Can you find area of Green triangle in the rectangle? | Easy explanation |

Learn how to find the area of the Green triangle in the rectangle. Important Geometry skills are also explained: area of the triangle formula; Pythagorean theorem; area of the rectangle formula. Step-by-step tutorial by PreMath.com
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[@jamestalbott4499]

Thank you! Appreciate the detailed explanation.

[@PreMath]

You are welcome! Thanks ❤️

[@angeluomo]

The value of x can be calculated simply by using a bit of trig: x = 6*tan(45°-arctan(2/6)) = 3. With x, you have the area of the rectangle = (2+3)*6 = 30. The green triangle is half of that (=15). Done.

[@gaylespencer6188]

Did it with trig also, but via different route.

[@montynorth3009]

Tan DEC = 6/2 = 3. DEC = 71.565 degrees. Therefore angle AEB = 180 - 71.565 - 45 = 63.435 degrees. Triangle AEB. Tan 63.435 = 6/AE. AE = 6/ tan 63.435. AE = 3. So AD = 5 = BC, and the area follows.

[@jimlocke9320]

Drop a perpendicular from E to BC and label the intersection as point F. Let

[@marcgriselhubert3915]

I did it this way.

[@GetMatheFit]

You rotate the triangle CDE counterclockwise. Half a square or an isosceles triangle is formed. If you extend EB, you hit E'. You get two similar triangles (E'D'B and BAE). 2 : (4 - x ) = 6 : x 2x = 24 - 6x 8x = 24 x = 3 So the rectangle has a length of 6 and a width of 5 (l=6; b=5) The area of ​​the green triangle corresponds to half the rectangle. Furthermore: A(Green) = 6*5/2 A(Green) = 15 FE Best regards Gerald

[@Waldlaeufer70]

I really like this solution because it avoids trigonometry and uses the strategy of "thinking outside the box"! 😊

[@GetMatheFit]

@@Waldlaeufer70 Thanks.

[@ChuzzleFriends]

Draw the height of △BCE. This altitude, EF, is perpendicular to the base, segment BC, and divides rectangle ABCD into two smaller rectangles with a pair of congruent triangles each as segments BE and CE are the diagonals of the small rectangles. Find the area of either rectangle CDEF or △CDE. You can determine the area of △CFE. A₁ = 1/2 * b * h = 1/2 * 6 * 2 = 6 * 1 = 6 Because △CDE ≅ △CFE, the area of △CFE is 6 square units. Find the area of either rectangle ABFE or △ABE. Label AE = x. You can determine the area of △FEB. A₂ = 1/2 * b * h = 1/2 * 6 * x = 3 * x = 3x Because △ABE ≅ △FEB, the area of △FEB is 3x square units. △CFE and △FEB combine to form △BCE. The area of △BCE is 3x + 6 square units. But then we can also find the area of △BCE by using the side-sine area formula. To do so, find BE and CE. Apply the Pythagorean Theorem twice. a² + b² = c² 2² + 6² = (CE)² (CE)² = 4 + 36 (CE)² = 40 CE = √40 = √(2 * 2 * 2 * 5) = 2√10 x² + 6² = (BE)² (BE)² = x² + 36 BE = √(x² + 36) A₃ = 1/2 * a * b * sinC, where a & b are the lengths of two given sides with an included angle measuring C degrees. = 1/2 * 2√10 * √(x² + 36) * sin(45°) = √10 * √(x² + 36) * sin(45°) = √[10 * (x² + 36)] * sin(45°) = √(10x² + 360) * sin(45°) = √(10x² + 360) * [(√2)/2] 2A₃ = √(10x² + 360) * √2 2(3x + 6) = √(10x² + 360) * √2 6x + 12 = √[2 * (10x² + 360)] = √(20x² + 720) (6x + 12)² = [√(20x² + 720)]² 36x² + 144x + 144 = 20x² + 720 16x² + 144x + 144 = 720 16x² + 144x - 576 = 0 x² + 9x - 36 = 0, This equation can be factored because 225 (discriminant) is a perfect square! x² + 12x - 3x - 36 = 0 x(x + 12) - 3(x + 12) = 0 (x + 12)(x - 3) = 0 x = -12 or x = 3 But then x = -12 causes negative areas and side lengths (e.g. A₃ = 3x + 6 = 3 * -12 + 6 = -36 + 6 = -30 u²) so we'll reject the value and accept x = 3. Plug x = 3 into 3x + 6. A₃ = 3(3) + 6 = 9 + 6 = 15 So, the area of the green triangle is 15 square units.

