 / 9 * S = 4
=> 10 S = 36
=> S = 3.6
Using similar triangles, you can determine that the area of the square is 6 times the area of the little right triangle having points (5,0) and (7,0) on its hypotenuse. Also using similar triangles and relating sides of the right triangle with sides of the square, you find the longest side of that small right triangle is 3 times the length of its shortest side. Call the short side b and the longer side h. By pythagorean theorem we know h^2+b^2=9b^2+b^2=4. This implies b is sqrt(2/5). h is 3 times that. So the area of the little triangle is (1/2)b(3b)=(3/2)b^2=3/5. The area of the square is six times that so 18/5.
Let s = the side length of the square. Let the x-axis intersection points be labeled C, E, G, and M in order left to right, and let the square's vertices be labeled PQRS clockwise from top left. This problem gives us 4 similar triangles: ERG (green and purple lines), CSG (red and green lines plus bottom side of the square), EQM (yellow and purple lines plus right side of the square), and CPM (yellow and red lines plus top/left sides of the square).
Let x be the length of ER (purple line). As CG = 2EG, then CS (red line) = 2x and as SR = s, SG = 2s. This means that RG (green line) = s.
ER/EG = CP/CM
x/2 = (2x+s)/10
2(2x+s) = 10x
4x + 2s = 10x
2s = 6x
s = 3x
ER² + RG² = EG²
x² + s² = 2²
x² + (3x)² = 4
x² + 9x² = 4
10x² = 4
x² = 4/10 = 2/5
x = √(2/5)
s = 3√(2/5)
Area = s² = (3√(2/5))² = 9(2/5) = 18/5 sq units
y/z=2 ; (x+z)/8=z/2 ; (y/2)*2 +x*2 =2*2=4 ; this gives x*2=36/10=3.6 (y=red dotted line, z=purple dotted line, x= square side length )
The dotted line that passes through 5 is parallel to one side of the large right triangle, so it divides the other two sides in the same proportion.
Ditto the dotted line that passes through 7.
Finally, apply Pythagoras to the three sides of the large right triangle, whose hypotenuse is given.
Three equations, three unknowns. Simple.
Let x be the angle between bottom left side and horizontal axis. Then,
side = (5-3)cosx = 2cosx and
side = (13-7)sinx = 6sinx
So 2cosx = 6sinx
As sin²x + cos²x = 1
→ sin²x + 9sin²x = 1
→ sin²x = 0.1
Area = side² = 36sin²x = 36×0.1 = 3.6
Let purple segment = x. Then, by similar triangles, red segment = 2x, and entire red side of the big triangle is 5x, so square is 3x per side. Also by similar triangles, extended green edge is 6x. and green segment is half of that, or 3x. Now, by Pythagoras, we have x^2 + (3x)^2 = 2^2. 10x^2 = 4, and x^2 = 2/5. But the area of the square is (3x)^2, or 9x^2, so 18/5.
I solved this in a much more complicated manner: I extended a line (parallel to the x axis) from the leftmost point of the square to the right side of the square, and noted that this line has length 2 (since it must be the same length as the distance from 3 to 5). Then considering the angle between this line and the bottom of the square to be θ, we have cos(θ) = x/2 (adjacent over hypotenuse). I then extended a line from the bottom of the square off to the left, and also drew another x-axis-parallel line leftward from the rightmost corner of the square, and found that the length from the rightmost corner to where those two new lines meet must have length 6 (again, same distance as from 7 to 13). The angle between the length-6 line and the rightmost side of the square is π/2-θ radians (since the sum of this angle and our earlier θ angle is π/2 radians, or 90 degrees), so we get cos(π/2-θ) = x/6. Now that I had two equations and two variables, I could plug them into an equation solver and get the value for x, which I squared to get the final answer.
Fourth solution:
1. make a vertical line at point 7, the distance on this line between the parallel lines must be 2.
2. Now we have a right triangle with sides 2, 6 and an hypotenuse.
3. The height relative to this hypotenuse is the size of the side of the square, x.
4. The inverse square of the height relative to the hypotenuse is equal to the sum of the inverse square of the sides of the right triangle.
1/x^2 = 1/2^2 + 1/6^2
1/x^2 = (9+1)/36
1/x^2 = 10/36 = 5/18
x^2 = 18/5 = 3.6
generalization
let 0
The 5-7 right triangle is repeated in the 3-5 section plus another 2 triangles together making a rectangle.. if we complete the triangle to make it a rectangle .. we can see that it isn’t a square yet.. but each 1 triangle high rectangle is 1/3rd of the way towards being a square …
Red dotted line being 2x/3
& blue dotted line being x/3
Long adjacent side is the X
The hypotenuse is the length between the 5-7.
