Can you find the area of each shaded region? | (Fun Geometry problem) |

[@wackojacko3962]

The results show quite good congruence! 🙂

[@PreMath]

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

[@ChuzzleFriends]

Label the side length of square BEDF as s. △ABC & △AFD share ∠A. △ABC & △DEC share ∠C. So, △ABC ~ △AFD ~ △DEC by AA. AC/AD = BC/DF 84/48 = BC/s BC/s = 7/4 BC = 7s/4 = 1.75s AC/CD = AB/DE 84/36 = AB/s AB/s = 7/3 AB = 7s/3 Use the Pythagorean Theorem on △ABC. a² + b² = c² (7s/4)² + (7s/3)² = 84² (49s²)/16 + (49s²)/9 = 7056 441s² + 784s² = 1016064 (Multiply by 144, LCM of 16 & 9, on both sides) 1225s² = 1016064 s² = 829.44 s = √(829.44) = 28.8 AB = 7s/3 = (7 * 28.8)/3 = 67.2 BC = 1.75s = 1.75 * 28.8 = 50.4 AF = AB - BF = 67.2 - 28.8 = 38.4 CE = BC - BE = 50.4 - 28.8 = 21.6 Find the areas of △AFD & △DEC. A = (bh)/2 = (38.4 * 28.8)/2 = (1105.92)/2 = 552.96 [Yellow] A = (21.6 * 28.8)/2 = (622.08)/2 = 311.04 [Green] Thus, the areas of the shaded regions are as follows: Yellow: 552.96 square units Green: 311.04 square units Pink: 829.44 square units

[@shadrana1]

At 0:36, Triangles ABC,DAF and CDE are smilar (all angles are equal) Let x=DE=EB=BF=FD, 48/x=36/CE CE=36x/48=3x/4 units. Consider triangle DCE, DC^2=DE^2+CE^2 (pythagoras) 36^2 =x^2+(3x/4)^2 36^2= 25x^2/16 Take square root, 36=5x/4 x= 144/5 units. IDEBFI=x^2=(144/5)^2 = (20736/25) sq.units. CE/DE=DF/AF (3x/4)/x= x/AF AF= 4x/3 units IADFI= (1/2)*(AF)(FD)=(1/2)*(4x/3)*x (2/3)*x^2=(2*20736)/(3*25) = (13824/25) sq.units. ICDEI = (1/2)DE*CE)=(1/2)*x*(3x/4)=(3/8)*x^2=(3/8)*(20736/25)= (7776/25) sq.units. AB =AF+FB =4x/3+4x/4 =(7x/3) units BC =CE+EB =3x/4+4x/4 =(7x/4) units. IABCI=(1/2)*AB*BC= (1/2)(7x/3)(7x/4)=49x^2/24=(49/24)*(20736/25)=(42336/25) sq.units IABCI=IDEBFI+IADFI+ICDEI hence the answers are correct. Note:- AC=84 units,BC=50.4 units and AB=67.2 units,therefore ABC was a (3,4,5) triangle. Thanks for the puzzle professor.

[@himo3485]

DE=DF=BE=BF=x 36 : 48 = 3 : 4 AF=4x/3 CE=3x/4 x²+(4x/3)²=48² 25x²/9 = 2304 x² = 20736/25 Pink Square area : x * x = x² = 20736/25 Green Triangle area : x * 3x/4 *1/2 = 3x²/8 = 3/8 * 20736/25 = 3*2592/25 = 7776/25 Yellow Triangle area : x * 4x/3 * 1/2 = 2x²/3 = 2/3 * 20736/25 = 2*6912/25 = 13824/25

[@MrPaulc222]

A quick inspection tells me that the yellow hypotenuse is 1/3rd larger than the pink hypotenuse. As the yellow and green triangles are similar, the yellow triangle's base is (4/3)x, due to the 48:36 ratio of the two hypotenuses. This makes AB=(7/3)x and BC = (7/4)x As 12 is the common multiple of 4 and 3, multiply everything by 12 Large triangle sides are 28x, 21x, which indicates the triangles are of 3,4,5 nature. 36 hypotenused triangle as a 3,4,5 is 7.2*5, 7.2*4 and 7.2*3 so (28.8 * 21.6)/2 = 311.04 for the green triangle. Pink square is 28.8^2 = 829.44 Yellow triangle is the green one times (16/9) 311.04 * (16/9) = 552.96 I got there a bit more clumsily than you, but at least I got there :) Thank you once again.

[@prossvay8744]

∆AFD~∆DCE (AA) 36/48=DE/AF=3/4 DE=3x ; AF=4x In ∆AFD AD^2=AF^2+DF^2 (4x)2+(3x)^2=(48)^2 x=48/5 AF=4(48/5)=192/5 Yellow triangle area=1/2(48/5)(3))(192/5)= 552.96 square units Pink square area=(144/5)^2= 829.44 square unitd CE=√36^2-(144/5)^2=21.6 units. Green triangle area=1/2(144/5)(21.6)=311.04 square units.❤❤❤ Thanks sir.