[@MrPaulc222]

[@giuseppemalaguti435]

Posto EA=a risulta (90-arctga/6)+45+(90-arctg2/6)=180...45=arctga/6+arctg1/3...applico tang...a=3...Ag=(3+2)6/2=15

[@PreMath]

Excellent! Thanks ❤️

[@Ramkabharosa]

Let θ = ∠DCE and |AE| = x. Then tan(θ) = 2/6 = 1/3 & ∠ABE = 45°- θ. So x = 6.tan(∠ABE) = 6.tan(45°- θ) = 6. {tan(45°) - tan(θ)}/{1+ tan(45°).tan(θ)} = 6{1 -1/3}/{1+ 1.(1/3)} = 3. ∴ area(green Δ) = (½).|BC|.|CD| = ½(6)(5) =15. Trig is KING!

[@Waldlaeufer70]

x = 6 / tan (180° - 45° - arctan (6/2)) x = 6 / tan (135° - arctan (3)) = 3 A(green) = 1/2 b h = 1/2 (2 + x) h = 1/2 (2 + 3) (6) = 1/2 (5) (6) = 15 square units

[@GetMatheFit]

Es geht auch ohne Trigonometrie. Meine Lösung findest du bei mir unter Community - wenn es dich interessiert. LG Gerald

[@Irishfan]

I found it simpler to work with tangent function to get the angles of the unshaded triangles and solve for X. Then, using the area of a triangle formula, finding the area of both of those triangles. Then, by inspection, I realized the area of the sum of the area of the two triangles equal to the area of the shaded triangle. However, I also could have subtracted the areas of the two triangles from the area of the rectangle to get the shaded triangle area.

[@MarieAnne.]

Since green triangle has base x+2 and height 6, then once you find x you can also calculate area of triangle directly using formula area = 1/2 * base * height = 1/2 * (x+2) * 6

[@michaelstahl1515]

I draw a parallel line EF to DC . Using trigonometrie I calculated the angle EFC and got arccos ( 6 / sqr (40) ) = 18, 43° . The angle BEF was 45° - 18,43° = 26,57 ° . With tan ( 26,57 °) = x/6 I got x = 6 * tan ( 26,57 ° ) = 3. The Area of the green triangle is A = 1/2 * 6 * 5 = 15 FE .

[@devondevon4366]

Calculate the hypotenuse of the triangle with sides 6 and 2: it is sqrt 40 or sqrt 2 * sqrt 20 Using the law of cosine the angle next to green is 18.435 degrees Draw a perpendicular line from the vertex of the green to form a 45 -45 -90 right triangle The length of this perpendicular line is sqrt 20 The triangle next to 45-45-90 have angle of26.565 (90 - 63.435 )(18.435 + 45), 90 and a side of sqrt 20. Using the law of sine its hypotenuse = 5 . Hence the length of the triangle = 5 Hence, the area of the green = 6 * 5* 1/2 = 15 Answer =15

[@tombufford136]

At a quick glance, Angle ECD = Tan^-1(6/2) = 71.6 degrees. Angle EAB = 180 - 45 -71.6 = 63.43. Then EB = 6/sin(63.43) = 6.71. Then EA = sqrt( 6.7^2 - 36) = 3. Then the area of the green triangle is 0.5 * base * height = 5/2 * 6 =15 area units.

[@phungpham1725]

Drop the height BH to the base EC. Because of the triangle BHC and CDE are similar so BH/HC =CD/DE=6/2=3 So BH= EH= 3/4.EC= 3/4 x 2sqrt 10= 3/2 x sqrt10 Area of the green triangle= 1/2 x 3/2 sqrt10 x 2sqrt10= 15 sq units

[@hongningsuen1348]

I like the idea of using simple geometric calculations only instead of solving quadratic equation. It is better to state that BHE is an isosceles triangle (base angles equal) such that BE = EH for easy understanding.

