Let E be the fourth vertex of the rectangle AEPQ according to the Pythagorean theorem AC=√85 and AB=6, and from it [AEC]=(2*9)/2=9,[ABC]=(6*7)/2=21,[AEPQ]=5*2=10, so the area of the red region is equal to 9+21-10=20
The shaded area is 20 units square. I feel like an idiot for missing out on some of the HL similarity problems. I have been celebrating Christmas and New Year's. I just find it quite wonderful that HL similarity is JUST as useful as the Pythagorean Theorem. Happy New Year 2025!!!
The diagonal AC is √85 and AB is 6. Rotate a copy of the rectangle by 180° about its center to form a red Rectangle, with a small white rectangle at its center with sides 1 and 2. Red area = 6*7 - 1*2 =40 Half of it is 20.
Let the copy have the same point designations but with a ' appended. To be more thorough, you need to prove that PQP'Q' is a rectangle with sides 1 and 2, and that P' lies on AQ as well as P lies on A'Q'.
@@jimlocke9320 Another way to see it - If AC is the diagonal of a rectangle whose sides pass through AQ and PC, then a 180° rotation takes PQ to Q'P', and forms the rectangle Q'PQP', with PQ=1 and Q'P= 2.
@@jimlocke9320Another way to see it - If AC is the diagonal of a rectangle whose sides contain AQ and PC, then a 180° rotation will take PQ to Q'P', forming a small rectangle in the middle with PQ=2 and Q'P=1.
@@jimlocke9320Another way to see it - If AC is the diagonal of a rectangle whose sides contain AQ and PC, then a 180° rotation will take PQ to Q'P', forming a small rectangle in the middle with PQ=2 and Q'P=1.
@jimlocke9320 Another way to see it - If AC is the diagonal of a rectangle whose sides contain AQ and PC, then a 180° rotation will take PQ to Q'P', forming a small rectangle in the middle with PQ=2 and Q'P=1.
Thank you for your presentation of this problem. Let x = AC and y = PC In this question, we are given x = 5 and y = 4 but if we generalise the problem we can look at other near cases. If x=y=4.5 the area would be half of the rectangle and we would still need the length of AB to solve the problem Drawing SR parallel to PQ and forming a small rectangle, PQRS, of unit area , then: twice the red area (where x=5 and y=4) is equal to ABCD - 2. or the red area is half ABCD minus one. Looking at the diagonal AC of this rectangle, it will go through O, the midpoint of SR, so that AO=OC In triangle OAR OA is the hypotenuse. OA.OA = AR.AR + RO.RO AR = 5- 1/2 RO = 1/2 of 2 OA.OA = ( 20 +1/4 ) + (1) = 21 +1/4 so AC.AC = 85 this is the square of the diagonal and equals AB.AB + BC.BC using Pythagoras' theorem again. 85 = AB.AB + 49 so AB = 6 AB.AC-2 = 6 .7 - 2 = 40 Half of this is 20. The red shaded area is twenty square units.
I called angle DCP theta and this led me to 9sin(theta) + 2 cos(theta) = 7. This comes to sin(theta) = 3/5 and cos(theta) = 4/5 and the width of the rectangle = 6. Drawing the rectangle's diagonal leads to PQ being divided into the lengths (8/9) and (10/9). From this the desired area is A = 21 + (16/9) - (25/9) = 20.
I always solve your problem on the preview in my head, and then check how you solved the problem. This time I thought for about 15 minutes, but did not come to a solution. I launched the video - and there, unlike the preview, the side of the square is indicated (7)! Don't do that again ))
I've found that it's never a good idea to try and solve based on thumbnails alone. There's always some vital piece of information missing. It's always best to start the video then pause and attempt.
Pero estoy de acuerdo que la imagen que muestra el video antes de iniciar la reproducción debería indicar tanto la medida de 7 como la condición de rectángulo. Es verdad que tampoco se dice que sea cuadrado! Error de interpretación, pero es un dato que se ofrece al inicio del video que no figura en pantalla de inicio. Saludos
AC²=2²+(5+4)²=85---> AB=√(85-7²)=6 ---> AC corta a PQ en E---> Razón de semejanza entre EQA y EPC, s=5/4---> Si EQ=h---> 2-h=4h/5---> h=10/9---> EP=8/9 ---> Área ABCPQ =7*6/2 +4*8/9*2 -5*10/9*2 =20 u². Gracias y saludos
Let E be the fourth vertex of the rectangle AEPQ according to the Pythagorean theorem AC=√85 and AB=6, and from it [AEC]=(2*9)/2=9,[ABC]=(6*7)/2=21,[AEPQ]=5*2=10, so the area of the red region is equal to 9+21-10=20
The shaded area is 20 units square. I feel like an idiot for missing out on some of the HL similarity problems. I have been celebrating Christmas and New Year's. I just find it quite wonderful that HL similarity is JUST as useful as the Pythagorean Theorem. Happy New Year 2025!!!
The diagonal AC is √85 and AB is 6. Rotate a copy of the rectangle by 180° about its center to form a red Rectangle, with a small white rectangle at its center
with sides 1 and 2.
Red area = 6*7 - 1*2 =40
Half of it is 20.