[@phungpham1725]

1/ Let a be the side of the square. Because the green and yellow triangles are similar so, DE/AF= 36/48= 3/4-> AF=4a/3 Consider the yellow triangle, by using Pythagorean theorem, we have: sq (4a/3) + sqa= sq 48--> a= 144/5 --> Area of the red square= 829.44 sq. units 2/AF=4a/3=(4x144)/15-> Are of the yellow triangle = 1/2 AF x DF= 552.96 3/ Area of the green/Area of the yellow= sq(3/4)=9/16 Area of the green= 9/16 x area of the yellow= 311.04 4/ Another alternative approach: Consider the yellow right triangle: We have: DF/AF= a/(4a/3)= 3/4 So, the yellow triangle is a 3-4-5 triple( the other, the green and ABC as well) --> DF/AD= a/48=3/5--> a= 3x48/5=144/5 Consider the green triangle: CE/DE= 3/4--> CE=3a/4 The rest is easy

[@montynorth3009]

Let side length of square = s. Tan alpha = 3/4 = 0.75. Alpha = 36.87 degrees. Beta will be 90 - 36.87 = 53.13. s = 48 sin Alpha = 28.80. Area Yellow = 1/2 x 48 x 28.80 x sin 53.13 = 552.96.. Area Green = 1/2 x 36 x 28.80 x sin 36.87 = 311.04. Area Pink = 28.80 x 28.80 = 829.44.

[@quigonkenny]

Assume ∠CAB = α and ∠BCA = β, where β = 90°- α. As ∠DAF = α, ∠FDA = β, and as ∠ECD = β, ∠CDE = α. Thus ∆AFD and ∆DEC are similar to ∆ABC. Let the side length of the square equal s, and the other unknown side lengths of the yellow and green triangles equal y (AF) and g (EC) respectively. s/48 = g/36 48g = 36s s = 48g/36 = 4g/3 ---> [1] s/36 = y/48 36y = 48s s = 36y/48 = 3y/4 ---> [2] Triangle ∆AFD: y² + s² = 48² = 2304 y² + [3y/4]² = 2304

[@andrepiotrowski5668]

You could have used 4 and 3 (instead of 48 [= 4 *12] and 36 [= 3*12] (much easier to calculate) and multiply the resulting areas by 144.

[@yalchingedikgedik8007]

Thanks Sir Thanks PreMath Very nice and knowledgeable ❤❤❤

[@humbertorodriguezperez1214]

Same rational, but I made the side of the square l = 4x, than the hipotenusa of the small triangle 36 = 5x so x = 7.2. The side l = 4 * 7.2, so l = 28.8 and the area of the square a1 = 829.44 The small triangle area a2 = 4x * 3x/2 So a2 = 14.4 * 21.6, a2 = 311.04 The base of the big triangle is B = (4/3) * l, B = 4 * 28.8/3, so B = 4 * 9.6, B = 38.4. The area of the big triangle is a3 = B * l/2, so a3 = 38.4 * 14.4, then a3 = 552.96 The area of the biggest triangle is the total sum of the 3 smaller areas Note: the side l is a different cathet in each of the different triangles, and the proporcion of the small cathets in the different triangles is 4:3. So is easy to conclude that the rectangle triangles are the 5:4:3 proporcion type

[@gelbkehlchen]

Solution: a = side of the square, b = horizontal side of the yellow triangle, c = vertical side of the green triangle. The two triangles have the same angles, are therefore similar and are right-angled. ⟹ (1) a/48 = c/36 (2) Pythagoras: c = √(36²-a²) |in (1) ⟹ (1a) a/48 = √(36²-a²)/36 |*36 ⟹ (1b) 36a/48 = √(36²-a²) |()² ⟹ (1c) 36²*a²/48² = 36²-a² |+a² ⟹ (1d) 36²*a²/48²+a² = 36² ⟹ (1e) (36²/48²+1)*a² = 36² ⟹ (1f) (36²+48²)/48²*a² = 36² |*48²/(36²+48²) ⟹ (1f) a² = 36²*48²/(36²+48²) = 829.44 = Area of ​​the square |in (2) ⟹ (2a) c = √(36²-a²) = √(36²-829.44) = 21.6 Side of the square = a = √829.44 = 28.8 ⟹ Area of ​​the green triangle = a*c/2 = 28.8*21.6/2 = 311.04 (3) Pythagoras: b = √(48²-a²) = √(48²-829.44) = 38.4 Area of ​​the yellow triangle = a*b/2 = 28.8*38.4/2 = 552.96

[@marcgriselhubert3915]

Yellow and green triangles are similar, so we have: c/36 = (sqrt(48^2 --c^2))/48, or 4.c = 3.sqrt(2304 -c^2) with c the side length of the squere. Then we have 16.c^2 = 20736 -9.c^2 and finally c^2 = 20736/25 which is the area of the square, and c = 144/5 The height EC of the green triangle is then sqrt(36^2 -c^2) = (sqrt(11664))/5 = 108/5, so the area of the green triangle is (1/2).(108/5).(144/5) = 7776/25. The basis AF of the yellow triangle is then sqrt(48^2 -c^2) = sqrt(36864/25) = 192/5, so the area of the yellow triangle is (1/2).(192/5).(144/5) = 13824/25.

