Draw a segment thru point C and a point E on base AB such that it is perpendicular to the bases of right trapezoid ABCD. This is the height CE of the trapezoid. This segment forms a square ADCE and a right △BEC (We know ADCE is a square because AD = CD). Label the side length of the square as s. Then, BE = AB - AE = (1 + √3) - s. Use the Pythagorean Theorem on △BEC. a² + b² = c² s² + [(1 + √3) - s]² = 2² s² + [(1 + √3)² - 2s(1 + √3) + s²] = 4 s² + [(1 + 2√3 + 3) - 2s - 2s√3 + s²] = 4 s² + 4 + 2√3 - 2s - 2s√3 + s² = 4 2s² + 2√3 - 2s - 2s√3 = 0 2s² - 2s√3 - 2s + 2√3 = 0 (We can try to group) 2s(s - √3) - 2(s - √3) = 0 (2s - 2)(s - √3) = 0 2s - 2 = 0 or s - √3 = 0 2s = 2 s = √3 s = 1 Both lengths are reasonable. BE = (1 + √3) - 1 = √3 or (1 + √3) - √3 = 1. Find the areas of the square and the triangle. It will be easier to find the area of the trapezoid this way. First, substitute s = 1. A = s² = 1² = 1 A = (bh)/2 = (√3 * 1)/2 = (√3)/2 So, one answer for the area of the trapezoid is 1 + (√3)/2 square units, a. w. a. (2 + √3)/2 square units (exact), or about 1.87 square units (approximation). Next, substitute s = √3. A = (√3)² = 3 A = (bh)/3 = (1 * √3)/2 = (√3)/2 So, another answer for the area of the trapezoid is 3 + (√3)/2 square units, a. w. a. (6 + √3)/2 square units (exact), or about 3.87 square units (approximation). So, the possible areas of the trapezoid are as follows: A = 1 + (√3)/2 u² = (2 + √3)/2 u² ≈ 1.87 u² A = 3 + (√3)/2 u² = (6 + √3)/2 u² ≈ 3.87 u²
Complete 🔺 ABC (√3+1)/sin ACB=2/sin BAC=2/sin45=2√2 >sin ACB =(√3 +1)/2√2=sin 75=sin 105 Then ACB = 75 or 105 Hence BDE =75-45=30 or 105 - 45=60 Hence 🔺 BDE is a 30-60-90 triangle whose angle BDE =30 or 60 When it is 30, BE =1and side of square is √3 When it is 60, BE =√3.side of square is 1
Let's name c = AD = DC. In triangle EBC we have (c^2) + (1 +sqrt(3) - c)^2 = 2^2, then (c^2) -(1 + sqrt(3)).c +sqrt(3) = 0, and the c = 1 or c = sqrt(3). The area of the trapezoïd is A = (1/2).(DC + AB).AD = (1/2).(c^2).(c + 1 + sqrt(3). If c = 1, then A = (2 + sqrt(3))/2; if c = sqrt(3) then A = (6 + sqrt(3))/2.
Sir, may I offer a conceptual solution of the problem keeping ur construction? The side of square will be 1or √3 If side is 1 BE= √3 and based on 🔺 BCE we may say CE =2^2 -√3^2=1 Hence assumption for taking the sides as1 unit is right. Again if the side is √3 then BE will be 1unit and based on 🔺 BCE we see CE =2^2-1=√3 Hence the assumption for taking the side √3 unit is correct. Hence parallel sides of trapezium is 1 and 1+√3(case 1) Or √3 and 1+√3(case 2) For case 1 Area of trapezium is 1/2(1+1+√3) sq units For case 2 Area of trapezium is 1/2(√3+√3+1) sq units
STEP-BY-STEP RESOLUTION PROPOSAL : 01) AD = DC = X 02) BC = 2 03) AB = (1 + sqrt(3)) 04) Drop a Vertical Line from Point C until AB. Find Point E. 05) AE = X and EB = (1 + sqrt(3) - X) 06) BC^2 = CE^2 + EB^2 07) 4 = X^2 + (1 + sqrt(3) - X)^2 08) Two Solutions : X = 1 and X = sqrt(3) 09) Solution #1 : a) AT = [1 + (1 + sqrt(3))] / 2 b) AT = (2 + sqrt(3)) / 2 c) AT = 1 + sqrt(3) / 2 d) AT ~ 1,87 10) Solution #2 : a) AT = [sqrt(3) + (1 + sqrt(3))] * sqrt(3) / 2 b) AT = (1 + 2sqrt(3)) * sqrt(3) / 2 c) AT = (sqrt(3) + (2 * 3)) / 2 d) AT = 3 + sqrt(3)/2 e) AT ~ 3,87 Therefore, OUR BEST ANSWER : Trapezoid Area equal to approx. 1,87 Square Units or approx. equal to 3,87 Square Units.
