A = 80cm² = ½b.h = ½.2h.h = h² h = √80 = 4.√5 cm b = 2h = 8√5 cm tan α = ½x / x = 1/2 α = 26,565° x = b cos α x = 16 cm ( Solved √ )

BQ=a Triangles BPQ and PAD are similar ( angle BPQ= angle PDA). therfore, AP/BQ=AD/PB or X/2/a= X//X/2 or a=1/4X. Express all triangles areas in terms of X and make the sum=X^2. you will get an equation 16a^2+4a^2 -320=0. a=4. X=4a=16. ( no Phythagoras).

x = 2L = side of the square. [Triangle] = a*b/2 where a=L*sqrt (5) and b = L*sqrt (5) /2 --> L^2 * 5/2/2 = 5 * L^2 / 4 = 80 --> L^2 = 80*4/5 = 64 --> L=8 --> x = 2L = 16

AP = 1/2 x DP= x√5/2 (by pythagoras) APD and PBQ are similar triangles By AA theorem So PQ/PB=DP/AD PQ=x/2*x√5/2x PQ=x√5/4 Area of triangle 80=5x^2/16 x^2=256 x= 16

Here is my version with trigonometry. Angle ADP = angle BPQ =a tan(a)=AP/AD=(x/2)/x=1/2 from which 1+tan(a)^2=sec(a)^2 -->sec(a)=sqrt(5)/2 --cos(a)=2/sqrt(5) DP*cos(a)=x hence DP=x/cos(a) PQ*cos(a)=x/2 hence PQ=x/(2*cos(a)) Area(DPQ)=1/2*DP*PQ=1/2*x^2/(2*cos(a)^2)=1/2*x^2*5/8=80 X^2=256 hence x=16

BQ=a..(√5x/2)^2+((x/2)^2+a^2)=x^2+(x-a)^2...a=x/4...(√5x/2)*√(5x^2/16)=160..(√5x/2)*(√5x/4)=160...5x^2=8*160..x^2=256..x=16

As ABCD is a square and AB = BC = CD = DA = x, AP = PB = x/2. Let ∠ADP = α, and ∠DPA = β, as ∠PAD = 90°, α and β are complementary angles that sum to 90°. As ∠QPD = 90°, ∠BPQ = 180°-(90°+ β) = 90°- β = α, so ∠PQB = β. ∆PAD and ∆QBP are therefore similar triangles by AAA. Triangle ∆PAD: PA² + AD² = DP² (x/2)² + x² = DP² DP² = x²/4 + x² = 5x²/4 DP = √(5x²/4) = √5x/2 Triangle ∆QBP PQ/BP = DP/AD PQ/(x/2) = (√5x/2)/x PQ = (x/2)(√5x/2)/x = √5x/4 Triangle ∆QPD: Aₜ = bh/2 = PD(QP)/2 80 = (√5x/2)(√5x/4)/2 80 = 5x²/16 x² = 80(16/5) = 16(16) x = 16 units

⊿DAP∞⊿PBQ AP=BP=x/2 BQ=x/4 QC=3x/4 DC=x ⊿DAP=x*x/2*1/2=x²/4 ⊿PBQ=x/2*x/4*1/2=x²/16 ⊿QCD=3x/4*x*1/2=3x²/8 ⊿DPQ=x²-(x²/4+x²/16+3x²/8)=5x²/16=80 5x²=1280 x²=256 x>0 , x=16

PQ can be obtained as under..... angle BQP will be beta(B) PQ/PD = PB/AD Putting the values we will get PQ =√5.X/4

AD=X→ AP=X/2=PB→ PD²=X²+(X/2)²=5X²/4→ PD=X√5/2 → Razón de semejanza entre QBP y PAD: s=PB/AD=1/2→ PQ=X√5/4→ PD*PQ=2*80→ X=16. Gracias y un saludo cordial.

My Third elementary solution 180° rotation PBQ in P.....BQ=X/4=a Area BPQ a*2a/2 + Area DAP 2a*4a/2 = Area DPQ= 80 a^2+4a^2 = 5a^2 =80 a^2=16 a=4 X=4*a=4*4=16

360°ABCD/80x =40.40 2^20.2^20 1^5^4.1^5^4 1^2^2.1^2^2 1^1.1^2 1^2 (ABCDx ➖ 2ABCDx+1)

φ = 30° → sin⁡(3φ) = 1; ∎ABCD → AB = AP + BP = a + a = x = BC = BQ + CQ = CD = AD = ? sin⁡(QPD) = 1 → (1/2)(QP)(DP) = 80; ADP = PDQ = BPQ = δ → sin⁡(δ) = √5/5 → cos⁡(δ) = 2√5/5 → tan⁡(δ) = sin⁡(δ)/cos⁡(δ) = 1/2 → PQ = a√5/2 → DP = a√5 → (1/2)(DP)(PQ) = 80 → a = 8 → 2a = x = 16

Here is an approach without using Pythagoras. Let _∠APD = θ_ In _ΔAPD: tanθ = 2_ ⇒ _tan(90° - θ) = ½_ In _ΔBPQ:_ _|BQ|/|BP| = |tan(∠BPQ)_ ⇒ _|BQ|/(½x) = tan(90° - θ) = ½_ ⇒ _|BQ| = (¼)x_ Area of _ΔAPD_ + Area of _ΔBPQ_ + Area of _ΔCDQ_ = Area of _□ABCD_ - Area of _ΔDPQ_ ⇒ _(¼)x² + (⅟₁₆)x² + (⅜)x² = x² - 80_ ⇒ _x² = 16²_ ⇒ *_x = 16_*

Pythagoras' theorem is not needed to solve this problem. ADP and BPQ are similar triangles so the angles ADP and BPQ are the same and you can calculate line BQ (which is x/4) Now you can calculate line CQ; Line CQ=x-x/4=3x/4 The square of x is equal to the four triangles inside it You can write this in a formula and calculate x ; x^2=1/2*((3x/4*x)+(x*x/2)+(x/2*x/4))+80 x^2-80=1/2*((3x^2/4)+(x^2/2)+(x^2/8)) 2*(x^2-80)=(6x^2/8)+(4x^2/8)+(x^2/8) 2x^2-160=(6x^2+4x^2+x^2)/8 2x^2-160=11x^2/8 8*(2x^2-160)=11x^2 16x^2-1280=11x^2 5x^2-1280=0 5x^2=1280 x^2=256 x=16

x=16

This explanation honestly *better* than all of the hearted comments. I hope that this means I can do this as a practice easily!!!

You could have found P by using similarity of triangles instead of doing the Pythagoras calculation again...

You didn't have to use the pythgorean thorem twice. Triangle PBQ is similar to APD

I did not read that abcd was a square in the original question. My bad

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X=16

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X=16 cm

But 16√5 also correct