All the right triangles formed are equal. One side is 4 and the other side is x. The hypotenuse is equal to 8-x. Therefore, x²+4²=(8-x)². Therefore, x=3. The area of the parallelogram formed is equal to 5*4=20.
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Si α es el ángulo entre la diagonal y el lado vertical del rectángulo→ tg α=1/2→ El rectángulo superpuesto ha sido girado un ángulo =2α→ La diagonal menor del rombo rojo forma un ángulo α con la base del rectángulo e interseca al lado vertical a una distancia (4/2)*(1/2)=1→ Los triángulos rectángulos situados arriba y abajo del rombo tienen catetos de longitudes (4-1=3) y 4→ Área del rombo rojo =8*4-(3*4)=32-12 =20 u² Gracias y saludos.
Solution: The resulting 4 black triangles are all congruent because they have the alternate angle, the right angle and the long side with the length 4 in common (sww). b = short side of these black triangles. Then the hypotenuse of these black triangles = h = √(4²+b²). Now: h+b = 8 ⟹ √(4²+b²)+b = 8 |-b ⟹ √(4²+b²) = 8-b |()² ⟹ 4²+b² = 8²-16b+b² |+16b-4²-b² ⟹ 16b = 8²-4² = 48|/16 ⟹ b = 3 ⟹ Red area = rectangle - 2*black triangles = 4*8-2*4*b*1/2 = 4*8-2*4*3*1/2 = 4*8-4*3 = 32-12 = 20
The ask area is ( in Greece ) a ROMVOS ( 4 SIDES EQUALS beacuse the 4 right triangles are equals) 2) The diagonius of romvos are right and 1 of the right triangles into the romvos is ΗΟΓ. 3) ΗΟΓ is similar with the big right triangle ΑΖΓ ( The start rectangle is ΑΒΓΔ and O is the center of romvos and ΑΒΓΔ). 4) Then we have ΗΟ=sq(5), ΟΓ=2*sq(5), ( ΗΟΓ )=5 and A=4*5=20 s.u. 5) Of course 4/ΗΟ=4*sq(5)/ΗΓ and after Py.Th 6) The second rectangle is ΑΖΓΕ and the common points of 2 rectangles are Η,Θ. Thanks NGE
@@jesseadamson1077 The question is very easy We first find the value of x, which is a Which is equal to 3 Then the space The area of the rectangle is 4 × 8 Minus the area of the two triangles A = 2 × ½(4 × 3) = 4 × 3 = 12 32 - 12 = 20
It’s also possible to visualize - if the slanted rectangle was very skinny, almost a line, it would fit if it was long enough . It would just need to rotate more
@@ThePhantomoftheMath He said that but it is not in the actual problem as given, it seems like he made an extra assumption. If the rectangles were supposed to be congruent they would need to be marked as such
d₁² = 8² + 4² ---> d₁ = 4√5 cm
d₂ = ½d₁= 2√5 cm
A = ½ d₁d₂ = 20 cm² ( Solved √ )
All the right triangles formed are equal. One side is 4 and the other side is x. The hypotenuse is equal to 8-x. Therefore, x²+4²=(8-x)². Therefore, x=3. The area of the parallelogram formed is equal to 5*4=20.
How do we know the triangles are equal? The problem does not state the two rectangles are congruent.
@@cm5754it doesn't visually (and it should), but he actively stated at the start of the video that the rotated rectangle was a clone of the first.
@@cm5754It's the same rectangle, we just rotated it.
@@cm5754It's the same rectangle, we just rotated it and we know that rotation preserves distances.
@@midnatheblackrobe That seems like he added an assumption to the problem
Fun! Good English-language skills also.
Isn’t this a rectangle of 32 minus a square of 12?
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Si α es el ángulo entre la diagonal y el lado vertical del rectángulo→ tg α=1/2→ El rectángulo superpuesto ha sido girado un ángulo =2α→ La diagonal menor del rombo rojo forma un ángulo α con la base del rectángulo e interseca al lado vertical a una distancia (4/2)*(1/2)=1→ Los triángulos rectángulos situados arriba y abajo del rombo tienen catetos de longitudes (4-1=3) y 4→ Área del rombo rojo =8*4-(3*4)=32-12 =20 u²
Gracias y saludos.
Solution:
The resulting 4 black triangles are all congruent because they have the alternate angle, the right angle and the long side with the length 4 in common (sww).
b = short side of these black triangles.
Then the hypotenuse of these black triangles = h = √(4²+b²).
Now:
h+b = 8 ⟹
√(4²+b²)+b = 8 |-b ⟹
√(4²+b²) = 8-b |()² ⟹
4²+b² = 8²-16b+b² |+16b-4²-b² ⟹
16b = 8²-4² = 48|/16 ⟹
b = 3 ⟹
Red area = rectangle - 2*black triangles
= 4*8-2*4*b*1/2 = 4*8-2*4*3*1/2 = 4*8-4*3 = 32-12 = 20
3rd Solution:
A (Rectangle) = 8 × 4 = 32
A (Trinagle) x 2 = 3 × 4 = 12
A (Diamond) = 32 - 12 = 20
The ask area is ( in Greece ) a ROMVOS ( 4 SIDES EQUALS beacuse the 4 right triangles are equals) 2) The diagonius of romvos are right and 1 of the right triangles into the romvos is ΗΟΓ.
3) ΗΟΓ is similar with the big right triangle ΑΖΓ ( The start rectangle is ΑΒΓΔ and O is the center of romvos and ΑΒΓΔ). 4) Then we have ΗΟ=sq(5), ΟΓ=2*sq(5), ( ΗΟΓ )=5 and A=4*5=20 s.u.
5) Of course 4/ΗΟ=4*sq(5)/ΗΓ and after Py.Th 6) The second rectangle is ΑΖΓΕ and the common points of 2 rectangles are Η,Θ. Thanks NGE
It’s possible the slanted rectangle has one side of 2 and the other of sqrt(76). The problem does not state the rectangles are congruent
a² + 4² = (8 - a)²
⇒a = 3
∴ area shaded
= 8 × 4 - 4 × 3
= 32 - 12
= 20 u²
Sorry how did you derive that equation?
@@jesseadamson1077
The question is very easy
We first find the value of x, which is a
Which is equal to 3
Then the space
The area of the rectangle is 4 × 8
Minus the area of the two triangles
A = 2 × ½(4 × 3) = 4 × 3 = 12
32 - 12 = 20
The problem only gave two lengths, it did not say the rectangles were congruent.
Aren’t they bound with the same diagonal?
@ the slanted one could have sides 2 and sqrt(76) and it would have the same diagonal
It’s also possible to visualize - if the slanted rectangle was very skinny, almost a line, it would fit if it was long enough . It would just need to rotate more
Hi! Check the beginning of the video please 0:14. I said "then we will clone this rectangle...".
@@ThePhantomoftheMath He said that but it is not in the actual problem as given, it seems like he made an extra assumption. If the rectangles were supposed to be congruent they would need to be marked as such