Solution #2 = beautiful.

Sweet! I never remember that two tangents from an external point theorem.

Wow!

It's easy if you're used to thinking about unit cells in crystals. Fill the plane with these unit cells, and you'll recognize that the four triangles can be translated through a distance of one unit cell length (distance = 10; one up, one down, one left, and one right (which this video gets to at 6:50). The total area isn't changed, but five congruent 'tilted' squares can be seen, including the one in the center and four adjacent squares, each sharing an edge with the center square. Since the area of the big outline square is 10² = 100, each of the five squares has an area of 20.

Let angle CBF = x => angle ABC = 45-x tan x = 1/3 and BC = 2.sqrt(10) tan (45-x) = (1-tan x)/(1+tan x) = 1/2 Area ABC = 1/2*BC* AC = 1/2 * BC*BC *tan(45-x) = 1/2 * 40 * 1/2 = 10

I used coordinate geometry- less creative but it works. Let the bottom left corner be the origin- the line from that point is y=1/2x. The 2 downward sloping lines have equations y=10-2x and y=20-2x. Find the points of intersection, then the distance between these, then square it for the answer (though you can save 2 steps at the end- just don't take the square root when finding the distance. You'll have d^2 which is the desired area).

He over complicated the solution. We know that the hypotenuse of the triangle made by one side of the big square is 10. We also know that halfway creates a triangle with half of the big triangle's hypotenuse is 5. We can create an equivalent triangle using the lower half of the big hypotenuse and we know then that the side of the red square is equal to the y in the figure he drew. So that y^2 + (2y)^2 = 10^2 = 100 => 5y^2 = 100 => y^2 = 100/5 = 20 Therefore the area of the red square is 20 because y is one side of the red square and the are is y^2.

I watch your videos like others watch football or hockey games! Thank you!

Yeah, it would have been helpfull, if the midpoints of the square had been shown in the diagram. Now it is only briefly mentioned at the introduction and I missed that.

Nice problem. Here is a nice extension: Generalise by replacing 8 by R. Show that R=4 is impossible. Why? What is the smallest possible value for R?

6.666666

Bonita forma de resolver. Parabéns!

Consider the tangent points of the circle, they divide the sides of the triangle into lengths a and b where x+y=8, x+1 and y+1. Thus the big triangle is divided into 3 little ones all of height 1 with bases of 8, x+1 and y+1. So area =1/2(8+x+y+2) =9 Not even a hint of Pythagoras...

Take some lessons to improve your ENGLISH

I solved it orally and got the area to be 20 square units. One big triangle had sides 5, 10, 5√5. So the ratio was 1:2:√5. For the small triangle Which is similar to the big, the hypotenuse was 5(or √5×√5), so using the ratio, I found the other sides to be 2√5 and √5. Add them and subtract from 5√5 to get 2√5, the side length of the red square. Square it to get 20.

Solution: s = side of a square = √4 = 2, A = top left corner of the left square, B = bottom right corner of the right square, C = top corner of the red triangle, D = top left corner of the right square, E = bottom left corner of the left square. AB = √(6²+2²) = √40, Angle ABE = arctan(2/6) = arctan(1/3), Angle CBA = 45°-arctan(1/3), AC = AB*tan(angleCBA) = AB*tan[45°-arctan(1/3)] Area of ​​the red triangle = AB*AC/2 = AB*AB*tan[45°-arctan(1/3)]/2 = AB²*tan[45°-arctan(1/3)]/2 = 40*tan[45°-arctan(1/3)]/2 = 20*tan[45°-arctan(1/3)] = 10

wrong the aide of the smaller square is 5. use the basic proportionality theorem

We can also solve it using Trigonometry (and Geometry). One of the acute angles (call it c) of the red triangle which is inside the 3rd square on the right is given by c = (45 - a) where a is the acute angle in the right triangle that is half of the rectangle formed by the 3 squares. This right triangle has two of its sides as 2 and 6 and so tan(a) = 2/6 = 1/3. Now, tan(45 - a) can be found by using tan(A-B) formula which comes out to be 1/2. Now, the side of the red triangle that is almost vertical is d tan(c) where d is the diagonal of the rectangle formed by the 3 squares and d = sqrt(40). Finally, the area of the red triangle is 1/2 * d * d tan(c) = 1/2 * d^2 * tan(c) = 1/2 * 40 * 1/2 = 10.