[@unknownidentity2846]

Let's find the areas: . .. ... .... ..... It is obvious that the right triangles ABC, ADF and CDE are similar. With s being the side length of the square we can conclude: AF/DE = AF/s = AD/CD = 48/36 = 4/3 ⇒ AF = (4/3)*s DF/CE = s/CE = AD/CD = 48/36 = 4/3 ⇒ CE = (3/4)*s Since the triangles ADF and CDE are right triangles, we can apply the Pythagorean theorem: AF² + DF² = AD² [(4/3)*s]² + s² = 48² (16/9)*s² + s² = 48² (25/9)*s² = 48² ⇒ s² = 48²*9/25 = 20736/25 ⇒ s = √(20736/25) = 144/5 DE² + CE² = CD² s² + [(3/4)*s]² = 36² s² + (9/16)*s² = 36² (25/16)*s² = 36² ⇒ s² = 36²*16/25 = 20736/25 ✓ AF = (4/3)*s = (4/3)*(144/5) = 192/5 CE = (3/4)*s = (3/4)*(144/5) = 108/5 Now we can calculate the size of all areas: A(yellow) = A(ADF) = (1/2)*AF*s = (1/2)*(192/5)*(144/5) = 13824/25 = 552.96 A(green) = A(CDE) = (1/2)*CE*s = (1/2)*(108/5)*(144/5) = 7776/25 = 311.04 A(pink) = A(BEDF) = s² = 20736/25 = 829.44 Best regards from Germany

[@Marcus-y1m]

This one was easy. Thanks sir

[@jimlocke9320]

At 7:15, you have a value for x. I don't understand why you use the yellow triangle to compute x again.

[@Martin-hi7wx]

I used a proportionality theorem and trig. Cos and tan. Shorter and easier to explain

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال x=DE Y=CE z=AF cosDAF =z/48 =x/36 x^2 +z^2=48^2 x=144/5 z=192/5 y=108/5 S(CDE)=311,04 S(ADE)=552,96 S(BEDF)=829,44

[@gervaischouinard9809]

Why doing Pythagorean 2 Times cause X must be the same everywhere. As soon as we calculate it one, we are able to use it to calculate all missing Side lengths.

[@LuisdeBritoCamacho]

@@someonespadre 😂😂😂😂😂

[@pranavvenktesh3718]

I think it was for verification that the value of x remains constant

[@AmirgabYT2185]

S(yellow)=552,96 S(green)=311,04 S(square)=829,44

[@jamestalbott4499]

Thank you!

[@LuisdeBritoCamacho]

Let's go adventuring!! 1) Angle CAB = Angle CDE = alpha 2) FB = DE = DF EB = X 3) sin(alpha) = X/48 4) cos(alpha) = X/36 5) (sin(alpha))^2 + (cos(alpha))^2 = 1 6) (X/48)^2 + (X/36)^2 = 1 ; X^2/2.304 + X^2/1.296 = 1 ; X = - 144/5 or X = 144/5. Let's take the Positive Solution! 7) sin(alpha) = 144/240 = 3/5 8) cos(alpha) = 144/180 = 4/5 9) tan(alpha) = 3/4 10) AC = 48 + 36 = 84 11) 3/5 = CB/84 ; CB = 252/5 ~ 50,4 12) 4/5 = AB/84 ; AB = 336/5 ~ 67,2 13) Total Area of Triangle [ABC] = 84.672/50 = 42.336/25 ~ 1.693,44 14) Square Area = (144/5)^2 = 20.736/25 ~ 829,44 15) EC = 252/5 - 144/5 = 108/5 ~ 21,6 16) AF = 336/5 - 144/5 = 192/5 ~ 38,4 17) Area (CDE) = 2A = 144/5 * 108/5 ; 2A = 15.552/25 ; A = 15.552/50 ; A = 7.776/25 ~ 311,04 18) Area (ADF) = 2A = 144/5 * 192/5 ; 2A = 27.648/25 ; A = 27.648/50 ; A = 13.824/25 ~ 552,96 19) Area of Yellow Triangle = 13.824/25 Square Units 20) Area of Green Triangle = 7.776/25 Square Units 21) Area of Pink Square = 20.736/25 22) Check with Total Area = 42.336/25 = 13.824/25 + 7.776/25 + 20.736/5 ; 42.336/25 = (13.824 + 7.776 + 20.736)/25 ; 42.336 = 42.336 23) THE END

[@giuseppemalaguti435]

48/l=36/√(36^2-l^2)...l(lato del quadrato)=144/5

[@CloudBushyMath]

Enjoyable

[@msafasharhan]

Sir ,can you solve this without 4/3 ratio

[@garypaulson5202]

clever

[@tontonbeber4555]

Answer is YES, I can.

[@himadrikhanra7463]

Yellow portion 144 square unit ? Green portion 81 square unit ? Pink square 1476 square unit ?

[@infoskipper1]

Idon't understand how you décidé that DE=4/3x.

[@FlavioGaming]

Because DF is 4x/3 because the triangles are similar and DE = DF