Let the copy have the same point designations but with a ' appended. To be more thorough, you need to prove that PQP'Q' is a rectangle with sides 1 and 2, and that P' lies on AQ as well as P lies on A'Q'.
@@jimlocke9320 Another way to see it - If AC is the diagonal of a rectangle whose sides pass through AQ and PC, then a 180° rotation takes PQ to Q'P', and forms the rectangle Q'PQP', with PQ=1 and Q'P= 2.
@@jimlocke9320Another way to see it - If AC is the diagonal of a rectangle whose sides contain AQ and PC, then a 180° rotation will take PQ to Q'P', forming a small rectangle in the middle with PQ=2 and Q'P=1.
@@jimlocke9320Another way to see it - If AC is the diagonal of a rectangle whose sides contain AQ and PC, then a 180° rotation will take PQ to Q'P', forming a small rectangle in the middle with PQ=2 and Q'P=1.
@jimlocke9320 Another way to see it - If AC is the diagonal of a rectangle whose sides contain AQ and PC, then a 180° rotation will take PQ to Q'P', forming a small rectangle in the middle with PQ=2 and Q'P=1.
Thank you for your presentation of this problem.
Let x = AC and y = PC
In this question, we are given x = 5 and y = 4 but if we generalise the problem we can look at other near cases.
If x=y=4.5 the area would be half of the rectangle and we would still need the length of AB to solve the problem
Drawing SR parallel to PQ and forming a small rectangle, PQRS, of unit area , then:
twice the red area (where x=5 and y=4) is equal to ABCD - 2.
or the red area is half ABCD minus one.
Looking at the diagonal AC of this rectangle, it will go through O, the midpoint of SR, so that AO=OC
In triangle OAR OA is the hypotenuse. OA.OA = AR.AR + RO.RO AR = 5- 1/2 RO = 1/2 of 2
OA.OA = ( 20 +1/4 ) + (1) = 21 +1/4
so AC.AC = 85 this is the square of the diagonal and equals AB.AB + BC.BC using Pythagoras' theorem again.
85 = AB.AB + 49 so AB = 6
AB.AC-2 = 6 .7 - 2 = 40 Half of this is 20.
The red shaded area is twenty square units.
(5)^2(2)^2(4)^2={25+4+16}=45 (7)^2=49 {45+49}=94{90°A90°B+90°C+90°D}=360°ABCD/94=3 .76ABCD 3.4^19 3.4^19^1 3.4^1^1 3.2^2^1 3.1^2^1 3.2 (ABCD ➖ 3ABCD+2).
Pytagorean theorem, twice
b²+7²=(5+4)²+2² --> b=6cm
Red shaded area:
A = ½bh -½bh + ½bh
A = ½ [ 6*7 - 5*2(5/9) + 4*2(4/9) ]
A = 21-25/9+16/9 = 20 cm² ( Solved √ )
I called angle DCP theta and this led me to 9sin(theta) + 2 cos(theta) = 7.
This comes to sin(theta) = 3/5 and cos(theta) = 4/5 and the width of the rectangle = 6.
Drawing the rectangle's diagonal leads to PQ being divided into the lengths (8/9) and (10/9).
From this the desired area is A = 21 + (16/9) - (25/9) = 20.
I always solve your problem on the preview in my head, and then check how you solved the problem. This time I thought for about 15 minutes, but did not come to a solution. I launched the video - and there, unlike the preview, the side of the square is indicated (7)! Don't do that again ))
I've found that it's never a good idea to try and solve based on thumbnails alone.
There's always some vital piece of information missing.
It's always best to start the video then pause and attempt.
I agree! @@Grizzly01-vr4pn
Pero estoy de acuerdo que la imagen que muestra el video antes de iniciar la reproducción debería indicar tanto la medida de 7 como la condición de rectángulo. Es verdad que tampoco se dice que sea cuadrado!
Error de interpretación, pero es un dato que se ofrece al inicio del video que no figura en pantalla de inicio.
Saludos
AC²=2²+(5+4)²=85---> AB=√(85-7²)=6 ---> AC corta a PQ en E---> Razón de semejanza entre EQA y EPC, s=5/4---> Si EQ=h---> 2-h=4h/5---> h=10/9---> EP=8/9 ---> Área ABCPQ =7*6/2 +4*8/9*2 -5*10/9*2 =20 u².
Gracias y saludos
So good Thanks 🎉
∎ABCD → AD = BC = 7; AB = CD = a → AC = √(49 + a^2)
AQ = 5; CP = 4; sin(AQP) = sin(CPQ) = 1; PQ = 2 = PM + QM = (2 - k) + k
CM = m; AM = n; MCP = MAQ = δ → tan(δ) = k/5 = (2 - k)/4 → k = 10/9 → 2 - k = 8/9 →
m = 4√85/9 → n = 5√85/9 → n + m = √85 → a = 6
tan(δ) = 2/9 → sin(δ) = 2√85/85 → (1/2)sin(δ)(4m - 5n) = -1 → 7a/2 - 1 = 20 = red area
АС=√(81+4)=√85
АВ=√(85-49)=6
S=21+(1/2)*4*2*(4/9)-(1/2)*5*2*(5/9)=21+(16/9)-(25/9)=20
Thumb without side with 7 unitis.
AP problem
a-d=0 is wrong
First