I got the purple area to be (b^3/2a). Which together with your answer implies that b = a*sqrt(3). Which implies that the triangles must be 30-60-90 triangles for this set-up to be possible.
Yep, you can see that G is the midpoint of AC, since FGCB is made up of two congruent triangles, each of area ab/2, meaning GC must be length b and AC is 2b. ∆ADC is thus a 30-60-90. By extension, all the other triangles in the diagram are 30-60-90 as well.
I recognized, that AF=FC, and that AC=2b. Hence AC/BC=2b/b=2. So we have the triangles to be special case 30,60,90 triangles. Asked area = 1.5*a*b. Thanks for sharing this interesting geometric puzzle 👍
Yes. The PreMath answer of 3ab/2 ONLY applies if the triangles are 30-60-90 right triangles. The area formula of A =(0.5)b(a + sqrt(a^2+b^2)) works for all values of a and b.
@@thewolfdoctor761 Yes it does. The area of triangle ADC is (0.5)b(a + sqrt(a^2+b^2)) and that is the same area as the stated expression i.e., b*[(a^2+b^2)]^(0.5)-(ab/2). If you equate these two expressions, you end up with b*sqrt(a^2 + b^2) = 2ab. If you substitute that in the stated expression, you get the area of the purple shaded area as 3ab/2.
The above answer is correct, and is the General solution for any value of a or b. The PreMath answer of 3ab/2 ONLY applies if the triangles are special 30-60-90 right triangles.
@@geraldgiannotti8364 Can you come up with any set-up that has "a" and "b" values for which all the requirements of the problem are met and for which b*[(a^2+b^2)]^(0.5)-(ab/2) is a solution and 3ab/2 is not a solution?
@@AdemolaAderibigbe-j8s Yes. The area formula of A =(0.5)b(a + sqrt(a^2+b^2)) works for all values of a and b. The solution of 3ab/2 is only obtained when b=sqrt(3)*a. This constraint of b/a = sqrt(3) makes all the triangles 30-60-90 right triangles
*Solução:* Por Pitágoras no ∆AQF: AF² = a² + b² → AF = (a² + b²)½ A área [AFED] = BC × AF *[AFED] = b(a² + b²)½* A área do triângulo [AFG] = ab/2 A área purple shaded é: [AFED] - [AFG] = = *_b[(a² + b²)½ - a/2]_*
Caso queira brincar mais nas equações, o professor provou que a área sombreada roxa é 3ab/2. Daí, b [(a² + b²)½ - a/2] = 3ab/2 (a² + b²)½ - a/2 = 3a/2 (a² + b²)½ = 3a/2 + a/2 = 2a a² + b² = 4a² → 3a² = b² → b=a√3. Portanto, a área sombreada roxa pode ser dada por : 3ab/2 = *3√3a²/2.*
Purple area = rectangle ADEF - half of the green rectangle = half of the rectangle ABCD b√(a^2 + b^2) - ab/2 = b(a + √(a^2 + b^2))/2 Let √(a^2 + b^2) = L bL - ab/2 = b(a + L)/2 ab = bL/2 L = 2a Purple area = b(2a) - ab/2 = 3ab/2
You can also express the answer in terms of a or b alone, since b = a.sqrt(3): Area = (3a^2.sqrt(3))/2. If you want to be even more picky, you could write: (3^(3/2).a^2)/2. Area = (b^2.sqrt(3))/2 The answer can also be expressed as follows: Area = b^3/ (2a) Substituting b^2 = 3a^2, one arrives at Premath's answer. Cheers!
This is a very good problem to test your ability to stick only to the information provided. Triangle AFQ is not necessarily a 30-60-90 triangle and could be anything. Quadrilateral AGFQ is not a rectangle in general. Any solutions that assume that AGFQ is a rectangle will just be valid for the special case of AFQ being a 30-60-90 triangle, but not correct in general. I get the purple area to be in general 0.75*b*sqrt(a^2+b^2) and not an invariant.
▲EPC переносим в равный ему ▲GFP. Видим, что получилась фигура, составленная из прямоугольника и дельтоида с равной площадью (составленные фактически каждый из пары треугольников-половинок), а лиловая площадь S(ACD)=S(АВСВ)/2, обозначим просто как S. Общая площадь 2ab. Видим, что 3 из 4 этих треугольников занимают вторую половину площади большого, равную искомой, т. е. S=¾*2ab=3ab/2.