1)Join AC Ang BAC = 45 degrees 2) Triangle BAC sin (Ang ACB) /(1+√3) = [sin (ang BAC)]/2 > sin (ang ACB) /(1+√3) =(sin 45)/2=1/2√2 Then sin(ACB)= (1+√3)/2√2=sin75 or sin 105 Hence ACB =75 or 105 degrees Then ang ABC is 180-45-105=30 Or 180-45-75=60 3) Drop perpendicular CE on AB Now triangle is rt triangle whose hypotenuse is 2 units Hence side opposite to 30 degree will be 1 unit and that of 60 degree will be √3 unit Hence CE =1 or √3 BE =√3 or 1 4) Hence the parallel sides of trapezium √3 and 1+√3 Or 1 and 1+√3 Area of trapezium =1/2(1+1+√3) Or 1/2(√3+√3+1) sq units
Method 2 Previous method part 1 Add AC =x √2 Previous method part 2 keep as it is The part 3 now be replaced by the bellow AC /sin ABC =2/sin45=2√2 >x √2/sin30=2√2 x = 1 Or AC /sin ABC =2/sin45 > x √2/sin60=2√2 x =√3 The 4th will be as it is
Solution: E = point exactly below C. AD = DC = CE = AE = x Pythagoras for the right-angled triangle EBC: (1+√3-x)²+x² = 2² ⟹ (1+√3)²-2*(1+√3)*x+2x² = 4 ⟹ 2x²-2*(1+√3)*x+4+2*√3 = 4 |/2 ⟹ x²-(1+√3)*x+2+√3 = 2 |-2 ⟹ x²-(1+√3)*x+√3 = 0 |p-q formula ⟹ x1/2 = (1+√3)/2±√[(1+√3)²/4-√3] = (1+√3)/2±√[(1+2*√3+3)/4-√3] = (1+√3)/2±√[1+√3/2-√3] = (1+√3)/2±√(1-√3/2) ⟹ x1 = (1+√3)/2+√(1-√3/2) = 1.7320508075688773 and x2 = (1+√3)/2-√(1-√3/2) = 1 ⟹ Checking the result for x1 = (1+√3)/2+√(1-√3/2) = 1.7320508075688773: left Side: (1+√3-1.7320508075688773)²+1.7320508075688773² = 1²+1.7320508075688773² = 4 right side: 2² = 4 everything OK Area of the trapezoid = 1²+1*√3/2 = 1+√3/2 ≈ 1.8660 Checking the result for x2 = 1: left side: (1+√3-1)²+1² = 3+1 = 4 right side: 2² = 4 everything OK Area of the trapezoid = (√3)²+1*√3/2 = 3+√3/2 ≈ 3.8660
I set up a triangle ABC. Angle at A is 45 degrees, Line AB is 2.73, Line CB is 2. Use the sine rule. Angle C becomes 74.84. Angle B becomes 60.16 degrees. Height of the triangle CE is 1.73. Area of the set up triangle ABC is thus 2.368. Add area of remaining figure, triangle ADC, which is 1.496. Total area of figure is 3.864. More or less equal to the narrator's suggested solution.
Love the algebra reinforcement! It's got a lot of rhythm and soul...😊
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Nailed it.
Draw a segment thru point C and a point E on base AB such that it is perpendicular to the bases of right trapezoid ABCD. This is the height CE of the trapezoid.
This segment forms a square ADCE and a right △BEC (We know ADCE is a square because AD = CD).
Label the side length of the square as s.
Then, BE = AB - AE = (1 + √3) - s.
Use the Pythagorean Theorem on △BEC.
a² + b² = c²
s² + [(1 + √3) - s]² = 2²
s² + [(1 + √3)² - 2s(1 + √3) + s²] = 4
s² + [(1 + 2√3 + 3) - 2s - 2s√3 + s²] = 4
s² + 4 + 2√3 - 2s - 2s√3 + s² = 4
2s² + 2√3 - 2s - 2s√3 = 0
2s² - 2s√3 - 2s + 2√3 = 0 (We can try to group)
2s(s - √3) - 2(s - √3) = 0
(2s - 2)(s - √3) = 0
2s - 2 = 0 or s - √3 = 0
2s = 2 s = √3
s = 1
Both lengths are reasonable. BE = (1 + √3) - 1 = √3 or (1 + √3) - √3 = 1.