it's just 4 times 2 equations with 2 unknown numbers: 10 print "the phantom of the math geometry dez2024":dim x(1,3),y(1,3) 20 l1=10:l2=l1/2:xg11=0:yg11=0:xg12=l1:yg12=l2:xg21=l2:yg21=0:xg22=0:yg22=l1 30 gosub 70:x(1,0)=xl:y(1,0)=yl:xg21=l1:yg21=0:xg22=l2:yg22=l1:gosub 70:x(1,1)=xl:y(1,1)=yl 40 xg11=0:yg11=l2:xg12=l1:yg12=l1:gosub 70:x(1,2)=xl:y(1,2)=yl 50 xg21=0:yg21=l1:xg22=l2:yg22=0:gosub 70:x(1,3)=xl:y(1,3)=yl 60 x(0,0)=0:y(0,0)=0:x(0,1)=l1:y(0,1)=0:x(0,2)=l1:y(0,2)=l1:x(0,3)=0:y(0,3)=l1:goto 160 70 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12) 80 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22) 90 a13=a131+a132:a23=a231+a232:ngl1=a12*a21:ngl2=a22*a11 100 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end 110 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 120 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 130 xl=zx/ngl:yl=zy/ngl:print "x=";xl;"y=";yl 140 return 150 xbu=x*mass:ybu=y*mass:return 160 masx=900/l1:masy=850/l2:if masx<masy then mass=masx else mass=masy 170 for a=0 to 1:x=x(a,0):y=y(a,0):gosub 150:xba=xbu:yba=ybu:for b=1 to 4 180 ib=b:if ib=4 then ib=0 190 x=x(a,ib):y=y(a,ib):gosub 150:xbn=xbu:ybn=ybu:goto 210 200 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 210 gosub 200:next b:next a 220 gcol 5:x=0:y=0:gosub 150:xba=xbu:yba=ybu:x=l1:y=l2:gosub 150:xbn=xbu:ybn=ybu:gosub 200 230 x=0:y=l2:gosub 150:xba=xbu:yba=ybu:x=l1:y=l1:gosub 150:xbn=xbu:ybn=ybu:gosub 200 240 x=0:y=l1:gosub 150:xba=xbu:yba=ybu:x=l2:y=0:gosub 150:xbn=xbu:ybn=ybu:gosub 200 250 x=l2:y=l1:gosub 150:xba=xbu:yba=ybu:x=l1:y=0:gosub 150:xbn=xbu:ybn=ybu:gosub 200 260 ls=sqr((x(1,0)-x(1,1))^2-(y(1,0)-y(1,1))^2):print "der eckenabstand des inneren quadrats=";ls the phantom of the math geometry dez2024 x=4y=2 x=8y=4 x=6y=8 x=2y=6 der eckenabstand des inneren quadrats=3.46410162 > run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=@zoom%*1.4" at the beginning for full screen graphics

I solved it by using method 3 and made it right.

Δ ABC ~ Δ CBE (all angles equal) (in the notations V stands for square root) => AC/CE = CB/EB => AC = CE*CB/EB = 2V2*2V10/4V2 = V10 A = CB*AC/2 = 2V10*V10/2 = 10

I guess, At 4:08 We could have also replaced all "s" with "2x" Because, The Bigger Triangle is similar to the smaller one, having 2 Sides which have ratio of 2:1 to the smaller triangle, so the Third Side will also be Double of Smaller Triangle. I.e. For Sides of Triangle 2 * (∆5,2x,x ) = (∆10, (2x+s), x) => ∆10,4x,2x = ∆10,(2x+s),x => 4x = 2x+s => s = 2x And value of x is easily calculated in the illustration by using Pithagoruss Theory which comes to √5

It's interesting that the length x in this geometric construction is always 1/2 the length of the small square. How about a geometric proof of this proposition. No algebra, no numbers, nothing but geometric proof using proven theorems.