Draw FC. As FG = BF = a, ∠CBF = ∠FGC = 90°, and FC is common, then ∆FGC and ∆CBF are congruent triangles. As GC = CB = b and AG = FQ = b, then AC = 2b. As ∠BAC = ∠GAF and ∠AGF = ∠CBA = 90°, then ∆AGF and ∆CBA are similar triangles. FA/GF = AC/CB FA/a = 2b/b = 2 FA = 2a The purple shaded area is equal to the area of the rectangle ADEF minus the area of triangle ∆AGF. Purple shaded area: A = lw - bh/2 A = FA(AD) - AG(GF)/2 A = 2a(b) - b(a)/2 A = 2ab - ab/2 [ A = 3ab/2 sq units ]
When you knew that purple area is 1/2 of area of rectangle ABCD, you can calculate the area of the rectangle ABCD. How? AB is equal AF + FB. Lenght of FB is equal a, and lenght of AF you can calculate from right triangle AQF. When you use of Pythagorean theroem lenght of AF is equal sqrt of (a^2 + b^2). So lenght of AB is equal [sqrt (a^2 + b^2) + a]. When you multiply it by 1/2 of b you will have an area of purple figure.
since area triangle PEC = area triangle PGF then answer is area triangle ADC its area is half height times base. since AD = BC then height is b since DC = AB then base is AF + FB FB is a and we can calculate AF using Pythagorean Theorem or (a^2+b^2)^(1/2) although my answer is more complicated it was derived faster and has fewer steps does it simplify to your answer?
AB = AF + FB = sqrt(a^2 + b^2) + a, so the big rectangle area is (sqrt(a^2 + b^2) + a).b and the purple shaded area is ((sqrt(a^2 + b^2) + a).b - (3/2).a.b This result is less "nice" than the result you found but it is immediate to obtain. Now let's see why these results are really the same: (sqrt(a^2 + b^2) + a).b -(3/2).a.b = (3/2).a.b is equivalent to sqrt(a^2 + b^2) + a = 3.a or sqrt(a^2 + b^2) = 2.a or a^2 + b^2 = 4.a^2 or b^2 = 3.a^2 or b = sqrt(3).a This is the condition other persons already found. The initial drawing must verify b = sqrt(3).a to be constructed. For example t = angleFCB must be equal to 30° (tan(t) = a/b = sqrt(3)/3).
So we could continue the math: if purple area = 3ab/2, then dimensions of rectangle are b and 3a, so AF=2a. Then with the triangle, sides of a and b, hypotenuse of 2a, only works if b=a√3, which means it must be a 30-60-90 triangle.
1/ The two triangles FGP and CEP are congruent--> Area of the purple region= 1/2 area of the big rectangle. 2/ Note that AF= FC-> the triangle AFC is an isosceles one, which means that the three triangles AFG=CFG=CFB -> Area of the purple region= 3 ab/2😅😅😅
Let's find the area: . .. ... .... ..... First of all we observe that the triangles AFG and CEP are similar (∠AGF=∠CEP=90° ∧ ∠ECP=∠FAG). So we can conclude: EP/CE = FG/AG EP/a = a/b ⇒ EP = a²/b The triangles CEP and FGP are congruent (CE=FG=a ∧ ∠CEP=∠FGP=90° ∧ ∠CPE=∠FPG). Therefore we know that CP=FP. Now we apply the Pythagorean theorem to the right triangle CEP: CP² = CE² + EP² FP² = CE² + EP² (EF − EP)² = CE² + EP² EF² − 2*EF*EP + EP² = CE² + EP² EF² − 2*EF*EP = CE² b² − 2*b*(a²/b) = a² b² − 2a² = a² b² = 3a² ⇒ b = √3a Now we are able to calculate the area of the purple region: AF² = AQ² + FQ² = a² + b² = a² + 3a² = 4a² ⇒ AF = 2a A(purple) = A(ADEF) − A(AFG) = AF*AD − (1/2)*AG*FG = (2a)*b − (1/2)*b*a = 2ab − ab/2 = 3ab/2 = (3√3/2)a² Best regards from Germany
