Let's use an adapted orthonormal (as usual !!) and let's note c the length of the square. D(0;0) C(c;0) B(c;c) A(0;c) M(c/2;c) VectorDM(c/2;c) is colinear to VectorU(1;2) and orthogonal to VectorV(-2;1) VectorV is directing (ME), which equation is: (x - c/2).(1) - (y -c).(-2) = 0 or x +2.y -(5/2).c = 0 Tne intersection of (ME) with (BC), which equation is x = c, gives point E =(c, (3/4).c). Now we have VectorME(c - c/2; (3/4).c - c) or VectorME(c/2;-c/4), giving ME^2 = (c^2)/4 + (c^2)/16 = (5/16).(c^2), and ME = (sqrt(5)/4).c We also have VectorDM(c/2;c), so DM^2 = (c^2)/4 + (c^2) = (5/4).(c^2) and DM = (sqrt(5)/2).C The area of the green triangle DEM is (1/2).DM.ME = (1/2). ((sqrt(5)/2).c). ((sqrt(5)/4).c) = (5/16).(c^2). Or it is 555, so we have c^2 = (16/5).555 = 1776 which is the area of the square.
The triangles AMD and DME are similar and the ratio of the relevant sides is 2/sqrt5 so the area of AMD/ area of DME =4/5 ----> Area of AMD= (555x4)/5= 111x4 Area of the square= 4x111x4= 1776 sq units
Your conclusion that triangles AMD (to be correct should be DAM) and DME are similar is wonderful. No explanation is given in your edited comment. I would like to try to give an explanation. Angle ADM = angle BME from the given picture. Let it by A. For triangles DAM & DME, DM = DA/cosA, ME = MB/cos = AM/cosA (MB=AM as given). With the inclusive right angle, similarity test of 2 sides proportional with inclusive angle is satisfied.
@@hongningsuen1348 it is very simple: Notice that hypotenuse of the triangle AMD= long leg of the triangle DEM It is the same with the relevant hypotenuse vs short leg. So, the 2 triangles are similar with the ratio of the relevant side=2/sqrt5
Thanks for the fun math puzzle. You don't need Pythagoras; the three triangle areas are x^2, x^2/4 and 3x^2/2 add them to 555. they equal the square area 4x^2. solve for x.
At a quick glance: MD^2=5*x^2 where AM=x. Triangles AMD,MBE,ECD and MED are similar MB= 0.5 * AD then BE = 0.5* MB= 0.5*x and EC = 3/4*x. Area of AMD= x^2. Area of MBE= 0.25*x^2 and area DEC = 3/4*x^2. ME^2= x^2+ 0.25*x^2= 5/4*x^2. ME= sqrt(5/4)*x and MD = sqrt(5)*x. Using triangle area formula , 0.5* sqrt(5)*x * sqrt(5/4)*x =555 then x^2=2*555/(5*sqrt(1/4)) , .x=21. Area of Square = 4*x^2 1776 cm^2.
Very clever! I suggest labeling 4 s as the side of the square. By doing so, [AMD]= 4sq s, [BME]= sq s. and [CDF]= 6 sqs-----> [MDE]=5 sq s. It is easier to understand your idea. Thank you so much!
También se puede encontrar la solución sin calcular el valor de AB.- N es el punto medio de DE y O el punto medio de EC. P es la proyección ortogonal de E sobre MN. La razón de semejanza entre los triángulos EBM y MAD es s=1/2 ; longitudes relativas: AD=4; AM=MB=2; BE=1=2/2; EO=OC=3/2; MN divide a EMD en dos mitades de igual superficie, 555/2; MBON=2*MEN=555; PEON=555*3/5=333 ; Área ABCD =2(MBON+PEON) =2(555+333)=1776. Interesante acertijo. Gracias y saludos cordiales.