We go across, to the other side of the diagram like you did in “solution 2” and all these lengths are tripled up , because of similar shapes with same angles.. And so..
- Pythagorean formula starts:
• A²+B² =C²
• 3x² + x² = 6²
↪️• 4x² = 36
• 4x² /4 = 36 /4
↪️• X² = 9
• √ X² = √ 9
↪️• X = 3
X = 3
Area of 🟦 = X²
X² = 3²
↪️• Area of 🟦 = X² = 9
🔸 Answer is = 9 🔸
1:36 What's the point of this insight? I don't think it's immediately intuitive why the triangle's sides correspond to the axis segments, and you don't seem to even need that fact in your solutions.
I used geometry to conclude the triangles were similar: The sides of a square are parallel so the traversals are also parallel which means the triangles so formed are similar. Thus, the sides of the second quadrilateral must also be a square of side x.
I find this super complicated. Here is my solution. Let x be the side of the square and y the little segment over 5. From similar triangles : (x+y)/(13-5) = y /(7-5) so x = 3 y.
From Thales on the bottom triangle 3-7, the line over 5 cuts the line at the square bottom in the middle. We can now use Pythagorean theorem on the little triangle 5-7: 4 = y^2 + x^2 = 10/9 x^2 so x^2 =3.6
Coordinate solution..?
would be interesting if someone provides... i'm too busy rn to do it
- take top corner (x1, y1), construct equation for yellow and red lines in terms of x1 nad y1; apply their perpindicularity - get x1 or y1, plug one into line equation to get another
- take left corner (x2,y2), do the same around red and green
- find distance between these corners, square it
in the meantime realise purple line is irrelevant as condition of shape being a square suffice. For more general rectangle case must have taken third corner
I did that. it worked
Just make 4 point slope formula for the line for the side of squares and use general variable for slope. Then get the distance of the sides with distance formula for 2 different sides, set them equal to find what the slope would have to be because it’s a square (1/3) then just find area
I did it
Called the slope of red and purple lines m, and the slope of green and yellow lines -1/m (because the product of slopes of 2 perpendicular lines is -1)
Found the intersection of purple line and green line in terms of m and called it point M
Found the intersection of red line and yellow line in terms of m and called it point N
Found the intersection of red line and green line in terms of m and called it point Q
Found the distance QM and QN in terms of m using distance between 2 points formula
Made QM and QN equal because the sides of a square are equal to each other
Found out that m=3
Which means the side of the square is (3sqrt10)/5
So the area of the square is 3.6
Could you tell me what software/website/programming language/... you use to make your videos?
Extend the lines to make a 5 by 1 rectangle then, for example, (3x)²+x²=36 and A=3.6 no doubt has been done already, nice puzzle.
If right angle is "A":
10SinA = x + 4*SinA
10CosA = x + 8*CosA
=>
SinA = x/6
CosA = x/2
=>
x^2/36 + x^2/4 = 1
=>
S = x^2 = 3.6
Use a rotation matrix to rotate the coordinate system so the x axis is parallel to the bottom of the square
..another approach.. using intersection points
..once u realised the tangent is 1/3 one can construct the lines intersecting the x axis
and a pependicular line (purple in the video)
line1(13 x axis intersect)
y1 = -(1/3)x +13/3
line 2(7 x axis intersect)
y2 = -(1/3)x + 7/3
Perpendicular line intersecting x axis at 5
y3 = 3x - 15
intersection pooints between line 1 and 3, and line 2 and 3 give the points
equation 1: -(1/3)x +13/3 = 3x - 15 -> x=5.8
equation 2: -(1/3)x + 7/3 = 3x - 15 -> x=5.2
put value of x into line3 equation ang get the y points (it intersects both lines)
1: 5.8, 2.4
2: 5.2, 0.6
a=Δx=0.6, b=Δy=1.8 [a²+b²=c², c in this case is the side of the square]
1.8² + 0.6² = 3.6 sq units
...tada...
ps, corrected an error -1/3x vs -(1/3)x ...
If the difference between the first two values is x and the difference between the last two values is y then the area of the square is xxyy/(xx+yy) or 3.6 in this case.
Solving this problem is very easy. I think the real challenge happens when you are only given the word description of the problem and you have to convert words into a diagram. You have to be careful and make sure your drawing is completely accurate. If given the diagram, this is a piece of cake.
