Alternative solution (no Pythagorean Theorem): ua-cam.com/video/I6tImV5q58g/v-deo.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
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%
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
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.
nice, that was solution presented in the video but I actually found a way to do the problem without the Pythagorean formula, using only similar triangles and a simple system of equations, I'll post a new video on it soon.
Alternative solution (no Pythagorean Theorem):
ua-cam.com/video/I6tImV5q58g/v-deo.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
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%
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
I got this one using Pythagorean Theorem!
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.
nice, that was solution presented in the video but I actually found a way to do the problem without the Pythagorean formula, using only similar triangles and a simple system of equations, I'll post a new video on it soon.