Solution: Triangle FGP is congruent to Triangle CEP Therefore: GP = EP FG = CE FP = CP Like that, we conclude that Purple Area is half of Rectangle ABCD Area Purple Area = ½ Rectangle ABCD Area ... ¹ The Triangles AGF and CEP are similar, so we are going to use proportions EP/CE = FG/AG EP/a = a/b EP = a²/b Now, applying the Pythagorean Theorem in Triangle CEP CE² + EP² = CP² But CP = FP and FP = EF - EP CE² + EP² = FP² CE² + EP² = (EF - EP)² CE² + EP² = EF² - 2 EF . EP + EP² CE² = EF² - 2 EF . EP a² = b² - 2 . b . a²/b a² = b² - 2a² 3a² = b² b² = 3a² b = a√3 AG² + FG² = AF² b² + a² = AF² AF² = (a√3)² + a² AF² = 3a² + a² AF² = 4a² AF = 2a AB = AF + BF AB = 2a + a AB = 3a Substituting in ¹ Purple Area = ½ length × width Purple Area = ½ AB × BC Purple Area = ½ 3a × b Purple Area = 3ab/2 Square Units ✅
I got the purple area to be (b^3/2a). Which together with your answer implies that b = a*sqrt(3). Which implies that the triangles must be 30-60-90 triangles for this set-up to be possible.
Yep, you can see that G is the midpoint of AC, since FGCB is made up of two congruent triangles, each of area ab/2, meaning GC must be length b and AC is 2b. ∆ADC is thus a 30-60-90. By extension, all the other triangles in the diagram are 30-60-90 as well.
Thanks for sharing Sir ❤❤❤
I recognized, that AF=FC, and that AC=2b.
Hence AC/BC=2b/b=2.
So we have the triangles to be special case 30,60,90 triangles.
Asked area = 1.5*a*b.
Thanks for sharing this interesting geometric puzzle 👍
Final: S=3ab/2 and b^2=3a^2
AF^2=a^2+b^2, S(ACD)=0,5*b*{sqr(a^2+b^2)+a}
Thank you!
There is some inconsistency evident here.
Yes. The PreMath answer of 3ab/2 ONLY applies if the triangles are 30-60-90 right triangles. The area formula of A =(0.5)b(a + sqrt(a^2+b^2)) works for all values of a and b.
b*[(a^2+b^2)]^(0.5)-(ab/2) ???
That's what I got, so does this equal 3ab/2 ?
@@thewolfdoctor761 Yes it does. The area of triangle ADC is (0.5)b(a + sqrt(a^2+b^2)) and that is the same area as the stated expression i.e., b*[(a^2+b^2)]^(0.5)-(ab/2). If you equate these two expressions, you end up with b*sqrt(a^2 + b^2) = 2ab. If you substitute that in the stated expression, you get the area of the purple shaded area as 3ab/2.
The above answer is correct, and is the General solution for any value of a or b. The PreMath answer of 3ab/2 ONLY applies if the triangles are special 30-60-90 right triangles.
@@geraldgiannotti8364 Can you come up with any set-up that has "a" and "b" values for which all the requirements of the problem are met and for which b*[(a^2+b^2)]^(0.5)-(ab/2) is a solution and 3ab/2 is not a solution?
@@AdemolaAderibigbe-j8s Yes. The area formula of A =(0.5)b(a + sqrt(a^2+b^2)) works for all values of a and b. The solution of 3ab/2 is only obtained when b=sqrt(3)*a. This constraint of b/a = sqrt(3) makes all the triangles 30-60-90 right triangles
FC=FA---> ABC=3ab/2---> ABCD=3ab---> AFED=2ab---> AGFED=2ab-(ab/2) =3ab/2.
Gracias y saludos.
Why ABC=3ab/2
ABC=FGA+FGC+FBC=3ab/2.