Let's name the Length of the Square, 4a. So, the Area of the Square is A = 16a^2. 1) Triangle [ADM] = (4a * 2a) / 2 = 8a^2 / 2 = 4a^2 2) Triangle [BEM] = (2a * a) / 2 = 2a^2 / 2 = a^2 3) Triangle [CDE] =(3a * 4a) / 2 = 12a^2 / 2 = 6a^2 Now: 16a^2 = 555 + 4a^2 + a^2 + 6a^2 16a^2 = 555 + 11a^2 16a^2 - 11a^2 = 555 5a^2 = 555 a^2 = 555 / 5 a^2 = 111 If: a^2 = 111 Then: 16 * a^2 = 16 * 111 Conclusion: A = 16 * 111 sq cm = 1776 sq cm Answer: The Area of the Square (wich is a Symbol of Free Masonry - 6 Squares = 1 Cube = House), is the same as the year of the Constitution of the United States of America; 1.776 (4 of July). P.S. - Quite interesting!!
Call the square's sides each 2x, as it might help to minimise fractions. It looks like this might one of those where you compare similar triangles. AMD is similar to DME, and I think EBM qualifies too. AMD is one quarter of the square and is and is (2x*x)/2, so x^2 I think EBM is similar, so that would have an area of (x*(1/2)x)/2 (so much for avoiding fractions :) ). That would be an area of (1/4)x^2 DCE (definitely NOT similar) has an area of ((2x*(3/2)x)/2 which is (3/2)x^2. This gives a total are for the square of x^2 + (1/4)x^2 + (3/2)x^2 + 555 x^2 + (1/4)x^2 + (3/2)x^2 + 555 = 4x^2 x^2 + (1/4)x^2 + (3/2)x^2 = (11/4)x^2 Therefore, 555 = (5/4)x^2 As (5/4)x^2 = 555, (11/4)x^2 = 1221 (this is calculated as 111 times the numerator). Area of square is 1221+555=1776 sq units. This solution turned out to be very unlike what I was expecting, and it does depend on me being correct about the triangle similarities. Now to watch the video. We took slightly different paths, and I saw ways of simplifying as I went. I could not have done these a year or so ago without your tuition so, once again, thank you.
Label AM = x. Then, by definition of midpoint, BM = x. Therefore, AB = AD = BC = CD = 2x by definition of a square. A = s² = (2x)² = 4x² So, square ABCD has an area of 4x² cm². Let α & β be the measures of complementary angles. Let m∠ADM = α. Then, because ∠A is a right angle by definition of squares, m∠AMD = β. Since ∠DME is a right angle, m∠BME = α. And ∠B is also a right angle by definition of squares, so m∠BEM = β. Thus, △ADM ~ △BME by AAA Similarity. There will be a proportion. Label BE = y. AM/AD = BE/BM x/2x = y/x 1/2 = y/x y = x/2 Therefore, BE = x/2 cm. Apply the Pythagorean Theorem twice. a² + b² = c² x² + (x/2)² = (EM)² x² + x²/4 = (EM)² (EM)² = 5x²/4 EM = √(5x²/4) = (x√5)/2 (2x)² + x² = (DM)² 4x² + x² = (DM)² (DM)² = 5x² DM = √(5x²) = x√5 A = 1/2 * b * h 555 = 1/2 * x√5 * (x√5)/2 = (x√5)/2 * (x√5)/2 = 5x²/4 2220 = 5x² 444 = x² x = √444 = √(2 * 2 * 3 * 37) = 2√111 So, x = 2√111 cm. Input into the square area formula. A = 4x² = 4(2√111)² = 4 * 444 = 1776 So, the area of square ABCD is 1776 square centimeters.
BTW, the _first sentence_ in the description is from a previous video. There is no golden shaded square, nor are there isosceles triangles. Just thought I'd point it out.