[@phungpham1725]

@@hongningsuen1348 Glad you like it. Thank you so much. I agree with you but sometimes it's more fun to discover the rest of the puzzle.

[@dannymeslier6658]

Consider right triangle EDC. Tangent of Ê is 3. Now consider right triangle EAB. Tangent ofÊ in this traingle is: [tan(135)-3]/[1+tan 135 * 3]= 2, thus AE = 3, thus BC = 5, thus the area of the green triangle is 5*6/2=15 square units.

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال tan tana=1/3وtan[a+45]= 2 ....... S=15

[@hcgreier6037]

I did this as follows: h = height of rectangle. Let d = √(2² + 6²) = √40 = 2√10. Draw a perpendicular from C to EB to meet at P. Then EP = PC due to 45° isoceles triangle. Legs e = d/√2 = 2√5. Triangles △ABE and △BCP are similar, so one gets e/h = 6/EB, where EB = √[(h - 2)² + 6²] = √[(h - 2)² + 36]. e²/h² = 36/[(h - 2)² + 36], with e² = (2√5)² = 20, so 20·[(h - 2)² + 36] = 36h². 20·[(h² - 4h + 4) + 36] = 36h² 20·[h² - 4h + 40] = 36h² 20h² - 80h + 800 = 36h² 16h² + 80h - 800 = 0 |:16 h² + 5h - 50 = 0, which factors to (h - 5)(h + 10) → h = -10 (rejected) and h = 5. Area of the green triangle then 6·h/2 = 6·5/2 = 15 square units.

[@GetMatheFit]

Echt schade, dass du keine Videos mehr hochladest. Deine Rätselvideos haben mir echt toll gefallen. Ich habe diese Aufgabe mit einem Strahlensatz (bzw. ähnliche Dreiecke) gelöst. Meine Lösung findest du bei mir unter Community - wenn es dich interessiert. LG Gerald PS: Ach ja, einige Beispiele habe ich von dir übernommen. Die waren einfach zu gut. Die musste ich nochmals präsentieren.

[@hcgreier6037]

@@GetMatheFit Danke für's Feedback! 👍Videos kommen bald wieder, hatte anderes zu tun. Schöne Seite hast du da, Respekt!

[@GetMatheFit]

@@hcgreier6037 Yeah. Super. Da freue ich mich schon drauf. Finde deine Art lässige Art und die Hulk Hogan Mütze klasse. LG Gerald

[@andydaniels6363]

An analytical-geometric approach: The triangle’s area is half that of the rectangle’s, so the problem reduces to finding the height of the rectangle. For convenience, reflect the figure across CD and place a right-handed coordinate system at D with DC and DA along the positive x- and y-axes, respectively. This makes all of the coordinates nonnegative. Equivalently, use a left-handed coordinate system, with positive y downwards along DA. (This makes all of the relevant coordinates positive, which eliminates a potential source of silly errors.) The idea is to rotate the extension of CE by 45 degrees and compute where that line intersects the line x=6, which is our point B. We can immediately write an equation for the extension of CE in intercept-intercept form, i.e., x/6+y/2=1. In homogeneous coordinates this can be represented by the row vector 𝜆 = [1/6,1/2,-1]. A rotation through 45 degrees about the point E=(0,2) can be decomposed into a rotation about the origin flanked by two translations, i.e., the matrix product T((0,2))R(𝜋/4)T((0,-2)). The rotated line is then 𝜆ʹ=𝜆T((0,2))R(𝜋/4)T((0,-2)). We can then either let B=[6,y,1]^T, set 𝜆ʹ B=0 and solve for y, or compute the cross product 𝜆ʹ ⨉ [1,0,-6] (the latter is the line x=6) to find the coordinates of B directly. Either way, we end up with the Cartesian coordinates of B as (6,5), ∴ the triangle’s area is 6*5/2 = 15. Note that the daunting-looking matrix product is easily computed by clever grouping of the products and keeping in mind what the individual matrices represent. E.g., since T((0,-2)) is a translation by -2 in the y direction, we can immediately write T((0,-2)) B = [6,y-2,1]^T without having to multiply the product out. Similarly, 𝜆T((0,2)) is just the line 𝜆 shifted downward by 2 units. This new line passes through the origin, so by inspection that product must equal [1/6,1/2,0]. Since we’re working in homogeneous coordinates, we can then substitute the vector [1,3,0] for this result to further simplify the remaining calculations.

