We note that ΔBCE is a 15°-75°-90° triangle with known ratio of sides, short-long-hypotenuse, of (√3 - 1):(√3 + 1):2√2. Therefore, length BE = length AD = (13)(√3 - 1)/(2√2) and length BC = (13)(√3 + 1)/(2√2). Extend CD and AB until they meet and label the intersection as point F. We note that ΔADF and ΔBCF are both special 30°-60°-90° right triangles and, if the long sides are considered their bases, their heights are 1/√3 times the base and area = (1/2)bh = (1/2√3)b². Length BC squared = ((13)(√3 + 1)/(2√2))² = (169)(3 + 2√3 + 1)/8 = (169)(2 + √3)/4 and length AD squared = ((13)(√3 - 1)/(2√2))² = (169)(3 - 2√3 + 1)/8 = (169)(2 - √3)/4. The area of the trapezoid is area ΔBCF - area ΔADF = (1/2√3)(BC)² - (1/2√3)(AD)² = (1/2√3)((169)(2 + √3)/4) - (169)(2 - √3)/4)) = (1/2√3)((169)(2√3)/4) = 169/4 = 42.25 square units, as PreMath also found.
Note that, starting at about 10:50, PreMath evaluates cos²(15°) - sin²(15°). An alternative method is to use the ratio of sides of the 15°-75°-90° triangle to derive cos(15°) = (√3 + 1)/(2√2) and sin(15°) = (√3 - 1)/(2√2). Then, cos²(15°) = (√3 + 1)(√3 + 1)/(2√2)² = (3 + 2√3 + 1)/8 = (2 + √3)/4 and sin²(15°) = (√3 - 1)(√3 - 1)/(2√2)² = (3 - 2√3 + 1)/8 = (2 - √3)/4. So, cos²(15°) - sin²(15°) = (2 + √3)/4 - (2 - √3)/4 = 2(√3)/4 = (√3)/2, as PreMath also found, but using the cosine angle sum formula (in this case, cosine double angle formula) instead.
At a quick glance, Form a rectangle ABCG where G is the intersection of perpendiculars from A and C. Area of trapezoid = Area of Rectangle - Area of triangle DCG. Angle CEB = 180 - 105 = 75. Angle BCE = 90-75 = 15. Then CB = Cos(15) * 13 = 12.56 and EB = Sin(15) * 13 = 3.365 Then DG = CB- DA = 12.56 - 3.365 = 9.2. Then Angle DCG = 90-15-15 = 60. Then DG/ CG = tan(60) then CG = DG/tan(60) = 9.2/tan(60) =5.3 . Area of rectangle ABCG = 5.3 * 12.56 =66.68 and area of triangle DCG = 0.5 * 9.2 * 5.3 =24.38. Then area of trapezoid = 66.68 - 24.38 = 42.3 area units.
Nothing better. I just add that if needed another time, we have sin(15°) = (sqrt(6) - sqrt(2))/4; cos(15°) = (sqrt(6) + sqrt(2))/4 and tan(15°) = 2 - sqrt(3)
Went another way, but with same result (as it is to be expected as long as i manage to omit slip errors :D ). Just created a helper point F at intersection of prolongation of CD with prolongation of BA, creating a rectangle with 30°/60°, then getting EB, AD, BC the same way as in the video, then getting BF by tan(30)*BC, then getting AF by tan(30)*AD, then getting AB as BF - AF, then the area by AB * average(AD, BC). It's nice to see the different thinking approaches by different people in the chat :D The problem gave a big degree of freedom of "roads to Rome".
Since ∠AEC = 105° then ∠CEB = 75° and since ∠CBE = 90° then ∠BEC = 15°. Since sin 15° = (√6 − √2) / 4 and sin 75° = (√2 + √6) / 4 then BE = AD = 13 (√6 − √2) / 4 and BC = 13 (√2 + √6) / 4. So DA + BC = 13√6 / 2. Since BE = AD then CF = 13 (√6 + √2) / 4 − 13 (√6 − √2) / 4 = 13√2 / 2. Since CDF is 30°-60°-90° then DF = AB = (13√2 / 2) / √3 = 13√6 / 6. Therefore, the area of the trapezoid is ½ (13√6 / 6) (13√6 / 2) = 42.25 square units.
I took 4 hours doing this question because I missed the words “basic trigonometry”. I ended up solving the question by extending AB and CD, using the angle bisector theorem, and proving that DEA is a 30-60-90. Thanks for the amazing question!
Looking back at the question, the area of EBC is equal to AECD as long as EC bisects DCB (the fact becomes obvious if you reflect EBC across EC and connect DE). This means that the trapezoid has the same area as an isosceles triangle with the equal lengths being EC and the angle between them DCB, meaning its area is equal to (EC^2)*sin(DCB)/2.