It looks like we all kind of ended up in the same spot. I set [𝒔 = 2], reflecting the division of the top horizontal into 2 equal halves. [1] each. Then solving the inner green △ was straight forward. DM = √(1² ⊕ 2²) DM = √5 Because △AMD is similar to △BME (having same angles), then by proportionality: ME = ½DM ME = √5/2 Next in the game of similar parts, need to figure the proportionate area of my re-scaled △DME: Area DME = ½ base • height Area DME = ½√5 • √5/2 Area DME = ⁵⁄₄ Almost done! Since my rescaled □ABCD had [𝒔 = 2], it then had [area = 2² = 4]. The keys to the final solution are in our hands! Therefore … the as-shown square which has a △DME area of 555 inside must have a proportionate area, too. Area □ABCD = 4 ÷ ⁵⁄₄ × 555 Area □ABCD = 4 × 4 × 111 Area □ABCD = 1,776 u² Tada. Done. By proportionality scaling. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
I absolutely love it! 🙂
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Area of the square ABCD=1776cm^2
Let's use an adapted orthonormal (as usual !!) and let's note c the length of the square.
D(0;0) C(c;0) B(c;c) A(0;c) M(c/2;c) VectorDM(c/2;c) is colinear to VectorU(1;2) and orthogonal to VectorV(-2;1)
VectorV is directing (ME), which equation is: (x - c/2).(1) - (y -c).(-2) = 0 or x +2.y -(5/2).c = 0 Tne intersection of (ME) with (BC), which equation is x = c, gives point E =(c, (3/4).c). Now we have VectorME(c - c/2; (3/4).c - c) or VectorME(c/2;-c/4), giving ME^2 = (c^2)/4 + (c^2)/16 = (5/16).(c^2), and ME = (sqrt(5)/4).c
We also have VectorDM(c/2;c), so DM^2 = (c^2)/4 + (c^2) = (5/4).(c^2) and DM = (sqrt(5)/2).C
The area of the green triangle DEM is (1/2).DM.ME = (1/2). ((sqrt(5)/2).c). ((sqrt(5)/4).c) = (5/16).(c^2). Or it is 555, so we have c^2 = (16/5).555 = 1776 which is the area of the square.
The triangles AMD and DME are similar and the ratio of the relevant sides is 2/sqrt5
so the area of AMD/ area of DME =4/5 ----> Area of AMD= (555x4)/5= 111x4
Area of the square= 4x111x4= 1776 sq units
Excellent!
Thanks ❤️
Your conclusion that triangles AMD (to be correct should be DAM) and DME are similar is wonderful. No explanation is given in your edited comment. I would like to try to give an explanation.
Angle ADM = angle BME from the given picture. Let it by A.
For triangles DAM & DME, DM = DA/cosA, ME = MB/cos = AM/cosA (MB=AM as given). With the inclusive right angle, similarity test of 2 sides proportional with inclusive angle is satisfied.
@@hongningsuen1348 it is very simple:
Notice that hypotenuse of the triangle AMD= long leg of the triangle DEM
It is the same with the relevant hypotenuse vs short leg.
So, the 2 triangles are similar with the ratio of the relevant side=2/sqrt5
this is absolutely helpful! thanks :))!!
Glad it was helpful!
You are very welcome!
Thanks ❤️
Thanks for the fun math puzzle. You don't need Pythagoras; the three triangle areas are x^2, x^2/4 and 3x^2/2 add them to 555. they equal the square area 4x^2. solve for x.
Give us lots of video aboutn Olymlaid math ❤❤you Professor.
Keep watching!
Thanks ❤️
Thank you! 📐
At a quick glance: MD^2=5*x^2 where AM=x. Triangles AMD,MBE,ECD and MED are similar MB= 0.5 * AD then BE = 0.5* MB= 0.5*x and EC = 3/4*x. Area of AMD= x^2. Area of MBE= 0.25*x^2 and area DEC = 3/4*x^2. ME^2= x^2+ 0.25*x^2= 5/4*x^2. ME= sqrt(5/4)*x and MD = sqrt(5)*x. Using triangle area formula , 0.5* sqrt(5)*x * sqrt(5/4)*x =555 then x^2=2*555/(5*sqrt(1/4)) , .x=21. Area of Square = 4*x^2 1776 cm^2.
