Justify your answer | Find Blue shaded Triangle area | (Rectangle) |

[@Buy_YT_Views_721]

This video is pure entertainment. ✨

[@PreMath]

Thanks for the feedback ❤️🌹

[@waheisel]

An out of the way formula is useful for this one: For two angles that add to 45 degrees (e.g. EAB and CDA), if tan A=a/b then tan D=(b-a)/(b+a). Here a=4 and b=12. So tan D=1/2 and x=6. Here are a couple nice proofs of the formula; kzbin.info/www/bejne/iqPKgn2DZr2enJo I love my exciting daily puzzle, Thanks PreMath!

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

[@prossvay8744]

Let BE=x Area of the triangle ABE=24cm^2 1/2(12)(x)=24 so x=4cm Tan(AEB)=12/4=3 So LAEB=71.57° LCED=180-(45+71.57°)=63.43° Tan(63.43°)=12/CE So CE=6cm Area of the Blue triangle=1/2(12)(6)=36cm^2.❤❤❤

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Irishfan]

This is how I would have done this problem right from the start.

[@ОльгаСоломашенко-ь6ы]

X/sin(45°)=4√10/ sin(180°-45°-< DAE). Синус суммы находим по формуле, используя соотношения в ∆AEB. AD=x=10.

[@marcgriselhubert3915]

EB = 48/12 = 4. Let t = angleAEB, In triangle ABE: tan(t) = AB/AB = 12/4 = 3 Let u = angleDEC, We have t + u + 45° = 180°, so u = 135° -t and tan(u) = tan(135° -t) tan(u) = (tan(135°) - tan(t))/(1 + tan(135°).tan(t)) = (-1 -3)/(1+ (-1).(3)) =-4/-2 = 2 In triangle DEC: tan(u) = 2 = DE/EC = 12/EC, sonEC =6 Finally the area of the blue triangle is (1/2).12.6 = 36.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

1/ EB= 4cm so AE= 4sqrt10 2/ From D drop the height DH to AE. The angle ADH=angle EAB ( the sides are perpendicular) so the two triangles ADH and EAB are similar. Label DH= h-> h/AH=AB/AE=12/4=3 Notice that the triangle DHE is an right isosceles so h= HE-> AH/1= HE/3=(AH+HE)/4=4sqrt10/4 -> h=3sqrt10 and AH=sqrt10 12:47 -> DA =sqrt(sqh+sqAH)=sqrt100=10 -> EC=6-> Area of the blue triangle=36 sq cm

[@PreMath]

Excellent! Thanks for sharing ❤️

[@ChuzzleFriends]

ABCD is a rectangle. By the Parallelogram Opposite Sides Theorem, AB = 12. A = (bh)/2 24 = (12 * h)/2 6h = 24 h = 4 So, BE = 4 cm. Label CE = x. Then BC = x + 4 cm. Draw an altitude of △AED from E to a point F on base AD. This divides rectangle ABCD into two smaller rectangles with a pair of congruent triangles each. Because △ABE ≅ △AFE, the area of △AFE is 24 cm². A = (bh)/2 = (12 * x)/2 = 6x Because △DCE ≅ △DFE, the area of △DFE is 6x cm². △AFE & △DFE combine to form △AED. So, the area of △AED is 6x + 24 cm². But then we can also find the area of △AED by using the side-sine area formula. Find AE & DE. Use the Pythagorean Theorem. a² + b² = c² 12² + 4² = (AE)² (AE)² = 144 + 16 = 160 AE = √160 = (√16)(√10) = 4√10 12² + x² = (DE)² (DE)² = x² + 144 DE = √(x² + 144) A = 1/2 * a * b * sinC = 1/2 * 4√10 * √(x² + 144) * sin(45°) = 2√10 * √(x² + 144) * sin(45°) = √40 * √(x² + 144) * sin(45°) = √[40 * (x² + 144)] * sin(45°) = √(40x² + 5760) * sin(45°) = √(40x² + 5760) * [(√2)/2] 2A = √(40x² + 5760) * √2 2(6x + 24) = √(40x² + 5760) * √2 12x + 48 = √[2 * (40x² + 5760)] = √(80x² + 11520) (12x + 48)² = [√(80x² + 11520)]² 144x² + 1152x + 2304 = 80x² + 11520 64x² + 1152x + 2304 = 11520 64x² + 1152x - 9216 = 0 x² + 18x - 144 = 0 (Factorable) x² + 24x - 6x - 144 x(x + 24) - 6(x + 24) (x - 6)(x + 24) = 0 x - 6 = 0 or x + 24 = 0 x = 6 x = -24 But x represents the length of a segment and it can't be negative, so x ≠ -24 & x = 6. Finally, find the area of △DCE. Substitute x = 6. A = (bh)/2 = (12 * 6)/2 = 72/2 = 36 So, the area of the blue triangle is 36 square centimeters.

