that was great ^^

Easy to see each of the big triangles represent a quarter. Thus fraction that is green = 1/2 + overlap. The overlapping triangle is similar to big triangles, and w.l.o.g we give the square side length of 2 so we can get the hyp of the big triangles to be sqrt5 and the hyp of the small triangle to be 1. If the side length is scaled down sqrt5 then the area is scaled down sqrt5^2 = 5 so the overlap is 1/5 of the big triangle, or one 20th of the whole square or 5% this leaves 55% that is shaded green. (11/20)

1) Let the side of the square be 2a. Its area will be 4a². 2) The legs of the two large triangles are 2a and a, making their areas = a² (i.e. 2a•a/2), so, both together would be 2a². 3) They overlap in a small triangle which is similar to the large triangles (left to the reader.) 4) The hypotenuse of the small triangle is a. The hypotenuse of the large triangles is a√5, making the ratio of the sides 1/√5... 5) Therefore the legs of the small triangle are a/√5, and 2a/√5. 6) therefore the area of the overlap is (2a²/5)/2, or a²/5. 7) So the colored area is given as: (area of square) - (sum of areas of large triangle - from #2 above) +(area of overlap - from #6 above). Or: 4a² - 2a² + a²/5 => 2a² + a²/5 => 11a²/5 8) Finally, to get the colored area's percentage of area of the square, we divide the area from #7 by the area of the square, or: (11a²/5)/4a² => 11/20 9) Converting 11/20 to a percentage yields: 55% So, that's the answer: 55% Now to watch the video to see if I got it right. -s.west

1. Orange + small blue on the right is a triangle where the base is the right side of the square. So A = b*h/2 = 50% 2. Equation of the line sloping down and to the right is y = .5 - .5 x = (1-x)/2 3. Equation of the perpendicular from the upper right corner to that sloping line is (y - 1) = 2 (x - 1), since it intersects (1,1) 4. y = (1-x)/2 = 2(x-1) + 1, 5. so 1 - x = 4x - 4 + 2; 5x = 3; x = 3/5 6. So the vertex of the right triangle is 3/5 of the way from the left vertex of the orange triangle to the lower right corner of the square. 7. So the area of the triangle id 3/5 * 50% = 30%.

tan (BEC) = tan (AED) = 2; let both angles be "a" since they are equal. Then tan (2a) = 2*2/ [1- 2^2] = 4/ [1-4] = -4/3. Now b=BEA = 180-2a -> tan (b) = -(-4/3) = 4/3. If I let all square sides be 4 (so half-side is 2), then BE^2 = 2^2 + 4^2 = 20 -> BE=2*sqrt (5). Now, tan (b) = 4/3 --> sin (b)=4/5 & cos (b) = 3/5. Now EF = BE*cos (b) = 2*sqrt (5) = 6/5 * sqrt (5) and BF = BE*sin (b) = 2*4/5 * sqrt (5) = 8/5 * sqrt (5). Now [BEF] = EF*BF/2 = 6/5 * 8/5 * 5/2 = 24/5. Now [ABCD] = 4^2 = 16 --> BEF % = 24/5/16 = 24/80 = 3/10 = 30%

Isosceles triangle

i fixed the side length at 2 and origin at the point labeled D in the video; then the line segment AE is on the line with equation y = 1 - (x/2) and the line segment BF is on the line with equation y = 2x - 2; they intersect at (6/5, 2/5) and hence AE has length 3/√5, BF has length 4/√5, the triangle has area 6/5, and since in the coordinate system i've chosen the square has area 4 the fraction of the square that is orange is 3/10

Alternative solution (no Pythagorean Theorem): kzfaq.info/get/bejne/f5ykfNCHmteZaJs.html Draw a segment from vertex C that intersects segment BE perpendicularly, let's called this intersection point G, then triangles BCG, CEG and BCE are all similar such that the ratio of similarity of triangles CEG and BCG is 1/2, so the ratio of their areas is 1/4. Let x be the area of triangle BCG, then the area of triangle CEG is x/4. Area ▲BCE = Area ▲BCG + Area ▲CEG = x + x/4 = 1/4 → x = 1/5 ▲BCG and ▲ABF are congruent (same angles and same hypotenuse) therefore Area ▲ABF = 1/5 The rest of the solution is the same as in the video

Nice problem! Here my way: Let s be the side length of the square, A it’s area. It‘s easy to show that the upper and lower right triangle each have are A/4. The third right triangle is similar to the other right triangles. It‘s hypotenuse has length s. The other right triangles have hypotenuse length sqrt(1 + (1/2)^2) = s*sqrt(5) / 2. The scaling factor therefore is sqrt(5) / 2. Since the area goes with the square of the scale factor, the area of the third right triangle has only 4/5 the area of the first or second one. Altogether these triangles have an area s^2*(1/4 + 1/4 + 1/5) = 7/10 * s^2. The remaining fourth triangle must therefore have area 3/10 * s^2.