Find the areas of the square and the triangle. It will be easier to find the area of the trapezoid this way.
First, substitute s = 1.
A = s²
= 1²
= 1
A = (bh)/2
= (√3 * 1)/2
= (√3)/2
So, one answer for the area of the trapezoid is 1 + (√3)/2 square units, a. w. a. (2 + √3)/2 square units (exact), or about 1.87 square units (approximation).
Next, substitute s = √3.
A = (√3)²
= 3
A = (bh)/3
= (1 * √3)/2
= (√3)/2
So, another answer for the area of the trapezoid is 3 + (√3)/2 square units, a. w. a. (6 + √3)/2 square units (exact), or about 3.87 square units (approximation).
So, the possible areas of the trapezoid are as follows:
A = 1 + (√3)/2 u² = (2 + √3)/2 u² ≈ 1.87 u²
A = 3 + (√3)/2 u² = (6 + √3)/2 u² ≈ 3.87 u²
I did the standard trapezoid area formula and it worked.
Also, I did factoring by grouping to solve for x
After you solved for x, you could just use the trapezoid area formula in each case: A = (x^2 + x + x*sqrt3)/2
Thank you!
Let's find the area:
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First of all we add point E on AB such that BCE is a right triangle. Since AD=CD, in this case ADCE is a square. Now we apply the Pythagorean theorem to the right triangle BCE:
BC² = BE² + CE² = (AB − AE)² + CE² = (AB − CE)² + CE²
2² = [(√3 + 1) − CE]² + CE²
4 = (√3 + 1)² − 2(√3 + 1)*CE + CE² + CE²
4 = 3 + 2√3 + 1 − 2(√3 + 1)*CE + 2*CE²
0 = 2*CE² − 2(√3 + 1)*CE + 2√3
0 = CE² − (√3 + 1)*CE + √3
⇒ Δ = [(√3 + 1)/2]² − √3 = (3 + 2√3 + 1)/4 − 4√3/4 = = (3 − 2√3 + 1)/4 = [(√3 − 1)/2]²
⇒ CE = (√3 + 1)/2 ± √Δ = (√3 + 1)/2 ± (√3 − 1)/2
First solution:
CE = (√3 + 1)/2 + (√3 − 1)/2 = (√3 + 1 + √3 − 1)/2 = √3
A(ABCD) = (1/2)*(AB + CD)*AD = (1/2)*(AB + CE)*CE = (1/2)*(√3 + 1 + √3)*√3 = (√3 + 6)/2
Second solution:
CE = (√3 + 1)/2 − (√3 − 1)/2 = (√3 + 1 − √3 + 1)/2 = 1
A(ABCD) = (1/2)*(AB + CD)*AD = (1/2)*(AB + CE)*CE = (1/2)*(√3 + 1 + 1)*1 = (√3 + 2)/2
Best regards from Germany
ABCº=60º---> AD=2√3/2=√3---> AB=√3+(2/2)---> ABCD =(√3)²+(1*√3/2) =3+(√3/2)
Gracias y saludos
Thanks so much sir ❤❤
You're very welcome! ❤️
Bom dia Mestre
Complete 🔺 ABC
(√3+1)/sin ACB=2/sin BAC=2/sin45=2√2
>sin ACB =(√3 +1)/2√2=sin 75=sin 105
Then ACB = 75 or 105
Hence BDE =75-45=30 or
105 - 45=60
Hence 🔺 BDE is a 30-60-90 triangle whose angle BDE =30 or 60
When it is 30, BE =1and side of square is √3
When it is 60, BE =√3.side of square is 1
Let's name c = AD = DC. In triangle EBC we have (c^2) + (1 +sqrt(3) - c)^2 = 2^2, then (c^2) -(1 + sqrt(3)).c +sqrt(3) = 0, and the c = 1 or c = sqrt(3).
The area of the trapezoïd is A = (1/2).(DC + AB).AD = (1/2).(c^2).(c + 1 + sqrt(3).
If c = 1, then A = (2 + sqrt(3))/2; if c = sqrt(3) then A = (6 + sqrt(3))/2.
Let AD = DC = x. Drop a perpendicular from C to E on AB. As ∠ADC = ∠DCE = ∠CEA = ∠EAD = 90° and adjacent sides AD and DC equal x, then ADCE is a square with side length x. BE = (1+√3)-x.