[@montynorth3009]

Green triangle area = 24. 1/2 x 12 x EB = 24. EB = 4. Tan BEA = 12 / 4 = 3. Angle BEA = 71.565 degrees. Blue triangle. Angle DEC = 180 - 71.565 - 45 = 63.435 degrees. Tan 63.435 = 12 / CE. CE = 12 / 2 = 6. Area = 1/2 x 6 x 12 = 36.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... The green triangle is a right triangle, so we can conclude: A(green) = A(ABE) = (1/2)*AB*BE ⇒ BE = 2*A(ABE)/AB = 2*(24cm²)/(12cm) = 4cm By applying trigonometry we obtain: tan(∠AEB) = AB/BE = (12cm)/(4cm)= 3 tan(∠BED) = tan(∠AEB + ∠AED) = [tan(∠AEB) + tan(∠AED)]/[1 − tan(∠AEB)*tan(∠AED)] = [3 + tan(45°)]/[1 − 3*tan(45°)] = (3 + 1)/(1 − 3*1) = 4/(−2) = −2 tan(∠CED) = tan(∠BEC − ∠BED) = [tan(∠BEC) − tan(∠BED)]/[1 + tan(∠BEC)*tan(∠BED)] = [tan(180°) − (−2)]/[1 + tan(180°)*(−2)] = (0 + 2)/(1 − 0*2) = 2 Now we are able to calculate the area of the blue right triangle: tan(∠CED) = CD/CE ⇒ CE = CD/tan(∠CED) = (12cm)/2 = 6cm A(blue) = A(CDE) = (1/2)*CD*CE = (1/2)*(12cm)*(6cm) = 36cm² Best regards from Germany

[@PreMath]

Excellent!👍 Thanks for sharing ❤️🇺🇸

[@jimlocke9320]

As phungpham1725 did, I dropped a perpendicular from D to AE, let's call the intersection F. From the Pythagorean theorem, after finding that BE = 4 (2:24 in the video), AE has length 4√10. We note that

[@PreMath]

Excellent! Thanks for sharing ❤️

[@binondokhaliljustinb.5709]

Another solution! If you extend the line DE, then name it point F, you can get the angle BEF, angle BEF is equivalent to angle DEC since they are alternate angle. To get angle BEF, you first need to get angle AEB. Then you can now get the length of EC.

[@PreMath]

Thanks for the feedback ❤️

[@michaelkouzmin281]

brute force (i.e. with trigonometry): 1. BE = 4cm; 2. tg (AEF) = 4/12 = 1/3; AEF = atan(1/3); 3. DAF = D45 - atan(1/3); 4. tg(DAF) = x/12; => x= 12*tg(DAF) = 12 * tg(45-atan(1/3)) = 12 * (tg(45)-tg(atan(1/3)))/(1+tg(45)*tg(atan(1/3)) = 12* (1-1/3)/(1+1/3)= 12 * (2/3)/(4/3) = 12* (2/4) = 6 cm; 5. Ablue = 12*x/2= 12*6/2= 36 sq cm.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cebongkatro4073]

Like minded

[@VictorLonmo]

This was a fun question PreMath! I thought I was original by solving it with trig but based on the other comments I am not the only one...

[@PreMath]

Excellent! Glad to hear that! Thanks for the feedback ❤️

[@adgf1x]

trgl ADE and sq ABCD are having with same baseAD and between same parllelsAD and BC.Area trgl ADE=1/2*area sq ABCD=144/2=72 sq cm.so ar of blue trgl=144-72-24=48 sq cm.ans

[@moroofoloruntola199]

this can be solved in an easier way than the method here. finding angle AEB =71. then get angle DEC, s 180-45-71. then use tan =Opposite over adjacent then you get x easily and find the blue area

[@PreMath]

Thanks for the feedback ❤️

[@calvinmasters6159]

I was with you up to EB=4 Then arc tan 12/4 = 71.6 deg 180 - 45 - 71.6 = 63.4 height blue triang = 12/tan 63.4 = 6 A(blu) = 6 * 12 / 2 = 36 Enjoyable. Send more.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@gaylespencer6188]

My exact approach.

[@gervaischouinard9809]

Did same with arctan and tan.

[@adgf1x]

h=4,CE=8,ar.blue tr.=12×8/2=48 cm^2

[@peterdavidsalamanca8404]

Non-geometry problems please!