[@plamenpenchev262]

Quite easier: Devide angle CEB with a horizontal line in two angles a and b (upper and lower angles). tan(a) = 2/6 = 1/3 tan(b) = tan(45 - a) = (tan(45) - tan(a))/(1 + tan(45)*tan(a)) = (1 - 1/3)/(1 +1/3) = 1/2 Then, height of the rectangle is h = 6×1/2 + 2 = 5, and so on. P.S. I am computer chemist and spectroscopist and used Wikipedia for tan(x - y) formula.

[@paulscheele623]

Tan (a+b+c) = ((tan a + tan b + tan c) - (tan a * tan b * tan c)) / (some factor) Tan a = 6/2 = 3 Tan b = tan 45 degrees = 1 Tan (a + b + c) = tan 180 degrees = 0 So 0 = 3 + 1 + tan c - 3*tan c So tan c = 2 But tan c = 6/x, so x=2 Area = 3*(x+2), so area = 15

[@wackojacko3962]

@ 7:47 the system of equations....comparing to solve for x is really nice.! 🙂

[@PreMath]

Excellent! Thanks ❤️

[@MarieAnne.]

My solution: Let ∠CED = θ. Then ∠AEB = 180−45−θ = 135−θ In △CDE, tan θ = CD/DE = 6/2 = 3 In △ABE, tan(135−θ) = AB/AE = 6/x Using trig identity, we get: tan(135−θ) = (tan 135 − tan θ) / (1 + tan 135 * tan θ) = (−1−tan θ)/(1+(−1)tan θ) tan(135−θ) = (tan θ + 1) / (tan θ − 1) 6/x = (3 + 1) / (3 − 1) = 4/2 = 2 x = 6/2 x = 3 Green triangle has base = x+2 = 5 and height = 6 Area = 1/2 * 5 * 6 = 15

[@jphilsol6459]

Go to trigonometry, easy to find DÊC vith 6/2 for the tangeant, and 3 tan-1 = 71.56° the angle AÊB = 180 - (71.56 + 45) = 63.43°, tan 63.43= 2 so the length called X = 6/2 = 3 Now we have the area of the rectangle = 30, the area of the triangle EDC = 6*2/2 = 6 and the area of triangle EAB = 6*3/2 =9 At the end, the green triangle = 30 - (6+9) = 30-15 = 15

[@johnpiggott7426]

Just do tangent of sum of two angles. Cut the triangle in two. The blank lower triangle has interior angles of 45+ theta and 45-theta where theta is the angle of the top most triangle on the right. Its tangent is 1/3. Tan 45 = 1. Tan (45-theta) = 1/2. 1/2 x 6 equals x = 3. 3+2 =5. 0.5x6x5 = 15. 🎉

[@AndreasPfizenmaier-y7w]

Draw a perpendicular from E to BC, the two parts of the triangle are the halves of two rectangles, so Area=30:2=15

[@alokranjan4149]

By quadratic equation, the other side of rectangle can easily be calculated as 5 unit. Then area of triangle can be calculated as 15 sq. unit.

[@alster724]

Squaring part near the end made it easy

[@marcello3621]

Extend the side DC in the vertice E of EBC triangle. We got a height in E relative to the side BC, with length 6, and the feet of the height is F. So we divided the triangle EBC in two parts: the first is a triangle EFC with EF = 6 and FC = 2, and the another is a triangle EFB, with FB = x and EF = 6. The sum of these areas gives us the green triangle's area: 2 × 6/2 + x × 6/2 = 6 + 3x (in terms of x) Also, by the area formula in terms of angle and adjacent sides: [EC × EB × sin (π/4)]/2 EC = √(2² + 6²) = √40 EB = √(x² + 36) [√40 × √(x² + 36) × √2/2]/2 √(80x² + 2880)/4 √(80x² + 2880)/4 = 6 + 3x (80x² + 2880)/16 = 36 + 36x + 9x² 80x² + 2880 = 576 + 576x + 144x² - 64x² - 576x + 2304 = 0 x = 3 So its area is 6 + 3x ---> 6 + 3(3) = *15 u.a*

[@PreMath]

Excellent! Thanks ❤️

[@santiagoarosam430]

Tomando cómo vértice común el punto D, sobre el rectángulo original trazamos un cuadrado de lado DC+DE=6+2=8 y luego, tomando como lado superior el segmento EC trazamos otro cuadrado inscrito en el anterior, la longitud EA se puede obtener por semejanza de triángulos → EA/AB = (6-2)/(6+2) → EA/6=4/8→ EA=3 → Área BCE =ABCD/2 =6(DE+EA)/2 =6(2+3)/2=15 Acertijo interesante. Gracias y un saludo cordial.

