Hi everyone , I noticed that several of you mentioned in the comments that I should use a 3-4-5 triangle. Let me clarify why this is NOT always the best assumption: If you have a right triangle with one leg of 4, it does NOT automatically mean it is a 3-4-5 triangle. Similarly, if the hypotenuse is 5, this doesn’t necessarily imply a 3-4-5 triangle either. In this video, I specifically adjusted the starting dimensions of the rectangle to deliberately form a 3-4-5 triangle, making the problem more mathematically elegant and visually clear for you. However, if the starting rectangle had a base of 4 and a height of 6 (instead of 8), we would end up with a right triangle where one leg is 4, the other leg is approximately 1.66, and the hypotenuse is about 4.33. As you can see, a 3-4-5 triangle is not a default outcome for right triangles. Thank you for watching, and stay safe!
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.
20 Let the distance of the red = p (all sides are the same ), then the remaining distance of the rectangle = (8-p) but (8-p) is also the base of the triangle with another base of 4 and a hypotenuse of p Hence, (8-p)^2 + 4^2 = p^2 64 + p^2 - 16p + 16 =p^2 80 = 16p 80/16 = p 5 =p Hence, the other side of the triangle is 3. Hence, its area = 4*3*1/2 = 6 But there are two triangles above the red-shaded region. Hence, total area = 12 ( 2*6) Since the area of the rectangle = 32, then the area of the shaded = 32-12 = 20 Answer
You can generalize this situation to any rectangle with dimensions a and b (b >= a). The red overlap area will be given by: a * [b - (b^2 - a^2)/2*b]. In the present case, a = 4, b = 8, so area = 20. Notice if the rectangle is a square, so b =a, you end up with an area of overlap of a*b = a^2. In other words, you just end up rotating the “clone” 90 degrees so it exactly covers the “original” shape.
This problem can be solved mentally with a little insight. The area of the vertical rectangle is thirty-two (32). The top triangle can be fitted to the bottom triangle making a small four (4) by three (3) rectangle of area twelve (12).. thirty--two (32) minus twelve (12) equals twenty (20).
I'm going to say 25 cause it's a 3/4/5 triangle, and I'm only guessing at that based on one side being 4 and the sum of the other two being 8, and 3/4/5 fits.
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.
I immediately jumped to the Pythagorean Theorem, and solved it. But the second method never occurred to me, and probably never would, were it not pointed out!
In The center of the figure we can see a smaller rectangle which is proportional to the original one. So, 8/4=4/x. Its smaller side is 2. Then, the area of the parallelogram is 5 × 4 = 20.
Seeing the problem involves numerous iterations of right angled triangles, because we haVE BEEN GIVEN THE VALUE OF FOUR FOR ONE OF THE SIDES, WHICH WE CAN SEE OBVIOUSLY IS NEITHER THE SHORTEST SIDE, NOR THE LONGEST, I immediately conclude the values for the other two sides of each triangle must be three, and five. I add together the volumes of the triangles. Then I add together the volumes of both rectangles. The difference between the two sums must be the solution. Am I correct?
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
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
@@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
Hi! 👋 If you're referring to 1:47, the number 4 is the hypotenuse of the upper triangle because that hypotenuse is the base of our rectangle. Since both rectangles are congruent, it is also 4 in length. I hope that answers your question!
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
Hi everyone ,
I noticed that several of you mentioned in the comments that I should use a 3-4-5 triangle. Let me clarify why this is NOT always the best assumption:
If you have a right triangle with one leg of 4, it does NOT automatically mean it is a 3-4-5 triangle. Similarly, if the hypotenuse is 5, this doesn’t necessarily imply a 3-4-5 triangle either.
In this video, I specifically adjusted the starting dimensions of the rectangle to deliberately form a 3-4-5 triangle, making the problem more mathematically elegant and visually clear for you. However, if the starting rectangle had a base of 4 and a height of 6 (instead of 8), we would end up with a right triangle where one leg is 4, the other leg is approximately 1.66, and the hypotenuse is about 4.33.
As you can see, a 3-4-5 triangle is not a default outcome for right triangles.
Thank you for watching, and stay safe!