[@PreMath]

Sure! Please keep watching... Thanks for the feedback ❤️

[@mattsta1964]

I solved it using trig rather than algebra. Nice algebra solution!

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@misterenter-iz7rz]

(a/12+4/12)/(1-a/36)=(a+4)/((36-a)/3)=1, 3a+12=36-a, 4a=24, a=6, therfore the answer is 12×6/2=36.😊

[@PreMath]

Excellent! Thanks for sharing ❤️

[@jarikosonen4079]

What happens if this is substituted to the Heron's formula as a=x+4, b=4*sqrt(10), c=sqrt(x^2+144), s=(a+b+c)/2 and set equal to 6x+24 ? (That could be required if angle is not 30°, 45°, 60° with no solution to the sin(angle)) Should the Heron's formula result to the same result 6x+24 and task would not get completed with it this case...?

[@quigonkenny]

Triangle ∆ABE: Area = bh/2 = AB(BE)/2 24 =12BE/2 = 6BE BE = 24/6 = 4 Let ∠BEA = θ and ∠DEC = α. α = 180°- 45°- θ = 135°- θ. tan(θ) = AB/BE = 12/4 = 3. tan(α) = tan(135°- θ) tan(α) = (tan(135°)-tan(θ))/(1+tan(135°)tan(θ)) tan(α) = (-1-3)/(1+(-1)3) tan(α) = (-4)/(1-3) = -4/(-2) = 2 CD/EC = 2 12/EC = 2 EC = 12/2 = 6 Triangle ∆ECD: Area = bh/2 = CD(EC)/2 Area = 12(6)/2 = 36 cm²

[@calvinmasters6159]

It's funny how what was tedious in high school can be enjoyable now. Like Shakespeare, which is wasted on highschoolers.

[@PreMath]

Very well said! Thanks for the feedback ❤️ Stay blessed🌹

[@giuseppemalaguti435]

CE=12/tg(180-45-arctg(12/4))=-12/tg(45+arctg3)=6...Ablue=12*6/2=36

[@PreMath]

Excellent! Thanks for sharing ❤️

[@anthonycheng1765]

By drawing EF//CD, it is much easier to handle this problem (use the captioned approach)

[@JLvatron]

I used trig to solve this. I was discouraged by the heavy square roots.

[@wackojacko3962]

I take the long way home too cuz it's the scenic route . 🙂

[@PreMath]

You are blessed😀 Right way to enjoy the life! Thanks for the feedback ❤️

[@josephsalinas6725]

Fiz utilizando a tangente. Bem mais simples.

[@olivierjosephdeloris8153]

Possible avec la trigo, mais à un moment on a : tan(45 - atan(1/3)) = 0,5 qu'il est impossible à deviner sans calculatrice

[@draketheduelist]

I was looking at the thumbnail and going "...wait a second! Drawing not to scale!" You can't know for sure the quadrilateral shown is a rectangle from what we're given because you would have to be given the condition that (1) DA and CB are parallel or (2) either DAB or CDA are 90-degree angles, making the question unsolvable unless you know for a fact you're dealing with a rectangle. I hate it when the thumbnail doesn't give you everything you need to solve the problem.

[@AmirgabYT2185]

S=36 square units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@adgf1x]

48cm^2=ar blue shade.

[@MrPaulc222]

I used trig, so there was a tiny rounding error, but I put 36 as it was too close to not be!

[@kalavenkataraman4445]

36, Sq. Units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Ashi-cq7tj]

Me first again 😂😂

[@PreMath]

Excellent! Thanks ❤️

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Angle DEC = alpha 02) AD = BC = X 03) CE = X - 4 04) EB = 4 ; 48 / 12 = 4 05) Area of Rectangle [ABCD] = A = 12 * X 06) arctan(12/4) = arctan(3) ~ 71,6º 07) 71,6º + 45º = 116,6º 08) Alpha = 180º - 116,6º ~ 63,435º 09) tan(63,435º) = 2 10) 12 / (X - 4) = 2 11) X = 10 12) Total Area [ABCD] = 12 * 10 = 120 13) Blue Area = (120 - 48) / 2 = 72 / 2 = 36 14) ANSWER : Area of the Blue Triangle equal to 36 Square Centimeters. Greeting from The Center for Studies of Ancient Mathematical Thinking, Knowledge and Wisdom - Cordoba Caliphate!!

[@PreMath]

Excellent!🌹 Thanks for sharing ❤️🙏

[@sergeyvinns931]

A=36!

[@PreMath]

Excellent! Thanks for sharing ❤️