[@misterenter-iz7rz]

a+b=45, tan a+tan b/1-tan a tan b=1, tan a+tan b=1-tan a tan b, now 6(tan a+tan b)×6=total area=36(1-tan a tanb) tan a=1/3, 1/3+tan b=1-tan b/3, 4/3 tan b=2/3, tan b=1/2, so the area is 30(1-1/6)=36×5/6=30, and the area of the rectangle is 30/2=15.😊

[@andrewstayne503]

Find all the triangle angle using trig. Then use the sine rule to fine the base (=5) then use the formula for the area of a triangle =15.

[@AndreasPfizenmaier-y7w]

Tan 2/6=18,43 degrees, 45- 18,43 degrees=26,56 degrees, tan 26,56=x/6, hence x=3

[@Copernicusfreud]

Yay! I solved the problem. I used the trig functions to calculate all the interior angles and the length of segment AE. Area = (1/2) *(b) * (h) = (1/2) * (5) *(6) = 15.

[@LuisdeBritoCamacho]

I'll try to prove it with Trigonometry. Angle CED = a degrees Angle AEB = b degrees The Angles "a" and "b" are different. Angle CEB = 45 degrees a + b + 45 = 180 a + b = 180 - 45 a + b = 135 tan (a) = 6 / 2 = 3 Arctan (3) ~ 71,5 degrees Tan [ 180 - (arctan (3) + 45) ] = 2 Length EA = x 6 / x = Tan [ 180 - (arctan (3) + 45) ] 6 / x = 2 x = 6 /2 x = 3 So, AD = 5 Green Triangle Area = (6 * 5) / 2 = 15 Answer : Green Triangle Area is equal to 15 su

[@guenternoack3481]

Thanks ! Alpha between EA and EB = 180 - 45 - arctan 6/2 . tan alpha =2 . 2/1 =6/x ; x = 3

[@stevenconnell3206]

26* 33’ 54” (tan) 6x = 3

[@zsoltszigeti758]

sin(α+ß) = sin α * cos ß + cos α * sin ß α+ß=45 sin(α+ß)=√2/2 sin α = 2/√40 cos α = 6/√40 sin ß = x/√(x^2+36) cos ß = 6/√(x^2+36) √2/2=(2/√40)*(6/√(x^2+36))+(6/√40)*(x/√(x^2+36)) => x=3

[@dhaniramnagdeve1331]

Good solution of tenth base

[@jan-willemreens9010]

.... Angle(DEC) = ARCTAN(6/2) = ARCTAN(3) ... Angle(AEB) = 180 - 45 - ARCTAN(3) = 135 - ARCTAN(3) ... I AE I = 6 / TAN( 135 - ARCTAN(3)) = 3 ... Area (green triangle) = 6 * (2 + 3) - 0.5 * 2 * 6 - 0.5 * 3 * 6 = 30 - 6 - 9 = 15 ... not very elegant solution; thank you for your solution presentation ... best regards, Jan-W

[@StephenRayWesley]

(2)^2 (6)^2=(4+36)=40° 2x(45°)=90°x (40°+90°x+45°) =✓175°x +(360°-175)=√185°x 2^√9 5^1 2^√3^√3 ,5^1 2^√1^√1 5^√1. 25 (x+2x-5)

[@Ibrahimfamilyvlog2097l]

very nice❤❤❤

[@prossvay8744]

Area of the green triangle =1/2(6.32)(6.71)sin(45)=15 square units. ❤❤❤thanks

[@PreMath]

Thanks ❤️

[@manojkantsamal4945]

16.17( approx), may be

[@unknownidentity2846]

