Thanks. You are carrying me back in the early 60's in my maths lessons under my dear teacher Mr.Khan. I am now 79 years but for the love of maths I am still following the subject in the U_Tube. And you teach just like my former teacher.

easy to show: {y=1, 1+y=2, x-y = 2} ==> x=3

Love it!!!!!!!!!!!

Very good problem. The most important clue is that the hypotenuse is divided into 2 equal parts. Value of x is 3 units.

Nice example! This goes without trig, my reasoning goes as follows: The midpoint E on the hypotenuse AC is the center of circumference of △ABC, since in right triangles the center of circumference always lies on half hypotenuse (Thales). Therefore, BE = AE = CE (= radius) and △ABE is isoceles. Draw from E perpendicular to BC => F, then EF = 1. Since △BEF is congruent to ABD, BF = AB = 2, and 2·2 = x + 1, hence x = 3.

Posto β=DEA,a=AE... risulta,per il teorema dei seni a:sin(135-arctg2)=√5:sinβ..a:sin45=x:sin(180-β)...divido le equazioni..x/√5=sin(135-arctg2)/sin45...x=(√5/sin45)sin(135-arctg2)=3

Решения длинньіе. Оставить проведение перпендикуляра и треугольник с углом 45 и стороной 1. Дальше параллельньіе линии к основанию, делящие его пополам (теорема Фалеса). И в принципе - ответ готов.

E is mid point . EF parallel to AB . So EF will be half of AB I. e. 1 . So DF will be 1. EF will bisect BC at F . So BF= 1+1 = 2 =FC . Now DC = DF+ FC = 1 + 2 = 3

Draw parallel AF to ED. angle AFB=45degree. FB=AB. x-1=2.

By Thales, EF = AB/2 = 2/2 = 1 BF = BC/2 = (x + 1)/2 DF = (x + 1)/2 - 1 = (x - 1)/2 = 1 (x - 1) = 2 *x = 3*

Mid-point theorem is useful. Since EF is parallel to AB, and E is mid point of AC, y=AB/2 =1, and BF =2, BF = FC= x-1, x just equals 3.

If you draw a line from point E parallel to BC that connects to line AB at E', you get a rectangle, and you can see that as E is the midpoint of AC, therefore E' is the midpoint of AB, and that BE' = E'A = 1. But as BE' = FE, therefore FE = 1. As FE = y, therefore y = 1.

The analysis should come first. We need to use the 45 degree angle to help us How can we do this?

First complete a rectangle. In the middle is a parallelogram and next to it two triangles with sides 1 and 2. Now take the big rectangle and remove two smaller rectangles from both ends. (sides 1 and 2) You are left with a rectangle with sides 2 and x-1. Since the diagonal is in 45 degrees, the rectangle is a square. Thus, x - 1 = 2, x = 3

Is AD different from CD?

Please make true diagram

Why don't you just take measurements sqrt(x^2+2x+5)/2 ÷ x = sin45⁰?

Rashidissa Sir . Similar is the case with me . I m 80 .

Draw 2 middle lines and see that x is equal to 3

Draw FE, where F is the point on AB where FE is perpendicular to AB. Let G be the point of intersection between FE and DA. Let CE = EA = y. As ∠AFE = ∠ABC = 90°, FE and BC are both perpendicular to AB and thus ∠FEA = ∠BCA, and ∠EAF and ∠CAB are the same angle, ∆AFE and ∆ABC are similar. Triangle ∆AFE: FE/EA = BC/CA FE/y = (1+x)/2y FE = y(1+x)/2y FE = (1+x)/2 AF/EA = AB/CA AF/y = 2/2y = 1/y AF = y/y = 1 As AF = 1 and AB = 2, FB = 1. Draw HE, where H is the point on BC where HE is perpendicular to BC. As FB = 1, FE and BH are parallel, and FB and HE are both perpendicular to BH, HE = 1. Triangle ∆DHE: tan(45°) = DH/HE 1 = DH/1 DH = 1 As FE and BH are parallel and FB and HE are identical and both perpendicular to BH and FE, then FE = BH. (1+x)/2 = (1+1) = 2 1 + x = 2(2) = 4 x = 4 - 1 = 3

FE et AB parallèles E milieu donc F milieu Y EST MOITIÉ DE AB Puis 1+y=x-y alors x=2

1/3 is not one of the answers?

Quá dài và phức tạp.

EF II AB, AE=EC => EF - ΔABC midsegment => EF=AB/2=1, BF=FC Isosceles ΔDEF: DF=EF=1 BF=FC=BD+DF=2, x=DF+FC=3

In triangle ABD, by pythag, AD = √5 Using the cosine (supplement rule… The cosine of an angle is minus the cosine of it’s supplement, In triangles AED and CED Cos AED = - Cos CED (a^2 + DE^2 -5)/(2*DE*a) = (x^2 - DE^2 -a^2)/(2*DE*a) 2DE^2 +2a^2 = 5 +x^2 triangle DEF, by pythag., DE= √2 4 +2a^2 = 5 +x^2 8 +4a^2 = 10 +2x^2************ In triangle ABC, by python., 4a^2 = 4 +(x+1)^2 combined with above.. 8 +4 +(x+1)^2 = 10 +2x^2 2 + x^2 + 2x +1 = 2x^2 2x +3 = x^2 x^2 -2x -3 = 0 (x -3)(x +1) = 0 x = 3

asnwer=1cm isit

@MathBooster

4 ай бұрын

Answer is X = 3