Angle BEC = 180º - 105º = 75º Angle ECB = 90º - 75º = 15º Angle ADC = 150º EB = AD = x sin (15º) = x / 13 ; x = 13*sin(15º) ~ 3,365 lin un cos (15º) = CB / 13 ; CB = 13*cos(15º) ~ 12,557 lin un Check: 12,557^2 + 3,365^2 = 13^2 ; 157,68 + 11,321 = 169 ; 169 = 169 Major Base = BC = 12,557 lin un Minor Base = AD = 3,365 lin un Finding AB = AE + 3,365 Equation of Straight Line passing points C and D: y = - tan(60º)x + 12,557 and Equation of Straight Line passing points A and D y = 3.365 Intercept of both equations: 3,365 = -tan(60º)*AB + 12,557 ; - 9,192 = -tan(60º)*AB ; AE = 9,192/tan(60º) ; AB = 9,192 / 1,732 ; AB ~ 5,307 AE = 5,307 - 3,365 = 19,942 lin un height = AB = 5,307 lin un Trapezoid Area = (B + b) * (h / 2) = (12,557 + 3,365) * (5,307 / 2) = 15,922 * 2,6535 ~ 42,25 sq un Answer: Trapezoid Area equal to approx. 42,25 Square Units.
We note that ΔBCE is a 15°-75°-90° triangle with known ratio of sides, short-long-hypotenuse, of (√3 - 1):(√3 + 1):2√2. Therefore, length BE = length AD = (13)(√3 - 1)/(2√2) and length BC = (13)(√3 + 1)/(2√2). Extend CD and AB until they meet and label the intersection as point F. We note that ΔADF and ΔBCF are both special 30°-60°-90° right triangles and, if the long sides are considered their bases, their heights are 1/√3 times the base and area = (1/2)bh = (1/2√3)b². Length BC squared = ((13)(√3 + 1)/(2√2))² = (169)(3 + 2√3 + 1)/8 = (169)(2 + √3)/4 and length AD squared = ((13)(√3 - 1)/(2√2))² = (169)(3 - 2√3 + 1)/8 = (169)(2 - √3)/4. The area of the trapezoid is area ΔBCF - area ΔADF = (1/2√3)(BC)² - (1/2√3)(AD)² = (1/2√3)((169)(2 + √3)/4) - (169)(2 - √3)/4)) = (1/2√3)((169)(2√3)/4) = 169/4 = 42.25 square units, as PreMath also found.
Note that, starting at about 10:50, PreMath evaluates cos²(15°) - sin²(15°). An alternative method is to use the ratio of sides of the 15°-75°-90° triangle to derive cos(15°) = (√3 + 1)/(2√2) and sin(15°) = (√3 - 1)/(2√2). Then, cos²(15°) = (√3 + 1)(√3 + 1)/(2√2)² = (3 + 2√3 + 1)/8 = (2 + √3)/4 and sin²(15°) = (√3 - 1)(√3 - 1)/(2√2)² = (3 - 2√3 + 1)/8 = (2 - √3)/4. So, cos²(15°) - sin²(15°) = (2 + √3)/4 - (2 - √3)/4 = 2(√3)/4 = (√3)/2, as PreMath also found, but using the cosine angle sum formula (in this case, cosine double angle formula) instead.
Watching how cos(2x) identity is derived to be: cos(2x) = cos^2(x) - sin^2(x). is great lesson.
At a quick glance, Form a rectangle ABCG where G is the intersection of perpendiculars from A and C. Area of trapezoid = Area of Rectangle - Area of triangle DCG. Angle CEB = 180 - 105 = 75. Angle BCE = 90-75 = 15. Then CB = Cos(15) * 13 = 12.56 and EB = Sin(15) * 13 = 3.365 Then DG = CB- DA = 12.56 - 3.365 = 9.2. Then Angle DCG = 90-15-15 = 60. Then DG/ CG = tan(60) then CG = DG/tan(60) = 9.2/tan(60) =5.3 . Area of rectangle ABCG = 5.3 * 12.56 =66.68 and area of triangle DCG = 0.5 * 9.2 * 5.3 =24.38. Then area of trapezoid = 66.68 - 24.38 = 42.3 area units.
Nothing better.
I just add that if needed another time, we have sin(15°) = (sqrt(6) - sqrt(2))/4; cos(15°) = (sqrt(6) + sqrt(2))/4 and tan(15°) = 2 - sqrt(3)
Went another way, but with same result (as it is to be expected as long as i manage to omit slip errors :D ). Just created a helper point F at intersection of prolongation of CD with prolongation of BA, creating a rectangle with 30°/60°, then getting EB, AD, BC the same way as in the video, then getting BF by tan(30)*BC, then getting AF by tan(30)*AD, then getting AB as BF - AF, then the area by AB * average(AD, BC). It's nice to see the different thinking approaches by different people in the chat :D The problem gave a big degree of freedom of "roads to Rome".
