Very Nice Geometry Problem | You should be able to solve this | 2 Different Methods

[@Irtsak]

@Math Boo. I liked too much the second solution !!!!! Another solution with areas : Let α= side of square ABCD. Draw the straight line segment EF and EG is the height of triangle DEF. The triangles ADE=DEG , so EG=AE=5. Area (ADE) + area (DEF) + area (BEF) + area (DFC) = area (ABCD) => ½ •5α +½ • 5x + ½ •(α-5)(α-3) +1/2• 3α = α² ………. 5x+15= α² => 5x+15= x²-9 => x²-5x-24=0 cause α²=x²-9 ……. x=8 cause x>0

[@ashugarg5475]

Third method. Join EF. COS theta = (x*2+DE*2-EF*2)/2X.DE EF*2=(a-3)*2+(a-5)*2 DE*2=5*2+a*2 X*2=3*2+a*2 Solving above equations gives cos theta =8a/X.DE From triangle DAE, cos theta=a/DE equating both cos theta 8a/X.DE=a/DE 8/X=1 X=8

[@juanalfaro7522]

Let theta = "a" just for convenience. Then tan (a) = 5/L and tan (90-2a) = 3/L --> tan (2a) = L/3. Now tan (2a) = 2tan (a) / [1-(tan^2 (a))] -> L/3 = 2*5/L / [1-((5^2)/L^2)] = 10L / (L^2 - 25) -> L^2-25 = 3*10 = 30 -> L^2 = 55 -> X^2 = L^2 + 3^2 = 55+9 = 64 --> L=8 units

[@Irtsak]

My third solution in a few words . Extend the line BA (to the left) and make segment GA=3 Obviously ∆ ADG = ∆ DFC => DG=DF=x < GDA = < FDC =90-2θ < GDE = x=8

[@jimlocke9320]

At about 3:10,

[@MarieAnne.]

I used the first method, but solved for a without ever solving for tan θ In △ADE, ∠DAE = 90°, ∠ADE = θ → tan θ = AE/AD = 5/a In △CDF, ∠DCF = 90°, ∠CDF = 90−2θ → ∠DFC = 2θ → tan 2θ = CD/CF = a/3 Using Double Angle Identity, we get: tan 2θ = 2 tan θ / (1 − tan²θ) a/3 = (10/a) / (1 − 25/a²) a/3 = 10a / (a² − 25) 1/3 = 10 / (a² − 25) a² − 25 = 30 a² = 55 Using Pythagorean Theorem in △CDF, we get: DF2 = CD2 + CF2 x² = a² + 3² x² = 55 + 9 = 64 x = 8

[@holyshit922]

If we assume that this is a square , not rectangle we can solve it in that way tan(theta) = 5/y tan(pi/2-2theta) = 3/y tan(theta) = 5/y tan(2theta) = y/3 2*(5/y)/(1-25/y^2) = y/3 (10/y)/((y^2-25)/y^2) = y/3 10y/(y^2-25) = y/3 10y/(y^2-25) - y/3 = 0 y(10/(y^2-25) - 1/3)=0 y(10-1/3*(y^2-25)) = 0 30-y^2+25 = 0 y^2 = 55 From Pythagorean theorem y^2+3^2=x^2 55+9 = x^2 x^2 = 64 x = 8

[@ai2657]

X=8 i found by finding the side of square i produced DF to AB and used angle bister theorem i found the side =√55 So x=8

[@michaelweiminchang3431]

Master ,here is my solution Super easy We can use cosine theorem in triangle ABE and EBF to solve this problem.

[@bablubhattacharjee2736]

equal angle opposite side is also equal. let side of the square is y. so, (y-5)^2+(y-3)^2=25 y= 7.39 so x = 7.98

[@manojkantsamal4945]

^=read as square Finally from the given facts we'll get a equation X^2-10x+16=0 So X=8 or 2

[@murdock5537]

φ = 30°; AB = 5 + (5 - a) = BC = 3 + (a - 3) = CD = AD = a ADE = EDF = θ; FDC = δ → 2θ + δ = 3φ → tan⁡(δ) = cot⁡(2θ) = 3/a → cot⁡(δ) = tan⁡(2θ) = a/3 tan⁡(θ) = 5/a → tan⁡(2θ) = 2tan⁡(θ)/(1 - tan^2(θ)) = 10a/(a^2 - 55) = a/3 → a^2 = 55 → x = √(55 + 9) = 8 → sin⁡(δ) = 3/8 → EF = 4√(9 - √55)

[@jan-willemreens9010]

... Good day, [ Let THETA = T ] ... (1) Triangle(ADE) .... TAN(T) = 5 / I AD I .... (2) Triangle(DCF) .... TAN(2T) = I DC I / 3 = I AD I / 3 ... TAN(2T) = 2 * TAN(T) / (1 - TAN^2(T)) .... so, I AD I / 3 = ( 2 * (5/I AD I) / ( 1 - [ 5/I AD I ]^2 ) ... after a few basic algebraic steps solving for I AD I or of course I DC I we obtain .... I AD I = I DC I = SQRT(55) .... finally computing the X- value in Triangle(DCF) by applying Pythagoras .... X^2 = 3^2 + (SQRT(55))^2 .... X = 8 ... thank you for your clear explanations sir ... best regards, Jan-W

[@ulieggermann4346]

The second method is unbelievable!

[@tgx3529]

let's construct a right-angled triangle ABC with angle theta at vertex C, opposite perpendicular BC of length 5. Let's construct the point K on the perpendicular BC so that |BK|=2. Triangle ABK has at vertex B angle 90-θ, opposite side x, at vertex A angle 3θ-90, opposite side 2, applying Son's theorem we get the requirement x=(2cosθ)/(-3cosθ) Let's apply the function sin(90-2θ)=3/x to triangle AKC, i.e. x=3/cos2θ We get a request for the corresponding x x=2cosθ/(-cos3θ)=3/cos2θ This will give the request cosθ+cos3θ=-3cos3θ so cosθ+4cos3θ=0 x=(2(cosθ+4cos3θ)-8cos3θ)/(-cos3θ)=(0-8cos3θ)/(-cos3θ)=8

[@User-jr7vf]

I did the first method, but the second method is so much more interesting. I wish I had the math mind to think outside the box

[@Irtsak]

Let Dxy an orthogonal axis system . D(0,0) , A(0,α) ,Ε(5,α) , F(α,3). ( α= side of square ABCD). The vectors DA=(0,α) , DE=(5,α) , DF=(α,3) . Norm |DA|=α , |DF|=√( α² +9), The scalar product of vectors : DA•DE = |DA|•|DE|•cosθ DΕ•DF= |DE|•|DF|•cosθ Deviding by term ; (DA•DE) / (DΕ•DF) = |DA|•/ |DF| => (0•5+ α² )/ (5•α +3•α) =α / √ ( α²+9) We can find easily √( α²+9)=8 , so x=|DF|=√ (a² +9)=8

[@幕天席地-w9c]

Alternatively, let side length of the square be a. Extend DE and CB, intersecting at G. DF=FG. x=FG=FB+BG=(a-3)+a(a-5)/5=(a^2-15)/5. x^2=a^2+3^2. a=sqrt(55), x=8

[@Irtsak]

Very nice solution !!!!!!! Fix the equality FG=FB+FG in FG=FB + BG and all are fine .

[@幕天席地-w9c]

@@Irtsak I have fixed the typo. Thanks!