Let's find the size of the green area: . .. ... .... ..... Let's assume that E is the center of the coordinate system and that AB and CD are parallel to the x-axis. Then the unit vector EC is given by: u(EC) = v(EC) / |v(ec)| = ( +6 ; +2 ) / |( +6 ; +2 )| = ( +6 ; +2 ) / √(6² + 2²) = ( +6 ; +2 ) / √40 = ( +3 ; +1 ) / √10 Since the angle between the unit vectors EC and EB is known to be 45°, we can conclude: u(EC) * u(EB) = |u(EC)| * |u(EB)| * cos(∠BEC) u(EC) * u(EB) = cos(45°) = 1/√2 ( +3 ; +1 )/√10 * ( x ; −√(1 − x²) ) = 1/√2 3x − √(1 − x²) = √5 3x − √5 = √(1 − x²) 9x² − (6√5)x + 5 = 1 − x² 10x² − (6√5)x + 4 = 0 x = {6√5 ± √[(6√5)² − 4*10*4]} / (2*10) = [6√5 ± √(180 − 160)] / 20 = (6√5 ± √20) / 20 = (6√5 ± 2√5) / 20 x₁ = (6√5 + 2√5)/20 = 8√5/20 = 2√5/5 x₂ = (6√5 − 2√5)/20 = 4√5/20 = √5/5 y₁ = −√(1 − x²) = −√(1 − 4/5) = −√(1/5) = −√5/5 y₂ = −√(1 − x²) = −√(1 − 1/5) = −√(4/5) = −2√5/5 Check of the possible solutions: ( +3 ; +1 )/√10 * ( x₁ ; y₁ ) = (3*2√5/5 + 1*(−√5/5))/√10 = (6√5/5 − √5/5)/√10 = √5/√10 = 1/√2 ✓ ( +3 ; +1 )/(√10) * ( x₂ ; y₂ ) = (3*√5/5 + 1*(−2√5/5))/√10 = (3√5/5 − 2√5/5)/√10 = (√5/5)/√10 = 1/√50 Therefore only the first solution is correct and EB has the slope (−√5/5)/(2√5/5)=−1/2. So EB intersects BC at: ( x = 6 ; y = (−1/2)*x = −3 ) Now we can conclude that BC=+2−(−3)=5. So the final result is: A(BCE) = (1/2)*BC*CD = (1/2)*5*6 = 15

[@unknownidentity2846]

The solution shown in the video is very nice. Best regards from Germany

[@GetMatheFit]

Es reicht ein Strahlensatz für die Berechnung von x. Rest ist easy going. Meine Lösung findest du bei mir unter Community - wenn es dich interessiert. LG Gerald

[@misterenter-iz7rz]

You rarely use complicated trigonometric formulas 😊

[@misterenter-iz7rz]

Finally, I successfully struggle for reach this video, 🎉Let s×6 be the rectangle, arctan 2/6+arctan (s-2)/6=arctan 1/3+arctan (s-2)/6, so 1=(1/3+(s-2)/6)/(1-(s-2)/18)=((2+s-2)/1)/(18-s+2)/3, 20-s=3s, s=5, therefore the area is 5×6/2=15.😊

[@murphygreen8484]

Is the answer 15?

[@MattGilliesCwnAnnwn]

Aren't you assuming ABCD is a rectangle?

[@thumper88888]

That was given

[@JSSTyger]

(1/2)(6)(2)+(1/2)(6)(3) = 15

[@PreMath]

Thanks ❤️

[@sai3e8w65]

in 8:51 you are actually assuming that the area of the green triangle is equal to the area of the rectangle you can instead use SOH CAH TOA to work out angle DEC then 180-(45+DEC) then using SOH CAH TOA again to find EB and then A = 1/2absinC where a is EC and b is EB But you are getting the same answer as mine, so i don’t know why 🤷‍♀️ For me, my working is more logical than yours Besides that, Keep making great videos

[@devondevon4366]

15

[@murdock5537]