Let ∠CBA = θ.
sinθ = EC/CB = x/2
x = 2sinθ
cosθ = BE/CB = ((1+√3)-x)/2
2cosθ = (1+√3) - 2sinθ
2sinθ + 2cosθ = 1 + √3
2(sinθ+cosθ) = 1 + √3
4(sin²θ+2sinθcosθ+cos²θ) = 1 + 2√3 + 3
4(1+sin2θ) = 4 + 2√3
4 + 4sin2θ = 4 + 2√3
4sin2θ = 2√3
sin2θ = (2√3)/4 = √3/2
2θ = 60° | 2θ = 120° --- 0° < θ < 90°
θ = 30° | θ = 60°
Trapezoid ABCD:
A₁ = x(x+1+√3)/2
A₁ = 2sin30°((2sin30°)+1+√3)/2
A₁ = 2(1/2)(2(1/2)+1+√3)/2
A₁ = (1+1+√3)/2 = (2+√3)/2
[ A₁ = 1 + √3/2 ≈ 1.866 sq units ]
A₂ = x(x+1+√3)/2
A₂ = 2sin60°((2sin60°)+1+√3)/2
A₂ = 2(√3/2)(2(√3/2)+1+√3)/2
A₂ = (√3)(1+2√3)/2 = (6+√3)/2
[ A₂ = 3 + √3/2 ≈ 3.866 sq units ]
(1+sqrt(3)-s)^2=4-s^2, 2-s^2=2+sqrt(3)-(1+sqrt(3))s, s^2-(1+sqrt(3))s+2=0,........😅
(2)^2=4 {1x+1x ➖}+{3x+3x ➖ }={4+2x^2+6x^2}=12x^4 {90°A+90°B+45°C+45°D}=180°ABCD/12x^4=15x^4ABCD 3^5x^4 3^2^3x^2^2 1^1^3x^1^2 3x^2 (ABCDx ➖ 3ABCDx+2).
1.866
Sir, may I offer a conceptual solution of the problem keeping ur construction?
The side of square will be 1or √3
If side is 1
BE= √3 and based on 🔺 BCE we may say CE =2^2 -√3^2=1
Hence assumption for taking the sides as1 unit is right.
Again if the side is √3 then BE will be 1unit and based on 🔺 BCE we see CE =2^2-1=√3
Hence the assumption for taking the side √3 unit is correct.
Hence parallel sides of trapezium is
1 and 1+√3(case 1)
Or √3 and 1+√3(case 2)
For case 1
Area of trapezium is 1/2(1+1+√3) sq units
For case 2
Area of trapezium is
1/2(√3+√3+1) sq units
1+1.732+1=3.732÷2×1=1.866
STEP-BY-STEP RESOLUTION PROPOSAL :
01) AD = DC = X
02) BC = 2
03) AB = (1 + sqrt(3))
04) Drop a Vertical Line from Point C until AB. Find Point E.
05) AE = X and EB = (1 + sqrt(3) - X)
06) BC^2 = CE^2 + EB^2
07) 4 = X^2 + (1 + sqrt(3) - X)^2
08) Two Solutions : X = 1 and X = sqrt(3)
09) Solution #1 :
a) AT = [1 + (1 + sqrt(3))] / 2
b) AT = (2 + sqrt(3)) / 2
c) AT = 1 + sqrt(3) / 2
d) AT ~ 1,87
10) Solution #2 :
a) AT = [sqrt(3) + (1 + sqrt(3))] * sqrt(3) / 2
b) AT = (1 + 2sqrt(3)) * sqrt(3) / 2
c) AT = (sqrt(3) + (2 * 3)) / 2
d) AT = 3 + sqrt(3)/2
e) AT ~ 3,87
Therefore,
OUR BEST ANSWER :
Trapezoid Area equal to approx. 1,87 Square Units or approx. equal to 3,87 Square Units.
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Thanks for sharing ❤️
X^2 + (1 + √ 3 - x)^2 = 2^2 = 4
2X^2 - 2(1 + √ 3)x + (1 + √ 3)^2 = 4
2X^2 - 2(1 + √ 3)x + 2√ 3= 0
X^2 - (1 + √ 3)x + √ 3= 0
X = 1 or √ 3
area = (1 + √ 3 + 1)/2 or (1 + √ 3 + √ 3)(√ 3)/2 = 1 + (√ 3)/2 or 3 + (√ 3)/2
My way of solution ▶
[AE]= [EC]= [CD]= [DA] = x
E ∈ [AB] and [AE] ⊥ [EC]
⇒
[AE]= x
[EB]= 1+ √3 -x
By considering the right triangle ΔEBC :
[CE]= x
[EB]= 1+ √3 -x
[BC]= 2
According to the Pythagorean theorem, we can write:
x² + (1+ √3 -x)²= 2²
x² + 1+ √3 - x + √3 + 3- x√3 - x - x√3 + x²= 4
2x²+4+2√3 - 2x -2√3 x = 4
boh sides divided by 2
x²-x-√3x +2+√3 = 2
x(x-1) - √3(x-1)= 0
(x-1)[x-√3]= 0
⇒
x₁= 1 ✅
x₂= √3 ✅
Both solutions are possible !