Un saludo
*Solução:*
Por Pitágoras no ∆AQF:
AF² = a² + b² → AF = (a² + b²)½
A área [AFED] = BC × AF
*[AFED] = b(a² + b²)½*
A área do triângulo [AFG] = ab/2
A área purple shaded é:
[AFED] - [AFG] =
= *_b[(a² + b²)½ - a/2]_*
Caso queira brincar mais nas equações, o professor provou que a área sombreada roxa é 3ab/2. Daí,
b [(a² + b²)½ - a/2] = 3ab/2
(a² + b²)½ - a/2 = 3a/2
(a² + b²)½ = 3a/2 + a/2 = 2a
a² + b² = 4a² → 3a² = b² → b=a√3. Portanto, a área sombreada roxa pode ser dada por :
3ab/2 = *3√3a²/2.*
Purple area = rectangle ADEF - half of the green rectangle = half of the rectangle ABCD
b√(a^2 + b^2) - ab/2 = b(a + √(a^2 + b^2))/2
Let √(a^2 + b^2) = L
bL - ab/2 = b(a + L)/2
ab = bL/2
L = 2a
Purple area = b(2a) - ab/2 = 3ab/2
You can also express the answer in terms of a or b alone, since b = a.sqrt(3):
Area = (3a^2.sqrt(3))/2. If you want to be even more picky, you could write: (3^(3/2).a^2)/2.
Area = (b^2.sqrt(3))/2
The answer can also be expressed as follows:
Area = b^3/ (2a)
Substituting b^2 = 3a^2, one arrives at Premath's answer.
Cheers!
That's what I got. Pre-math's conclusion was in complete in my opinion. b = a * sqrt3. Therefore the area is A = a^2*3/2*sqrt3.
I believe the triangles are 30°, 60°, 90°. Therefor, the solution can be Area = (√(3)/2)*b².
∆AQF is Similar to ∆ADC. Hence [∆ADC]=(b/a)² *[∆AQF]=b³/2a
This is a very good problem to test your ability to stick only to the information provided. Triangle AFQ is not necessarily a 30-60-90 triangle and could be anything. Quadrilateral AGFQ is not a rectangle in general. Any solutions that assume that AGFQ is a rectangle will just be valid for the special case of AFQ being a 30-60-90 triangle, but not correct in general. I get the purple area to be in general 0.75*b*sqrt(a^2+b^2) and not an invariant.
Bom dia Mestre
Forte Abraço aqui do Rio de Janeiro
▲EPC переносим в равный ему ▲GFP. Видим, что получилась фигура, составленная из прямоугольника и дельтоида с равной площадью (составленные фактически каждый из пары треугольников-половинок), а лиловая площадь S(ACD)=S(АВСВ)/2, обозначим просто как S. Общая площадь 2ab. Видим, что 3 из 4 этих треугольников занимают вторую половину площади большого, равную искомой, т. е. S=¾*2ab=3ab/2.
Draw FC. As FG = BF = a, ∠CBF = ∠FGC = 90°, and FC is common, then ∆FGC and ∆CBF are congruent triangles.
As GC = CB = b and AG = FQ = b, then AC = 2b. As ∠BAC = ∠GAF and ∠AGF = ∠CBA = 90°, then ∆AGF and ∆CBA are similar triangles.
FA/GF = AC/CB
FA/a = 2b/b = 2
FA = 2a
The purple shaded area is equal to the area of the rectangle ADEF minus the area of triangle ∆AGF.
Purple shaded area:
A = lw - bh/2
A = FA(AD) - AG(GF)/2
A = 2a(b) - b(a)/2
A = 2ab - ab/2
[ A = 3ab/2 sq units ]
30, 60, 90 triangles is the (square root of 3/2)*b^2
That was AAS congruency, not ASA, not that it matters too much
That is only true if the triangle AQF is a triangle with angles 30°- 60°- 90°
When you knew that purple area is 1/2 of area of rectangle ABCD, you can calculate the area of the rectangle ABCD. How? AB is equal AF + FB. Lenght of FB is equal a, and lenght of AF you can calculate from right triangle AQF. When you use of Pythagorean theroem lenght of AF is equal sqrt of (a^2 + b^2). So lenght of AB is equal [sqrt (a^2 + b^2) + a]. When you multiply it by 1/2 of b you will have an area of purple figure.
since area triangle PEC = area triangle PGF then answer is area triangle ADC
its area is half height times base.
since AD = BC then height is b
since DC = AB then base is AF + FB
FB is a and we can calculate AF using Pythagorean Theorem or (a^2+b^2)^(1/2)
although my answer is more complicated it was derived faster and has fewer steps
does it simplify to your answer?