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
Isn’t this a rectangle of 32 minus a square of 12?
Yes. That's how I did it, too. Much easier.
It is a rectangle of 32sq units but you subtract 2 of 345 rectangles 6sq units each and it leaves you with 20sq unit rhombus
Almost, Carlos. 3x4 is not a square. James Faizi is correct.
20
Let the distance of the red = p (all sides are the same ), then the remaining distance of
the rectangle = (8-p)
but (8-p) is also the base of the triangle with another base of 4 and a hypotenuse of p
Hence, (8-p)^2 + 4^2 = p^2
64 + p^2 - 16p + 16 =p^2
80 = 16p
80/16 = p
5 =p
Hence, the other side of the triangle is 3.
Hence, its area = 4*3*1/2 = 6
But there are two triangles above the red-shaded region. Hence, total area = 12 ( 2*6)
Since the area of the rectangle = 32, then the area of the shaded = 32-12 = 20 Answer
Very nice!
You can generalize this situation to any rectangle with dimensions a and b (b >= a). The red overlap area will be given by: a * [b - (b^2 - a^2)/2*b]. In the present case, a = 4, b = 8, so area = 20. Notice if the rectangle is a square, so b =a, you end up with an area of overlap of a*b = a^2. In other words, you just end up rotating the “clone” 90 degrees so it exactly covers the “original” shape.
Really nice observation!
This problem can be solved mentally with a little insight. The area of the vertical rectangle is thirty-two (32). The top triangle can be fitted to the bottom triangle making a small four (4) by three (3) rectangle of area twelve (12).. thirty--two (32) minus twelve (12) equals twenty (20).
Fun! Good English-language skills also.
@@pseudotonal Thank you. I'm glad you liked it!
3-4-5 right triangle.
The base of the shaded area is 8-3 and the height is 4. 5x4 = 20.
d₁² = 8² + 4² ---> d₁ = 4√5 cm
d₂ = ½d₁= 2√5 cm
A = ½ d₁d₂ = 20 cm² ( Solved √ )
I'm going to say 25 cause it's a 3/4/5 triangle, and I'm only guessing at that based on one side being 4 and the sum of the other two being 8, and 3/4/5 fits.
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.
I immediately jumped to the Pythagorean Theorem, and solved it. But the second method never occurred to me, and probably never would, were it not pointed out!
In The center of the figure we can see a smaller rectangle which is proportional to the original one. So, 8/4=4/x. Its smaller side is 2. Then, the area of the parallelogram is 5 × 4 = 20.
Hello. How do you know that the rectangles are porportional? I used that rectangle to solve the problem, but I will spare you the ugly details.
Where did you see this inner rectangle??
@@Cyber_Cheese By tracing two horizontal lines.
Could someone explain _why_ the ratio 8:4 = d1:d2?
20 kite. ( √80 * √80/2)/2
Seeing the problem involves numerous iterations of right angled triangles, because we haVE BEEN GIVEN THE VALUE OF FOUR FOR ONE OF THE SIDES, WHICH WE CAN SEE OBVIOUSLY IS NEITHER THE SHORTEST SIDE, NOR THE LONGEST, I immediately conclude the values for the other two sides of each triangle must be three, and five. I add together the volumes of the triangles. Then I add together the volumes of both rectangles. The difference between the two sums must be the solution. Am I correct?
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
Isn't A(rectangle) - 2A(triangle) simpler?
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
@@gelbkehlchen Nice!!!
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
just use the 3, 4, 5 rule
Hi! Thank you for watching! ❤ Please check my comment above where I explained why I chose not to use a 3-4-5 triangle.
Where did the 4 come from?
Hi! 👋 If you're referring to 1:47, the number 4 is the hypotenuse of the upper triangle because that hypotenuse is the base of our rectangle. Since both rectangles are congruent, it is also 4 in length. I hope that answers your question!
@ thank you 🙏
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
I set up an equation to solve for the sides of the triangle, then do'h it's just 3 4 5🤦♂️.
20
3rd Solution:
A (Rectangle) = 8 × 4 = 32
A (Trinagle) x 2 = 3 × 4 = 12
A (Diamond) = 32 - 12 = 20