بسیار عالی
ADM♾️BME
ADM=4s BME=s CDE=6s
square ABCD=ADM＊4=4s＊4=16s
MDE=16s-(4s+6s+s)=5s=555
s=111
area of square ABCM : 16＊111=1776cm^2
Very clever! I suggest labeling 4 s as the side of the square. By doing so, [AMD]= 4sq s, [BME]= sq s. and [CDF]= 6 sqs-----> [MDE]=5 sq s. It is easier to understand your idea. Thank you so much!
También se puede encontrar la solución sin calcular el valor de AB.-
N es el punto medio de DE y O el punto medio de EC. P es la proyección ortogonal de E sobre MN.
La razón de semejanza entre los triángulos EBM y MAD es s=1/2 ; longitudes relativas: AD=4; AM=MB=2; BE=1=2/2; EO=OC=3/2;
MN divide a EMD en dos mitades de igual superficie, 555/2; MBON=2*MEN=555; PEON=555*3/5=333 ;
Área ABCD =2(MBON+PEON) =2(555+333)=1776.
Interesante acertijo. Gracias y saludos cordiales.
شكرا لكم على المجهودات
يمكن استعمال
tanADM = 1/2
tanBME=1/2
BE=x/2
CE= 3x/2
(2x)^2=مجموع مساحات المثلثات
داخل ABCD
نجد الحل 1776
Let's name the Length of the Square, 4a. So, the Area of the Square is A = 16a^2.
1) Triangle [ADM] = (4a * 2a) / 2 = 8a^2 / 2 = 4a^2
2) Triangle [BEM] = (2a * a) / 2 = 2a^2 / 2 = a^2
3) Triangle [CDE] =(3a * 4a) / 2 = 12a^2 / 2 = 6a^2
Now:
16a^2 = 555 + 4a^2 + a^2 + 6a^2
16a^2 = 555 + 11a^2
16a^2 - 11a^2 = 555
5a^2 = 555
a^2 = 555 / 5
a^2 = 111
If: a^2 = 111
Then: 16 * a^2 = 16 * 111
Conclusion: A = 16 * 111 sq cm = 1776 sq cm
Answer:
The Area of the Square (wich is a Symbol of Free Masonry - 6 Squares = 1 Cube = House), is the same as the year of the Constitution of the United States of America; 1.776 (4 of July).
P.S. - Quite interesting!!
Call the square's sides each 2x, as it might help to minimise fractions.
It looks like this might one of those where you compare similar triangles.
AMD is similar to DME, and I think EBM qualifies too.
AMD is one quarter of the square and is and is (2x*x)/2, so x^2
I think EBM is similar, so that would have an area of (x*(1/2)x)/2 (so much for avoiding fractions :) ).
That would be an area of (1/4)x^2
DCE (definitely NOT similar) has an area of ((2x*(3/2)x)/2 which is (3/2)x^2.
This gives a total are for the square of
x^2 + (1/4)x^2 + (3/2)x^2 + 555
x^2 + (1/4)x^2 + (3/2)x^2 + 555 = 4x^2
x^2 + (1/4)x^2 + (3/2)x^2 = (11/4)x^2
Therefore, 555 = (5/4)x^2
As (5/4)x^2 = 555, (11/4)x^2 = 1221 (this is calculated as 111 times the numerator).
Area of square is 1221+555=1776 sq units.
This solution turned out to be very unlike what I was expecting, and it does depend on me being correct about the triangle similarities.
Now to watch the video.
We took slightly different paths, and I saw ways of simplifying as I went.
I could not have done these a year or so ago without your tuition so, once again, thank you.
MH ⟂ DE. AMD = MDH, BME = HME. AMD + BME = DME = 1/4 + 1/16 = 5/16 = 555.
S(ABCD) = 16/5 × 555 = 1776.
If ∠ADM = α and ∠DMA = β, where β = 90°- α, thwn as ∠EMD = 90°, ∠BME = α, which means ∠MEB = β and ∆EBM and ∆MAD are similar. Let s be the side length of the square.