AB = AF + FB = sqrt(a^2 + b^2) + a, so the big rectangle area is (sqrt(a^2 + b^2) + a).b and the purple shaded area is ((sqrt(a^2 + b^2) + a).b - (3/2).a.b
This result is less "nice" than the result you found but it is immediate to obtain.
Now let's see why these results are really the same: (sqrt(a^2 + b^2) + a).b -(3/2).a.b = (3/2).a.b is equivalent to sqrt(a^2 + b^2) + a = 3.a
or sqrt(a^2 + b^2) = 2.a or a^2 + b^2 = 4.a^2 or b^2 = 3.a^2 or b = sqrt(3).a This is the condition other persons already found. The initial drawing must
verify b = sqrt(3).a to be constructed. For example t = angleFCB must be equal to 30° (tan(t) = a/b = sqrt(3)/3).
area AQFG- 1/2 ab is the answer, AQFG area =( root of a square + b square( Pythagorus))Xb, answer is b( root {a square+ b square}-1/2 a}?
So we could continue the math: if purple area = 3ab/2, then dimensions of rectangle are b and 3a, so AF=2a. Then with the triangle, sides of a and b, hypotenuse of 2a, only works if b=a√3, which means it must be a 30-60-90 triangle.
You are right, there are lots of 30° and 60° angles in the given drawing.
1/ The two triangles FGP and CEP are congruent--> Area of the purple region= 1/2 area of the big rectangle.
2/ Note that AF= FC-> the triangle AFC is an isosceles one, which means that the three triangles AFG=CFG=CFB
-> Area of the purple region= 3 ab/2😅😅😅
Let's find the area:
.
..
...
....
.....
First of all we observe that the triangles AFG and CEP are similar (∠AGF=∠CEP=90° ∧ ∠ECP=∠FAG). So we can conclude:
EP/CE = FG/AG
EP/a = a/b
⇒ EP = a²/b
The triangles CEP and FGP are congruent (CE=FG=a ∧ ∠CEP=∠FGP=90° ∧ ∠CPE=∠FPG). Therefore we know that CP=FP. Now we apply the Pythagorean theorem to the right triangle CEP:
CP² = CE² + EP²
FP² = CE² + EP²
(EF − EP)² = CE² + EP²
EF² − 2*EF*EP + EP² = CE² + EP²
EF² − 2*EF*EP = CE²
b² − 2*b*(a²/b) = a²
b² − 2a² = a²
b² = 3a²
⇒ b = √3a
Now we are able to calculate the area of the purple region:
AF² = AQ² + FQ² = a² + b² = a² + 3a² = 4a² ⇒ AF = 2a
A(purple) = A(ADEF) − A(AFG) = AF*AD − (1/2)*AG*FG = (2a)*b − (1/2)*b*a = 2ab − ab/2 = 3ab/2 = (3√3/2)a²
Best regards from Germany
Solution:
Triangle FGP is congruent to Triangle CEP
Therefore:
GP = EP
FG = CE
FP = CP
Like that, we conclude that Purple Area is half of Rectangle ABCD Area
Purple Area = ½ Rectangle ABCD Area ... ¹
The Triangles AGF and CEP are similar, so we are going to use proportions
EP/CE = FG/AG
EP/a = a/b
EP = a²/b
Now, applying the Pythagorean Theorem in Triangle CEP
CE² + EP² = CP²
But CP = FP and FP = EF - EP
CE² + EP² = FP²
CE² + EP² = (EF - EP)²
CE² + EP² = EF² - 2 EF . EP + EP²
CE² = EF² - 2 EF . EP
a² = b² - 2 . b . a²/b
a² = b² - 2a²
3a² = b²
b² = 3a²
b = a√3
AG² + FG² = AF²
b² + a² = AF²
AF² = (a√3)² + a²
AF² = 3a² + a²
AF² = 4a²
AF = 2a
AB = AF + BF
AB = 2a + a
AB = 3a
Substituting in ¹
Purple Area = ½ length × width
Purple Area = ½ AB × BC
Purple Area = ½ 3a × b
Purple Area = 3ab/2 Square Units ✅