A Very Nice Geometry Problem | 2 Different Methods

[@ناصريناصر-س4ب]

Let's assume that the length of the side of the square is x so tan(30)=x/(6-x)=1/√3, so x=3√3-3, and the area of the quadrilateral ABDE is (3√3-3+6)*((3√3-3)/2=9

[@prossvay8744]

Area of ABED1=1/2(3√3-3+6)(3√3-3)=9

[@RAG981]

Your second method seemed wild, but it is brilliant.

[@kateknowles8055]

Tan 60° = 3^½ = the square root of three CE = BC × tan 60º = BC × 3^½ DE = 6 = BC + CE = (1+3^½)BC so BC = 6/(1 + 3^½) = 6 (3^½-1)/(3^½-1)(3^½+1) = (3^½-1)×6/(3-1) =3×(3^½-1) ½×(DE+AB)×BC = area of this trapezium. This is ½×3× ( 3^½ +1)×(3 ×( 3^½ -1))= ½×3×3 ×(3-1) = 3×3 =9 ABED has an area of 9 square units. Thank you for the puzzle and your solutions.

[@PrithwirajSen-nj6qq]

Rt 🔺 BEC is a 30-60-90 triangle. BC is opposite to 30 degree angle. Hence if BC is x then angle opposite to 60 degree angle will be x √3 The sides of the square is x Now x + x √3 =6 > x =6/(1+√3)=2.196 (approx) Area of Trapezium = 1/2*(6+2.196)*2.196 =8.196/2 *2.196 = 4.098 *2.196= 8.9992 sq units (approx)

[@raghvendrasingh1289]

Let side of square = a then a+a√3 = 6 a = 3(√3-1) = 3√3-3 Area of trapezium = (6+3√3-3)(3√3-3)/2 = 9

[@marcelowanderleycorreia8876]

The second method is something... Great!

[@MdShahriarHossain-l1n]

The first method was easy but the second method was amazing! I love this problem!

[@michaeldoerr5810]

I really agree. I thought that the second method was just more of a proof of the Pythagorean Theorem. And the answer is x =9 units.

[@quigonkenny]

Let x be the side length of square ABCD. As ∠CFB = 30° and ∠BCF = 90°, ∆BCF is a special 30-60-90 right triangle and as BC = x, CE = √3x. √3x + x = 6 (√3+1)x = 6 x = 6/(√3+1) x = 6(√3-1)/(√3+1)(√3-1) x = 6(√3-1)/(3-1) = 6(√3-1)/2 x = 3(√3-1) Trapezoid ABED: A = h(a+b)/2 A = 3(√3-1)(3(√3-1)+6)/2 A = (3√3-3)(3√3-3+6)/2 A = (3√3-3)(3√3+3)/2 A = (27-9)/2 = 18/2 = 9 sq units

[@saronohandoyo4958]

i use your 1st method for solution, but your 2nd method is cooler n mind blowing

[@marioalb9726]

tan 30° = s / (6-s) s = (6-s) tan30 = 6/√3 - s/√3 s+s/√3 = 6/√3 s (1+1/√3) = 6/√3 s = 6/[√3(1+1/√3)] = 6/(√3+1) s = 2,196 cm A = s² + ½ s (6-s) = ½s² + 3s A = 9 cm² ( Solved √ )

[@Emerson_Brasil]

*Solution:* Let AB=BC=DC=AD=x. [ABED] = [ABD] + [DBE] *[ABED] = x.x/2 + 6.x/2 = x²/2 + 3x* Now, BE = BC/sin 30°= 2x and, Furthermore, BD=x√2. By the law of cosines ∆BED: BD² = BE² + DE² - BE× DE×cos 30° (x√2)² = 6² + (2x)² -2.6.2x.√3/2 2x² = 36 + 4x² - 12√3 x 2x² - 12√3 x + 36 = 0 (÷2) x² - 6√3 x + 18 = 0 ∆ = (-6√3)² - 4 × 18 = 108 - 72 = 36 x = (6√3 ± √∆)/2 = (6√3 ± 6)/2 x = 3√3 ± 3, as BE < DE, i.e., x < 3. Therefore, *x= 3√3 - 3.* Like this, [ABED] = x²/2 + 3x [ABED] = (3√3 - 3)²/2 + 3(3√3 - 3) [ABED] = 9(√3 - 1)²/2 + 9(√3 - 1) [ABED] = 9(4 - 2√3)/2 + 9(√3 - 1) [ABED] = 9(2 - √3)+ 9(√3 - 1) [ABED] = 9(2 - √3 + √3 - 1) *[ABED] = 9 square units*

[@wasimahmad-t6c]

4×2÷2+2×2=8

[@AmirgabYT2185]

S=9

[@yakupbuyankara5903]

9

[@ZiaTiger-j9r]

Mistake.3 into under root 3 minus 1 😂you can check

[@MathBooster]

Under root 3 minus 1 is in bracket so if you multiply it with 3 then it will become 3 times under root 3 minus 3

[@MegaSuperEnrique]

At 3:30, he is rationalizing the denominator, so he multiplied top and bottom so that the denominator wouldn't have a radical (root) in it.

[@adamhanna9940]

Your solution is unnecessary, long excessively, you just need to elongate the time of your presentation. All what you should do: is to put the dimensions of the right angle triangle as : x (opposite to 30), square of 3. X to the adjacent and the hypotenuse = 2x. Therefore : x +square root of (3).x =6 ..(1) That gives: x^2= 18(2-square root(3)) A= x^2 +(square root(3)/2.(x^2) =9 square units