Why would you make far too complex a solution to a problem that has a much simpler solution. Find the length CB of hypotenuse of triangle CBF using Pythagorean theorem, You have the base of 6 and the height of 2, [root 40]. Also find angle CBF. You know angle DBF is 45 degrees. Subtract angle CBF from angle DBF and this gives you angle ABC [26.56]. With base CB and angle ABC you can find length AC. Now having height AC and base CB, you have all you need to calculate for area of triangle ACB. The units are not mentioned. Inches or centimetres? But the answer is 10 units sq.

CB squared is equal to the sum of CF squared and FB squared. This is very much Pythagoras theorem. Your statement that we would avoid Pythagoras theorem is not true.

Субтитры мешают!

Oh 19 was so clever

Solution: The red area is clearly a rhombus, so we need the two diagonals to calculate its area. The long diagonal is easy to calculate with the given values, using pythagora: 4² + 8² = e² 16 + 64 = e² e² = 80 e = 4√5 The second diagonal is a bit trickier. First we realize, that all the triangles HAVE to be identical - if mirrored - because of symmetry. As such, we know that their side lengths are 4 and x and their hypotenuse is 8 - x. We can therefore use pythagoras again to calculate x: 4² + x² = (8 - x)² 16 + x² = 64 - 16x + x² |-x² -16 +16x 16x = 48 |:16 x = 3 We now know, that the side lengths of all the triangles are 3, 4 and 5 (since this is a pythagorean triple). As the second to last step, we draw a new triangle: The missing diagonal is the hypotenuse, the short leg of it is part of the rectangle and the long side is parallel to the short side of the same rectangle. This triangles legs are 4 and 2. The later is just the calculated 5 minus the calculated 3. With that we get the pythagors equation 2² + 4² = f² 4 + 16 = f² f = 2√5 As the last step, we just multiply the diagonals and half them (area formula for a rhombus): A = ef/2 = 4√5 * 2√5 / 2 = 40/2 = 20

The lower right corner = a The right-side corner of the red triangle = 45° - a Sin(a) = 1/√10 Sin(45° - a) = Sin(45°) x Cos(a) - Cos(45°) x Sin(a) = (1/√2)(3/√10) - (1/√2)(1/√10) = 1/√5 Cos(45° - a) = 2/√5 Red area = (2√10 x √10)/2 = 10

CD = x + 3x = 4x CD = 2 + 2 = 4 4x = 4 x = 1 AC² = x² + (3x)² AC² = 1² + 3² AC² = 1 + 9 AC² = 10 AC = √10 BC² = CF² + BF² BC² = 2² + 6² BC² = 4 + 36 BC² = 40 BC = 2√10 [ABC] = (1/2)·BC·AC [ABC] = (1/2)·(2√10)·(√10) [ABC] = 10

The thumbnail picture doesn’t give enough information for it to be solved. Only by playing the video and listening to the explanation is the fact that the vertices of the square are connected to the opposite side mid points. 0:10

10×10=100÷5=20

4.4731×4.4721=20

Another way to solve this is to set up cartesian space & solve for coordinates of point A. Assign point B as (0,0). Equation for AB is Y = -X. Slope of CB is -1/3, so slope of CA is 3. Equation for CA is Y = 20 + 3X. Solving intersection we get Y = 5 & X = -5. From this we get length CA = √10 & CB = √40. Area ABC = (1/2)(√10)(√40) = 10.

別解法　ＡからＣＤに垂線を引き交点をＯとすると、三角形ＣＢＢ’と三角形ａｃＯは相似となり、ＣＯをａと置くと、ＡＯ＝３ａ三角形ＡＯＤは直角二等辺三角形となり、ＯＤ＝３ａ、ＣＤ＝４ａ＝４、ａ＝１。三角形ＡＣＤは４＊３／２＝６。下の三角形は6*2/2-2*2/2=4。求める三角形の面積は和となる。

Solution: Cut out the triangles and rotate them 180° around the mid point of their respective side. You end up with a cross made of 5 equal squares, the middle one is red. Therefore the red area is exactly 1/5 of the total area. Since the whole big square is 10² = 100 units², the red area is 20 units².