Where does the 100x come from? The other numbers - 900, 2,500 & 400 - are fairly obvious, if someone is watching the video they probably don't understand were the 100x came from; if they do they are probably not watching the video.

Nice solution

if the sides are equal, and the angle is 90 degrees, you can rotate one triangle to get a rectangle

Let the side of the square be a. a*sin(alpha)=3. ….(1) a*sin(belta)=2 or a*cos(alpha)=2. …..(2) Dividing (1) by (2) tan(alpha)=3/2 from which sec(alpha)=sqrt(1+(3/2)^2)=sqrt(13)/2 hence cos(alpha)=2/sqrt(13) From (2) a*2/sqrt(13)=2 then a=sqrt(13) Area=a*a=13

(30)u^2

There's different ways to find DC. One of these, for example, is that CH:HD=3:1. Once we have CD=10 shaded area = 1/2*CD*BC = 30 squ

There's a significant mistake, as in the provocative Ramanujan's demonstration (1+2+3+..... infinite = -1/12): integer numbers properties are valid only for a finite value of n.

Upper line segment, x, is 10 (18^2 + 6^2 = x^2) The oblique segment, y, is 6×sqrt(10) (y^2 = 6^2 + 18^2) Sin(t) = 6/(6×sqrt(10)), then 2S = 10×6×sqrt(10)×1/(sqrt(10))

There are 3 different methods to find the top side of trapezium x = 10: 1. In triangle ADE, by Pythagoras theorem, x^2 = 6^2 + (18-x)^2 2. Triangle ABC similar to triangle CHD, by corresponding sides proportionality equation x/6sqrt10 = 3sqrt10/18 (DH perpendicular bisector of isosceles triangle ACD) 3. In triangle CHD, x = 3sqrt10/cosB (In triangle ABC, cos B = 18/6sqr10)

Easier is to create a rectangle ABCF where F is 6 units above A. Then note that the blue area is (6x18)/2 minus the area of triangle AFD = 6 * (18 - x) / 2 => Area 6x/2 = 30

Once you find that the legs of the shaded triangle have length 10, there's a much simpler way to finish the solution. The shaded area is the trapezoid area less the right triangle at its bottom. This works out to (1/2)(6)(10+18)-(1/2)(6)(18), which is (1/2)(6)(10) = 30.

I reflected each triangle along their hypotenuse and added two more on the other sides of the square. You end up with a 5*5 square, area 25. If you subtract 4 of the triangles, 4*1/2*2*3 = 12 you get 13.

or... you could have half a brain, take one look at the drawing and realize that the two triangles are equal and know that the area of the square is 2 squared plus 3 squared.

You don't need all those constructions! As soon as you determine the congruency of the 2 triangles, you know both are 2 - 3 - √13 triangles. Side of the square is √13, so its area is 13.

very obvious

Very good 👍

The area of the triangle is BH/2. We know that the Base is s. The Height of triangle B is half the height of triangle A, by the similar triangle argument you put forward. Combine with the fact that the heights of A and B add up to S, we get that the height of triangle A is 2S/3. So (S)(2S/3)/2 = S^2/3, and therefore, area of triangle / area of square = 1/3.

Join right lower corner to midpoint of top line. A trapezium is formed. The area ratios for the 4 triangles in the trapezium are 1:2:2:4 with total of 9 parts. The area ratio of (square minus trapezium) to (trapezium) is 1:3 hence (square minus trapezium) has 9/3 = 3 parts. Hence fraction of red triangle is 4/(1+2+2+4 +3) = 4/12 = 1/3. The trapezium area ratio can be derived from properties of similar triangles between 2 parallel lines. The general form is T^2:TB:TB:B^2 where T (1 in this problem) is top side ratio and B (2 in this problem) is bottom side ratio, T^2 (1 in this problem) is top triangle area ratio, TB (2 in this problem) is lateral triangles area ratio (equal on both sides), B^2 (4 in this problem) is bottom triangle area ratio. These ratios are very useful in solving area ratio problems.

I have been reading this book from a few days

How to purchase this book

Bring it to me in Tamil

One of my favorite books, thank you!

Pas mal

4

How you make your visualization video ?

Cool visual demo BUT wish you hadn't shown the answer prior to the demo.