Відео

Thank you!

Here's a different kind of solution. The locus of the right angle of the yellow triangle is a semicircle inside the square whose diameter is the left side of the square. If we assume that the area of the green triangle is determined uniquely by the available data, then we can safely place the right angle at the center of the square, making the green triangle an isosceles right triangle with legs 2 and area (1/2)(2)(2), which is 2. The solutions given in the video confirm the above assumption.

2^3^4

This is too hard

If F is the projection of E onto AB, let y be the length of EF. The area G of the green triangle is 1/2*x*y. Triangle AEF is similar to triangle DAE. Thus 2/x=y/2 => y=4/x so G=1/2*x*(4/x)=2.

DE is not given so the solution must be the same for all valid values of DE and we can solve special cases. monthnorth3009 chose DE = 2 and found the area of the green triangle to be 2. A simpler special case is DE = 0. ΔADE is now a line segment equal to the side of the square, so AD = AB = 2 and green triangle has area 2. However, we no longer really have a triangle. To make our case more legitimate, let DE be extremely small but >0. As we make DE smaller, we can make AD as close as we want to AE but never make them equal as long as DE > 0. Eventually, we can make DE small enough to treat the difference between AD and AE as negligible. In mathematics, AD = AE is called the limit as DE approaches 0.

I like this problem very much! Here’s my solution : let x be the side length of the square. draw a perpendicular to AD from E, let the intersection at AD be labelled as G. Let GE = h. draw another perpendicular to AB from E and label its intersection at AB as H. HE = sqrt(2^2 - h^2) = sqrt(4 - h^2) by Pythagoras. The area of triangle AED is either (AD)(GE)/2 or (AE)(ED)/2 so (AD)(GE)/2 = (AE)(ED)/2, so ED = xh/2. By Pythagoras, ED^2 + AE^2 = AD^2. So (xh/2)^2 + 2^2 = x^2. Solving for h^2 gives h^2 = 4 - 16/x^2. Now, HE = sqrt(4 - h^2) = sqrt(16/x^2) = 4/x. Finally, area of green triangle equal to (AB)(HE)/2 = 1/2 * x * 4/x = 2 units squared! When you can’t solve with similar triangles, look for Pythagoras. When you can’t solve by Pythagoras, solve by trigonometry. When you can’t solve by trigonometry, solve by coordinate geometry 😂

Both methods are wonderful.

Holly shit! Beautiful question!!!

DE can also be 2 units which fully satisfies the 90 degree angle requirement, and then each yellow corner angle will be 45 degrees. Yellow and green will be congruent with areas 1/2 x 2 x 2. 2 units.

Usual trick, we suppose the diagonal is 4, side is 4/sqrt(2), area is 1/4×(4/sqrt(2))^2=2.😂😂😂

1/Drop the height BH to AE (extended) We have: angle AEE = angle BAH ( having sides perpendicular) so the two triangles ADE and HAB are congruent so, AE=BH=2 --> Area of the green triangle= 1/2 x2x2= 2 sq units😅

Pretty killer

Trig identities are more fun! 🙂

Super Duper👍

1.extend AE→AH丄BH △ADE,△BAH congruent(AAS) →AE=BH=2 2.△AEB=AE·BH/2=2😊

Nice! ∎ABCD → AB = AF + BF = BC = CD = AM + DM = r + r; AE = 2; sin⁡(EFA) = 1; EF = h; ADE = FAE = δ → sin⁡(δ) = 2/2r = 1/r = h/2 → hr = 2 → h = 2/r → rh = 2 = area ∆ AEB

Так это тот же болт. только вид с боку! h/2 = 2/x=sina, тоже тригонометрия.

Интересно. какой второй метод. Первый решается устно!