Here’s my solution,
Extend the square so that it becomes a rectangle which has its vertice on the point 13 . Form there you can use the three triangles , one with base 57 other with 35 another with 7 13 to solve the problem
If you move the square to the x axis you get 2 similar triangles. By comparing the hypotenuse of the smaller (2) and the bigger (6), you know that the samller is 1/3 of the bigger. With a hypotenuse of 2, and one cathetus of x and the other of 1/3 x (because of the similar triangles), you get the equasion :
(1/3 x)^2 + x^2 = 2^2
x = 6/(10^0.5)
A=x^2
therefore A=3.6
okay basically the same solution1 you did
Pretty easy figured out right away via similar triangle property.
Just by trying to figure it out in my head, I knew it was going to involve similar triangles, but without writing it down and labeling everything, I kind of got lost in the algebraic manipulation. I was too lazy to grab a piece of paper and a pen(cil), so I just gave up just to see the conclusion.
Another method of solving this would be to use systems of linear equations however, we only had enough information for each line to know 1 point and not 2. We know where they cross the x-axis or when y is 0, solutions to the lines, but we didn't have any more information to construct any of the general forms of the linear equations. We don't know their slopes nor their y-intercepts so trying to solve these for these lines and using their slopes to find other points on the lines we couldn't easily determine the actual vertices or corner points of the square preventing us from using the distance formula for finding the length of one of its sides. It is another method of solving this, but not enough information was given to use this method. Another possible way to solve it would be to use the inscribed angles of a circle, but again, we still don't have enough information for that technique. So, using the properties of self-similar triangles appears to be the only valid approach towards solving it. My initial attempt of using self-similar triangles was the correct approach.
I did it by writing equations of line green, yellow, purple and red, in terms of slope m of line green.
Then equated the distance between line green and yellow to be equal to the distance between line red and purple since, they are sides of a square & got the value of m = ± 1/3
then I put m = ± 1/3 in one of the above equations to get Area = 3.6
I have a fun question for you.
Q. I will provide you a side fragment of a circular dish and you have to find the radius of that original dish. Please show in several ways
Let X be the side length of the square
Let Y be the length of the purple line
We know that the red-yellow-axis triangle is similar to the red-green-axis triangle and the purple-green-axis traingle
For Purple-Green-Axis:
X²+Y²=2²
For Red-Green-Axis, we double it since the hypotnuse is doubled
(2X)²+(2Y)²=4²
For Red-Yellow-Axis, we multiply by 5 since the hypotnuse goes from 2 to 10
(5X)²+(5Y)²=10²
We can now look at the Red side of Red-Yellow-Axis
Its length is 5Y
But it also has segments of X along the square and 2Y on the red dashes
Thus X=3Y
Subbing in we get
X²+(X/3)²=2²
10/9 • X² =4
X² = 36/10 or 18/5
The easiest problems to solve are those that have several approaches because you just need to visual one of several. This is one of those problems. I solved it differently than Presh and got the right answer. My solution was to get the red and purple lines in terms of the side length of the square and use pythagorean theorem. Really a blend of what he did.
The hardest problems to solve are those that only have one approach that leads to the correct answer and what makes the problem hard, the approach is not obvious.
I just found out the purple line in terms of x and then used Pythagoras.
I found it by using the *other* 2 similar triangles in the diagram.
This puzzle seemed to be easy, and I quickly found the solution. But then I wondered: how can I initially prove that this figure is even possible? I mean, I have a feeling that there are too many inputs to be correlated: the given figure is a square and all its faces lie on lines intersecting the x-axis at given points. Could it be that some of the inputs are redundant? What if the left coordinate is 2.9 instead of 3? Will we get the solution which won't make sense or will just the θ be different so that the figure remains square? More generally, is it true that for any tuple of four differing x coordinates the right value of θ forming the square exists (and if yes, how can it be found)?
This problem can be simplified in terms of the input data, for example, we can move the x-axis up so that the lower angle of the square lies on the x-axis and then reduce the number of parameters to just two lengths of segments on x-axis. Then we can ask ourselves: what is the condition making this problem consistent? First of all, we can compose some function of, lets say, θ describing the ratio of the rectangle (formed with four lines) faces, find its extremum and prove that there is indeed some value of θ (expressed in any useful terms) with which the formed rectangle is a square so we can make further conclusions about this particular problem consistency (as well as for the general case with different lengths).
It's just an assumption of a possible way, I didn't do it myself, but I expected such discussion to be in the video...