Triangle ∆MAD:
MA² + AD² = DM²
(s/2)² + s² = DM²
s²/4 + s² = DM²
DM² = 5s²/4
DM = √(5s²/4) = √5s/2
ME/BM = DM/AD
ME/(s/2) = (√5s/2)/s
ME(s) = (s/2)(√5s/2) = √5s²/4
ME = (√5s²/4)/s = √5s/4
Triangle ∆EMD:
A = bh/2 = MD(EM)/2
555 = (√5s/2)(√5s/4)/2
555 = 5s²/16
s² = 555(16/5) = 1776
Square ABCD:
A = s² = 1776 cm²
Area of the square ABCD=(4√111)^2=1776 cm^2.❤❤❤ Thakns
Excellent!
You are very welcome!
Thanks ❤️
Solve this question
Square ABCD has area 100
E is mid point of AB and F is midpoint of BC. AF and DE meet at G . Find the area of triangle DFG .
That was easy
Label AM = x. Then, by definition of midpoint, BM = x.
Therefore, AB = AD = BC = CD = 2x by definition of a square.
A = s²
= (2x)²
= 4x²
So, square ABCD has an area of 4x² cm².
Let α & β be the measures of complementary angles.
Let m∠ADM = α. Then, because ∠A is a right angle by definition of squares, m∠AMD = β.
Since ∠DME is a right angle, m∠BME = α.
And ∠B is also a right angle by definition of squares, so m∠BEM = β.
Thus, △ADM ~ △BME by AAA Similarity. There will be a proportion. Label BE = y.
AM/AD = BE/BM
x/2x = y/x
1/2 = y/x
y = x/2
Therefore, BE = x/2 cm.
Apply the Pythagorean Theorem twice.
a² + b² = c²
x² + (x/2)² = (EM)²
x² + x²/4 = (EM)²
(EM)² = 5x²/4
EM = √(5x²/4)
= (x√5)/2
(2x)² + x² = (DM)²
4x² + x² = (DM)²
(DM)² = 5x²
DM = √(5x²)
= x√5
A = 1/2 * b * h
555 = 1/2 * x√5 * (x√5)/2
= (x√5)/2 * (x√5)/2
= 5x²/4
2220 = 5x²
444 = x²
x = √444
= √(2 * 2 * 3 * 37)
= 2√111
So, x = 2√111 cm. Input into the square area formula.
A = 4x²
= 4(2√111)²
= 4 * 444
= 1776
So, the area of square ABCD is 1776 square centimeters.
BTW, the _first sentence_ in the description is from a previous video. There is no golden shaded square, nor are there isosceles triangles. Just thought I'd point it out.
x est le coté du carré X =x^2 X-(X/4+X/16+X.3/8) = 555 X=1766 😀
This would make a nice puzzle for Americans on Independence Day.
Right on!
Glad to hear that!
Thanks ❤️
asnwer=1500cm isit
1776
Thanks ❤️
I will say 1776 square units.
⊿ADM ~ ⊿BME. AD = 2AM ⇒ MD = 2ME ⇒ ½ ME 2ME = ME² = 555 ⇒ ME = √555 ⇒ BE = √555 / √5 = √111 ⇒ AB = 4√111 ⇒ area □ABCD = (4√111)² = 1776 cm²
Thanks ❤️
1776; very patriotic!😊
It looks like we all kind of ended up in the same spot. I set [𝒔 = 2], reflecting the division of the top horizontal into 2 equal halves. [1] each. Then solving the inner green △ was straight forward.
DM = √(1² ⊕ 2²)
DM = √5
Because △AMD is similar to △BME (having same angles), then by proportionality:
ME = ½DM
ME = √5/2
Next in the game of similar parts, need to figure the proportionate area of my re-scaled △DME:
Area DME = ½ base • height
Area DME = ½√5 • √5/2
Area DME = ⁵⁄₄
Almost done! Since my rescaled □ABCD had [𝒔 = 2], it then had [area = 2² = 4]. The keys to the final solution are in our hands!
Therefore … the as-shown square which has a △DME area of 555 inside must have a proportionate area, too.
Area □ABCD = 4 ÷ ⁵⁄₄ × 555
Area □ABCD = 4 × 4 × 111
Area □ABCD = 1,776 u²
Tada. Done. By proportionality scaling.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
1776