Can you find the area of each shaded region? | (Fun Geometry problem) |
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- Опубліковано 15 вер 2024
- Learn how to find the area of the Yellow and Green triangles along with the pink square area. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of a triangle formula; similar triangles; right triangles; area of a square formula. Step-by-step tutorial by PreMath.com
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The results show quite good congruence! 🙂
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
∆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.
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
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
Thanks Sir
Thanks PreMath
Very nice and knowledgeable
❤❤❤
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.
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
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
S(yellow)=552,96
S(green)=311,04
S(square)=829,44
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
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
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
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
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.
This one was easy. Thanks sir
At 7:15, you have a value for x. I don't understand why you use the yellow triangle to compute x again.
I used a proportionality theorem and trig. Cos and tan. Shorter and easier to explain
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.
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.
@@someonespadre 😂😂😂😂😂
I think it was for verification that the value of x remains constant
Answer is YES, I can.
Typical puzzle of square inscribed in right-angled triangle, but change known and unknown figures, let s×s be the square, a×s be the upper triangle and s×b be the lower triangle, then a/s=s/b=(a+s)/(s+b)), so ab=s^2, a^2+s^2=36^2, s^2+b^2=48^2, (a+s)^2+(s+b)^2=84^2😅😅😅😅😅
Continued 😮, a/s=s/b=3/4, so rewrite upper triangle be 3a×3b, lower triangle be 4a×4b, where 3b=4a, so 36^2=(3a)^2+(3b)^2=(3a)^2+(4a)^2=25a^2, a^2=(36/5)^2, a=36/5, b=(4/3)(36/5)=48/5, therefore the yellow area is 1/2×4×(36/5)×4×(48/5)=(8×36×4 8)/25=13824/25, Green area is 1/2 ×3×(36/5)×3×(48/5)=(9×36×48)/50=15552/50, pink area is (3×(48/5))^2=(9×48^2)/25=20736/25.😊😊😊😊😊
48/l=36/√(36^2-l^2)...l(lato del quadrato)=144/5
Enjoyable
Thank you!
clever
Sir ,can you solve this without 4/3 ratio
Idon't understand how you décidé that DE=4/3x.