Green area =2 square units.❤❤❤

Let's use an orthonormal center A and first axis (AD) We have A(0; 0) D(c; 0) and E(2.cos(t); 2.sin(t)) with c the side length of the square and t = angleEAD Then VectorAE(2.cos(t); 2.sin(t)) and Vector(ED(2.cos(t) -c; 2.sin(t) These vectors are orthogonal, so we have: (2.cos(t)).(2.cos(t) -c) +(2.sin(t)).(2.sin(t)) = 0, or 4 -2.c.cos(t) = 0 or c.cos(t) = 2 The basis of the green triangle is AB = c, the corresponding height is the abscissa of E: 2.cos(t), so its area is (1/2).(c).(2.cos(t)= = c.cos(t) = 2

S=4

If 2 sides are missing then it cannot be a triangle. My teacher taught me that a triangle has 3 sides. 😂

You are the best ever💚

1 Area = (pi r^2)/4 Circumference = (pi 2r)/4 pi r^2/4 = pi 2r/4 Since it was given that the area of the quarter circle= its circumference pi r^2 =pi 2r multiply both sides by 4 r = 2 divide both sides by pi r Since the diagonal of the square= the radius, then Pythoagrean Theorem can be employed Let the side of the square = n , then n^2 + n^2 = 2^2 2 n^2 = 4 n^2 = 2 Since the triangle if one-half the square , then area of triangle = 1

2*2=1*(2r-1)===>r=5/2

Good problem, regardless of slight discrepancy in nomenclature. That's why it's always a good idea to 'REVIEW THE SETUP', before you dive in.

Well, one cannot equal the other, because they are different kinds of measurements, but I assume you mean the same number describes them. 2r + (2*pi*r)/4 = (pi*r^2)/4 Try some simplifying, but I will need to wing it a little: (*4) 8r + 2*pi*r = pi*r^2 Factorise: r(8 + 2pi) = r(pi*r) 2pi + 8 = pi*r Although I dislike doing it, this can be shifted to decimal notation for a very close result: 14.284 = 3.142 * r 14.284/3.142 = r r = 4.546 OE = CA = r Hypotenuse of the isosceles right triangle is 4.546 CB and CA are each 4.546/sqrt(2) = 3.2146... 3.2146^2 = 10.3337... Divide by 2 for 5.167 un^2 (rounded)

I am puzzled too! As dansouth asks how can you equate quarter circle area to quarter circle perimeter since their units are not the same?

Interesting problem, starting with no numbers, yet the answer is not a variable.

You are the best!

Perimeter is in units, and area is in units squared. How are these comparable?

For me there is an error in the formulation of the question

well done explaining

2222

336😊

x=0,5 or 1/2

I did not understand. But after I reallized he was referring to the perimeter. Quite cool question.

blue area = (2/π)(π + 4) (for perimeter) or: r = 2 → blue area = 1(circumference)

The answer is 1. The circumference of the quarter circle is 2πr/4 and does not include the straight sides!

.@1:54...yes ..you should have labeled it..Perimeter....instead of circumference

Circumference = r*Pi/2+2r Area = Pi/4 *r*r Set em equal Pi/2=Pi/4 r Pi/4+2=r We have the triangle is half of the square so r^2/4 is it's area (Pi/8+1)^2=Pi^2/64+Pi/4+1 is the answer

Thank you!

h=32(4/13)unit.ans

Excelente día señor, gracias por sus aportes, por favor corregir 2πr/4 +2r es diferente de 2πr/4 Es igual cuando r=0 Que no es nuestro caso Desde BOGOTÁ D.C. COLOMBIA

Seems like the initial problem statement should have used “perimeter” vice “circumference”

Alternately, on the hunch of a Pythagorean triangle and leveraging Euclid's parametrization of Pythagorean triples, look for two squares adding to 277: 277 = 256 + 21 - no good = 225 + 52 - no good = 196 + 81 - Bingo! That suggests (14^2 - 9^2, 2*14*9, 14^2 + 9^2) = (115, 252, 277) Verify that 277 + 252 + 115 = 644 - Done; and we have our solution. The exercise of searching for sum of two squares is so rewarding when successful, and such a small expense with these small numbers, that it really should be the first test. Euclid (in modern terminology) showed that all Pythagorean triples are uniquely parametrized by natural numbers k, m, n where m, n are coprime and not both even. Then by expansion it's clear that: (2kmn)^2 + (k(m^2 - n^2))^2 = (k(m^2 + n^2))^2. Since 277 is prime, k must be 1 in this problem.

CDE -Area also ist 0.5*8*8 and CBE 10* sqrt 128*0.5 and finished. We dont Need to think outside the box!

2√3

Another way to compute the area of ΔABC once length AC is known to be r: 4 triangles congruent to ΔABC may be assembled into a square with sides of length r, therefore area r². So, the area of ΔABC is r²/4, as found at 8:10, but by a different method. Proceed to substitute r = (2 + 8/π) to find the area, as done in the video.