My thoughts exactly! How can we be sure that for every set of numbers a square can be formed?
Turns out that there's always a θ that can make the obtained rectangle a square.
If the y-axis is shifted to the right to coincide with the first point on the x-axis, then the problem is reduced to 3 parameters, say a, b and c, where a
I have a much faster solution!
If the tilt angle is â and side s:
(5-3)cos â=s
(13-7)sin â=s
Then
cos â=s/2
sin â=s/6
cos²+sin²=1
Then
s²/4+s²/(4*9)=1
10s²/36=1
s²=36/10
😊
Great ! Congratulation 🎉
Anyway my method is slightly different --- I just use the small triangle at the bottom center -- (x/3)^2 + x^2 = 4 which will give us x^2 right away.
How to make this amazing animation,
Please reply if you know,
Please
I tried assuming the area of the big triangle 6x8 because of the 10 hypothenuse , then try to subtract the area of 24 with the small triangle + trapezoid to get the area of the square , but my answer is 3,2 something
Probably make a mistake in my calculation but nice video
🤔Just complete the full rectangle (area=30), and subtract the lower strip (area=12). The remainder is 5 squares (area=18), so each square has area=18/5.
Square?
Yani kare , o zaman 4 kenarı da birbirine eşit.
Bu durumda 2×2= 4
Eğer mavi alanın alanını deseydiniz, o zaman kare olarak düşünmeyiz. Square deyince ben " kare" anlıyorum.
Acaba yanılıyor muyum?
I was too lazy to look for pen and paper, so I solved it in my head. Took whole 3 minutes.
2025 AYT'de çıkarsa başımızı yakacak soru.
Just one minor adjustment @8:05. Instead what you could do, is use the Pythagorean theorem to show that (x/3)^2 + x^2 = 4. It’s just a tad bit easier I think
7x13 = and intersects at 9
I did it using theorems of similar triangles. Did it faster by taking ratios of sides of triangle
When I saw this I was thinking we needed proof that it is a square.
The area of the square is in the lower left corner of the graph.
#NailedIt
Yup got the same answer using similar triangles.
Very nice one!
Why are the numbers from pythagorian triplets?
I solved using trigonometry. Really simple. 🙂
I have snatched the pebble from MindYourDecision's hand! Here's a simpler solution, sort of a solution by observation.
The x axis is a hypotenuse. Call the scaling factor to the long leg of the biggest right triangle x.
That makes the hypotenuse 10 and the longest leg 10x.
The ratio of the whole x axis segment to the smallest hypotenuse is 10:2, making the smallest right triangle's long leg 2x.
Add a perpendicular at that intersection on the x axis 6 units from the right. That divides the topmost, longest leg into segments of 6x and 4x. We know that littlest triangle has a long leg of 2x, so the distance along the longest leg between the perpendicular we added and the inset square is also 2x.
That long leg is 10x. Subtract 6x for the part above the segment of length 6, and subtract 2x for the length just determined, and the side of the square is 2x.
Now we can look at the rightmost inner triangle and see 6x^2 + 2x^2 = 36, or 40x^2 = 36.
x^2=36/40 and x = approximately 0.95. The sides of the square are 2x, so (2x)^2 is 3.6.
Woo-hoo!
Proud of myself. Got that before I hit play!
I am not sure if you are aware of this but I have a friend who is a nurse and she is a nurse who is a nurse at the hospital and she is a nurse in the hospital and she is a nurse that is a nurse for the hospital and she is a nurse practitioner and she is a nurse
Good One
It is disheartening when I can only remember I knew this stuff and now can't.
You had me until "Hey, this is Presh Talwalkar". 🤭
I think I need to do a lot more studying. 😁
3...5....7...? I swear I would put 2² in the answer and move on.
VIDEO IDEA: There’s one meme that’s making its way with an interesting question:
How is the percentage to get this question right?
25%
50%
60%
25%.
18/5or3.6???
3*7?
3.6 u^2
I was way off...
I wanted to come up with analytical equations for all lines, given the parallel and perpendicular properties of a slope but it didn;t give me enough to go on
I suck at math and I am a med student (Ofc I didn't took med stream because I sucked at math) But IDK I miss this subject so this channel is my daily dose of it..... YK physics is good but not good as math.
Once again, the axes must be oriented towards the positive values only.
x^2
We do have different definitions of a "fun puzzle" :) :) :)
Using similar triangle relation and the Pythagorean theorem, I calculated the area to be 2/5. Now, let's watch the video.
Ahh it seems i forgot to multiply by 9 on the other side