An error at 7:44 with DF length divided by 1.73
Since ∠AEC = 105° then ∠CEB = 75° and since ∠CBE = 90° then ∠BEC = 15°.
Since sin 15° = (√6 − √2) / 4 and sin 75° = (√2 + √6) / 4 then BE = AD = 13 (√6 − √2) / 4 and BC = 13 (√2 + √6) / 4. So DA + BC = 13√6 / 2.
Since BE = AD then CF = 13 (√6 + √2) / 4 − 13 (√6 − √2) / 4 = 13√2 / 2. Since CDF is 30°-60°-90° then DF = AB = (13√2 / 2) / √3 = 13√6 / 6.
Therefore, the area of the trapezoid is ½ (13√6 / 6) (13√6 / 2) = 42.25 square units.
Very nice video sar❤❤❤
Thanks ❤️
I took 4 hours doing this question because I missed the words “basic trigonometry”. I ended up solving the question by extending AB and CD, using the angle bisector theorem, and proving that DEA is a 30-60-90. Thanks for the amazing question!
Looking back at the question, the area of EBC is equal to AECD as long as EC bisects DCB (the fact becomes obvious if you reflect EBC across EC and connect DE). This means that the trapezoid has the same area as an isosceles triangle with the equal lengths being EC and the angle between them DCB, meaning its area is equal to (EC^2)*sin(DCB)/2.
This is amazing, many thanks, Sir!
φ = 30°; area ABCD = ? → AB = AE + BE = c + a; BC = BF + CF = a + b; CE = 13; sin(ABC) = 1; ECD = φ/2
AEC = 7φ/2 → CEB = 5φ/2 → cos(5φ/2) = cos(3φ/2 + φ) = cos(3φ/2)cos(φ) - sin(3φ/2)sin(φ) =
(√2/2)(√3/2) - (√2/2)(1/2) = (√2/4)(√3 - 1) = a/13 → a = (13√2/4)(√3 - 1) →
sin(5φ/2) = sin(3φ/2 + φ) = sin(3φ/2)cos(φ) + sin(φ)cos(3φ/2) = (√2/2)(√3/2) + (1/2)(√2/2) = (√2/4)(√3 + 1) =
(a + b)/13 → a + b = (13√2/4)(√3 + 1) → b = (a + b) - a = 13√2/2
CEB = 5φ/2 → BCE = 3φ - 5φ/2 = φ/2 → FCD = φ → (a + c) = b/√3 = 13√2/2√3 = 13√6/6 →
c = (a + c) - a = (13√6/12)(√3 - 1)
∆ BCT → CT = CD + DT → BT = TA + AE + BE = a√3/3 + c + a → sin(TAD) = 1 → ADT = BCD = φ →
area ∆ BCT = (√3/6)(a + b)^2
area ∆DAT = (√3/6)a^2 →
area ABCD = area ∆ BCT - area ∆ DAT = (√3/6)(b^2 + 2ab) = (1/6)(√3/2)(13^2)√3 = (13/2)^2
btw: cos(θ) = AB/BD = (√74/37)√(8 + 3√3) → θ ≈ 32,37°
Good morning sir I am a farmer just 58 years age but I like mathematics from my school student life till running life
Wow! CEB and CEAD have the same area! And this is true for any similar construction where AD = BE and ∠BCE = ∠DCE 🤩
I'll say 42.25
Angle BEC = 180º - 105º = 75º
Angle ECB = 90º - 75º = 15º
Angle ADC = 150º
EB = AD = x
sin (15º) = x / 13 ; x = 13*sin(15º) ~ 3,365 lin un
cos (15º) = CB / 13 ; CB = 13*cos(15º) ~ 12,557 lin un
Check:
12,557^2 + 3,365^2 = 13^2 ; 157,68 + 11,321 = 169 ; 169 = 169
Major Base = BC = 12,557 lin un
Minor Base = AD = 3,365 lin un
Finding AB = AE + 3,365
Equation of Straight Line passing points C and D:
y = - tan(60º)x + 12,557
and Equation of Straight Line passing points A and D
y = 3.365
Intercept of both equations:
3,365 = -tan(60º)*AB + 12,557 ; - 9,192 = -tan(60º)*AB ; AE = 9,192/tan(60º) ; AB = 9,192 / 1,732 ; AB ~ 5,307
AE = 5,307 - 3,365 = 19,942 lin un
height = AB = 5,307 lin un
Trapezoid Area = (B + b) * (h / 2) = (12,557 + 3,365) * (5,307 / 2) = 15,922 * 2,6535 ~ 42,25 sq un
Answer:
Trapezoid Area equal to approx. 42,25 Square Units.
1st like 😂
Legend!
Thanks ❤️