Happy New Year! I really love the problems you present. I did it a little different... I came up with formulas for the lines that go through each side of the triangle, assuming your point F is at (0,0): y = 3 x + 2 y = -1/3 x + 2 y = -x + 6 From this, I was able to get the corners as (0,2), (6,0), and then compute the third corner as (1,5). That makes it easy to find the lengths of the two sides adjacent to the right triangle, sqrt(10) and sqrt(40), making the area = 10.

You may be a genius in mathematics, but you have no right to put (X) in the face of Pythagoras. It is disrespectful..!!

Треугольник вправо от С прикладываем к квадрату с верхним левым углом В, но тут же вспоминаем, что столько же придётся отнять внизу (малый влево от В). В среднем квадрате остаётся закрашенной только половина. Поскольку мы уже маленький посчитали, смело прибавляем ещё половину квадрата с диагональю DB - получаем целый квадрат 4. Ещё равная площадь 4, подсчитанная сверху, всего 8. Кстати, такой же результат можно получить движением т. В влево, на стык оснований первого и второго квадратов, по теореме о сохранении площади при движении вершины параллельно основанию, получить диагонально построенный квадрат двойной площади.

d=s√2 это частный случай Теоремы Пифагора - не считается.

Second solution was very nice, thought about it for a little while but couldn't do it without Pythagoras. I used a coordinate system (origin at point F) to find the length of AC as the height of the triangle (with BC as base): Gradient of AB=gradient of BD=-1. AB through point B(6,0) gives AB:y=-x+6 BC has gradient -1/3 --> gradient of AC is 3. AC through point C(0,2) gives AC:y=3x+2 AB=AC --> -x+6=3x+2 --> x=1, y=-1+6=5 --> point A(1,5) AC=√((1-0)²+(5-2)²)=√10 BC=√(6²+2²)=2√10 Area triangle=1/2*BC*AC=1/2*2√10*√10=10 Happy 2025

φ = 30°; ∎ABCD → AD = BC = 2 = AB/3 → AB = 6 = CD = CE + DE = 2 + 4; ∆ ABD → sin⁡(DAB) = 1; BD = 2√10; ∆ BFD → sin⁡(FDB) = 1; ABD = δ; DBF = θ → δ + θ = 3φ/2 → tan⁡(δ) = 1/3 → tan⁡(θ) = 1/2 → area ∆ BFD = 10

I solved it using sines and cosines. First, by Pithagoras BC = 2√10. Let the angle CBF be theta. So sin theta = 1/√10 and cos theta = 3/√10. The angle DBF is 45º, therefore the angle DBC is 45º - theta. Then I calculated cos (45 - theta) = cos 45. cos theta + sin 45.sin theta = 2/√5, and AB = BC / (2/√5) = 5√2. Then AC² + (2√10)² = (5√2)², therefore AC = √10. Finally, the area is AC.BC/2 = √10. 2√10 / 2 = 10.

I used coordinate geometry and fact that perpendicular lines have slopes that multiply to -1.

Who WAS (!) Pythagoras? Now he is dead, but his philosophy is still "alive" in our minds (at least in most of ours)!

I used a combination of geometry and trigonometry. Step 1 is to calculate the smallest angle in the right triangle at the bottom of the array of squares. It's just the arctan of 1/3. Step 2 is to subtract this angle from 45 degrees to get the smallest angle in the red triangle (it's about 26.565 degrees). Step 3 is to calculate the base of the red triangle using the Pythagorean Theorem: √40. Step 4 is to calculate the height of the red triangle (the opposite side) using the tangent function, since we know the angle and the adjacent side (the base). Step 5 is just to compute the area of the red triangle since we now have its base and height.

10

5:00 ΔAEC ~ ΔCEB (AA) => AE/CE=CE/BE => AE=vʼ2; AB=5vʼ2 [ABC]=½AB•CE=½(5vʼ2)2vʼ2=10 sq.un.

Beautiful method. At 4:17, a perpendicular line could have been drawn from CD to vertex A, forming two triangles, ACP and ADP, in which ACP is similar to BCF. If CP= n, then DP =4-n, but since ACP is similar to BCF, and the relationship between the base and height is one is three times the others, the 4-n = 3n Hence, n =1 Hence, 4-n =3 (height Hence area = 3 (height) * 4 (base) * 1/2 = 6 6 + 4 = 10 answer