Nice! φ = 30°; ∎ABCD → AB = CD = 6; BC = AD = AE + DE = AE + 2 sin⁡(DAB) = sin⁡(ABC) = 1 → CE = 2√10 → sin⁡(δ) = CD/CE = 3√10/10 → cos⁡(δ) = √(1 - sin^2(δ)) = √10/10 CEB = 3φ/2 → sin⁡(3φ/2) = cos⁡(3φ/2) = √2/2 → sin⁡(δ + 3φ/2) = sin⁡(δ)cos⁡(3φ/2) + sin⁡(3φ/2)cos⁡(δ) = (3√10/10)(√2/2) + (√10/10)(√2/2) = 2√5/5 → cos⁡(δ + 3φ/2) = -√(1 - sin^2(δ + 3φ/2) ) = -√5/5 → θ = BEA = 6φ - (3φ/2 + δ) → sin⁡(θ) = sin⁡(3φ/2 + δ) = 2√5/5 → cos⁡(θ) = -cos⁡(δ + 3φ/2) = √5/5 → tan⁡(θ) = sin⁡(θ)/tan⁡(θ) = 2 → AE = AB/2 = 3 → area green ∆ = (1/2)6(5) = 15 = (1/2) area ∎ABCD or: φ = 30°; tan⁡(δ) = CD/DE = 3; tan⁡(3φ/2) = 1; tan⁡(6φ) = 0 tan⁡(δ + 3φ/2) = (tan⁡(δ) + tan⁡(3φ/2))/(1 - tan⁡(δ)tan⁡(3φ/2)) = -2 → δ + 3φ/2 + θ = 6φ → -2 + θ = 0 → θ = 2 = AB/AE → area green ∆ = 15

[@GetMatheFit]

Die richtige Aufgabe für dich. Ich hab ein Dreieck rotiert und dann den Strahlensatz verwendet um x zu berechnen. Meine Lösung findest du bei mir unter Community - wenn es dich interessiert. LG Gerald

[@murdock5537]

Danke, habe mir die Lösung in der Community angeschaut - sehr clever und elegant!@@GetMatheFit

[@GetMatheFit]

@@murdock5537 Danke. Darf ich fragen, warum du immer mit sin, cos bzw. tan arbeitest. LG Gerald

[@murdock5537]

Ich finde diese Methode unterhaltsamer, insbesondere wenn man die Additionstheoreme nutzen kann. Der Reiz besteht auch darin, dass man meist keinen TR benötigt. So kann man z. B. sin(x) oder cos(x) - mittels Bogenmaß - über die Taylor-Approximation erhalten, und das bis zu einer durchaus eindrucksvollen Winkelgröße (s. Bronstein). Vieles wird einfacher, siehe z. B. meine zweite Lösung oben. Zu Deinem Lösungsansatz: Wenn man das neue Rechteck voll aufspannt (8 x 6), erkennt man sofort, dass tan(θ) = 2 ist. Ich kombiniere rein geometrische Lösungswege gerne mit etwas Trig. 🙂 @@GetMatheFit

[@GetMatheFit]

@@murdock5537 Danke für die Info. In den trigonometrischen Sachen bist echt eine Koryphäe. Immer wieder erfrischend deine Lösungswege. LG Gerald

[@josephsalinas6725]

Fiz pela lei dos senos !

[@robertlynch7520]

Dunno… if you are going to use sin( ) and other trigonometric functions, why not just got for a direct trigonometric solution? tangent( angle ) = opposite/adjacent = 2 ÷ 6 = 0.33333333 atan( 0.333… ) = 18.435° = θ The ∠DEC is (90 - θ) = 71.565°. Since there is 180° around a point on a line, ∠AEB = (180° - (45° + 71.565°)) = 63.435° Armed with that, again working with tangents tan( 63.435 ) = 2 exactly Which means the opposite (AB) is 2× the adjacent (AE), or conversely (2 AE = AB) → (AE = AB/2) , and we know AB = 6 so, AE = 6 ÷ 2 = 3 The rest is obvious. Area of triangle is (½ BH) = (½ (3 ⊕ 2)( 6 )) = 15 u² ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

[@quigonkenny]

Since the only angle given, 45°, is one where the standard trig function values are well known and easily expressed in exact values, it's ideal to limit trig usage to only that one sine function for the area of a triangle. No need to use a calculator to find the angles of the other two triangles or to worry about precision errors from calculations performed on the calculator outputs, especially since we're only using it to get an alternate equation for the area of the triangle so we can solve for x. If the angle were something like 18° or 57°, we'd have to use trig functions and deal with the imprecision, but since it's 45° we don't.