For x= 1
A(AECD)= x²
A(AECD)= 1
A(ΔEBC)= (1+ √3 -1)*1/2
A(ΔEBC)= √3/2 square units
A(ABCD)= A(AECD) + A(ΔEBC)
A(ABCD)= 1+ √3/2
A(ABCD) ≈ 1,866 square units ✅
For x= √3
A(AECD)= (√3)²
A(AECD)= 3
A(ΔEBC)= (1+ √3 -√3)*√3/2
A(ΔEBC)= √3/2 square units
A(ABCD)= A(AECD) + A(ΔEBC)
A(ABCD)= 3+ √3/2
A(ABCD) ≈ 3,866 square units ✅
Therefore we have two solutions !
1)Join AC
Ang BAC = 45 degrees
2)
Triangle BAC
sin (Ang ACB) /(1+√3)
= [sin (ang BAC)]/2
> sin (ang ACB) /(1+√3)
=(sin 45)/2=1/2√2
Then sin(ACB)= (1+√3)/2√2=sin75 or sin 105
Hence ACB =75 or 105 degrees
Then ang ABC is 180-45-105=30
Or 180-45-75=60
3)
Drop perpendicular CE on AB
Now triangle is rt triangle whose hypotenuse is 2 units
Hence side opposite to 30 degree will be 1 unit and that of 60 degree will be √3 unit
Hence CE =1 or √3
BE =√3 or 1
4)
Hence the parallel sides of trapezium
√3 and 1+√3
Or
1 and 1+√3
Area of trapezium
=1/2(1+1+√3)
Or 1/2(√3+√3+1) sq units
Method 2
Previous method part 1
Add AC =x √2
Previous method part 2 keep as it is
The part 3 now be replaced by the bellow
AC /sin ABC =2/sin45=2√2
>x √2/sin30=2√2
x = 1
Or AC /sin ABC =2/sin45
> x √2/sin60=2√2
x =√3
The 4th will be as it is
Solution:
E = point exactly below C.
AD = DC = CE = AE = x
Pythagoras for the right-angled triangle EBC:
(1+√3-x)²+x² = 2² ⟹
(1+√3)²-2*(1+√3)*x+2x² = 4 ⟹
2x²-2*(1+√3)*x+4+2*√3 = 4 |/2 ⟹
x²-(1+√3)*x+2+√3 = 2 |-2 ⟹
x²-(1+√3)*x+√3 = 0 |p-q formula ⟹
x1/2 = (1+√3)/2±√[(1+√3)²/4-√3] = (1+√3)/2±√[(1+2*√3+3)/4-√3]
= (1+√3)/2±√[1+√3/2-√3] = (1+√3)/2±√(1-√3/2) ⟹
x1 = (1+√3)/2+√(1-√3/2) = 1.7320508075688773 and
x2 = (1+√3)/2-√(1-√3/2) = 1 ⟹
Checking the result for x1 = (1+√3)/2+√(1-√3/2) = 1.7320508075688773:
left Side:
(1+√3-1.7320508075688773)²+1.7320508075688773² = 1²+1.7320508075688773² = 4
right side: 2² = 4 everything OK
Area of the trapezoid = 1²+1*√3/2 = 1+√3/2 ≈ 1.8660
Checking the result for x2 = 1:
left side:
(1+√3-1)²+1² = 3+1 = 4
right side: 2² = 4 everything OK
Area of the trapezoid = (√3)²+1*√3/2 = 3+√3/2 ≈ 3.8660
I'm second
I’m here to be first.
I set up a triangle ABC. Angle at A is 45 degrees, Line AB is 2.73, Line CB is 2. Use the sine rule. Angle C becomes 74.84. Angle B becomes 60.16 degrees. Height of the triangle CE is 1.73. Area of the set up triangle ABC is thus 2.368. Add area of remaining figure, triangle ADC, which is 1.496. Total area of figure is 3.864. More or less equal to the narrator's suggested solution.