Join P and C to divide the yellow area into triangle APC and triangle CPQ . Area APC = ½*AP*CD = ½ab Area CPQ = ½*CQ*DP = ½ab Yellow Area=Area APC+Area CPQ = ab = 2√3
This method also works when the rectangles don't have the same side lengths! 👍 And anyway, the diagram did not define that the two rectangles were identical...🤔
An alternative solution is to construct a straight line between C and P. This will divide the yellow region into two triangles each with area equals to sqrt(3). Therefore the required area is 2*sqrt(3).
Since the rectangles' sides weren't given, one can just assume the long side is 2 and short side is rad 3. This makes the solution much easier than the given one.
Nice and basic solution. This was my first thought but then i removed from my logical solving the whole rectangle. If I may share the idea Just use the triangle ACD minus the triangle PQD. Reduces the number of calculations and complexity . Best wishes
I've got the same result, but without assuming that the measurements of the base and height of the two rectangles are equal (although they are reversed). So I did the calculations with 4 unknowns. B1*H1=2SQRT3 B2*H2=2SQRT3 and then I calculated the yellow area by subtracting the area of triangle PQD from the area of triangle ACD 1/2*H1(B1+B2) - 1/2*B2(H1-H2) 1/2*(H1B1+H1B2-H1B2+H2B2) 1/2*(2SQRT3+2SQRT3)= 2SQRT3
I did not consider that both rectangles had the same side lengths as it was not mentioned but got the very same answer, using the same method (area of the big rectangle - area of both right triangles) 😊
Here's one simple way, using your labelling. Spoiler alert. We can imagine extending AP and CQ to meet at point D, such that ABCD is a rectangle with side-lengths of a and a + b, whose area is a(a + b) = a² + ab. Then we need to find the area of triangle ADC, and subtract from it the area of triangle PDQ. Area of triangle ABC (visible) = a² / 2 + ab / 2 = area of triangle ADC. Area of triangle PDQ (invisible) = a² / 2 − ab / 2. Yellow area = a² / 2 + ab / 2 − (a² / 2 − ab / 2) = ab / 2 + ab / 2 = ab. But ab is also the area of each of our two initial rectangles, so the yellow area is 2√3 square units. Ah, it turns out that this is pretty much exactly what you did. Something of your teaching must have rubbed off on me. Thanks, as always.
How I solved it: Let M be the point along BC where the two rectangles meet. Let QC = x and AB = y Area ACQP = Area ABCQP - Area triangle ABC = (Area ABMP + Area Trapezoid PMCQ) - Area Triangle ABC = 2√3 sqcm ( since area of PMCQ = area triangle ABC = 0.5(x+y)y )
Let's assume that this puzzle should have only one valid solution. Since we can't find the value of a and b, let's assume they are equal. Having two squares forming a domino, point D is equal to point Q. Since AC is a diagonal of the domino, ABC is half the area of ABCD.
I did the same. When a question lacks information (and we assume from the question that there is a correct solution) simply adjust the lengths within the constraints of the information given, to give the simplest possible example, which is then easy to solve. 😀
@@ajbonmg According to the diagram, the two rectangles were not defined as being identical. With identical rectangles, we already have a selection of special cases. Of these, the case where both rectangles are squares is unique and easy to visualize. This is good enough for a quick assessment of the problem, but the real proof for the general case is even simpler...😉
I solved it from the preview image before watching the video, so I didn't know that the two rectangles had the same dimensions. It doesn't matter, the two could be different and the area is still 2 root 3.
Also an odd puzzle, challenging😅. Let a,b be the dimension of the rectangle, then the area of ABCQ is (a+b)^2/2, the area of ABC is (a+b)b/2, thus the answer is (a+b)a/2=(a^2+ab)/2=(a^2+2root 3)/2, so undetermined, for unknown a.🤔
The diagram makes it look as though 2*sqrt3 is the area of each of the blue *trapezoids*, not rectangles. Not until I started watching the video did I realize that the number is supposed to be the area of each entire rectangle -- even though the corner of each rectangle is yellow, not blue.
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I know you're going to solve it, but I can't help thinking you bit off a lot to chew here. That's mind bending stuff right there. I'm still trying to figure out how it works. Lol. A little bit advanced for me, I have to admit.
Very nice. Like always.
LG Gerald
Glad you think so!
Thanks for your continued love and support!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
Zajímavý příklad, elegantní řešení.
You come up with incredible problems and your explanation is easy to follow, love the colors!!
Join P and C to divide the yellow area into triangle APC and triangle CPQ .
Area APC = ½*AP*CD = ½ab
Area CPQ = ½*CQ*DP = ½ab
Yellow Area=Area APC+Area CPQ
= ab = 2√3
This method also works when the rectangles don't have the same side lengths! 👍
And anyway, the diagram did not define that the two rectangles were identical...🤔
@@ybodoN he stated that the rectangles are identical.
Yeah, most of his problems can be solved much more easily by just figuring out how to create triangles with known heights. Kudos, Hari!
Realy great, that the solution is the area of one rectangles. Very smart. Thank you!
An alternative solution is to construct a straight line between C and P. This will divide the yellow region into two triangles each with area equals to sqrt(3). Therefore the required area is 2*sqrt(3).
A Great solution! Thank you!
Surprising result!🧠 A beautiful exercise.
Great. Thank you.
Since the rectangles' sides weren't given, one can just assume the long side is 2 and short side is rad 3. This makes the solution much easier than the given one.
Great 👍👍
Thanks for sharing😊
Nice and basic solution. This was my first thought but then i removed from my logical solving the whole rectangle.
If I may share the idea
Just use the triangle ACD minus the triangle PQD.
Reduces the number of calculations and complexity .
Best wishes
Very cleverly thought out problem, very interesting and great exercise...👍🏻
Thanks for video.Good luck sir!!!!!!!!!!!!
Good constructed question
Elegant solution 🥂👍❤
Glad to hear that!
Thanks for your continued love and support!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
Si suponemos que ambos rectángulos son congruentes》APQC =APC+CQP =(AP×AB/2)+(CQ×AB/2) =2sqrt3
Sorprendente puzle. Gracias y saludos.
Area yellow region = triangle ADC - triangle PDQ =
[a(b+a)/2] - [a(a-b)/2] =
(ab + a²)/2 - (a² - ab)/2 = 2ab/2 =
ab = 2sqrt3
I've got the same result, but without assuming that the measurements of the base and height of the two rectangles are equal (although they are reversed). So I did the calculations with 4 unknowns.
B1*H1=2SQRT3
B2*H2=2SQRT3
and then I calculated the yellow area by subtracting the area of triangle PQD from the area of triangle ACD
1/2*H1(B1+B2) - 1/2*B2(H1-H2)
1/2*(H1B1+H1B2-H1B2+H2B2)
1/2*(2SQRT3+2SQRT3)= 2SQRT3
I did not consider that both rectangles had the same side lengths as it was not mentioned but got the very same answer, using the same method (area of the big rectangle - area of both right triangles) 😊
Yay, I solved it.
Let T be the unlabelled point at the inside of the L-shape:
APQC = ACD - PQD
= ABCD/2 - PTQD/2
= (ABCD - PTQD)/2
= (2 2√3)/2
= 2√3
Area of triangle APC=1/2•AP•AB
=sqrt3
Area of QCP=1/2QC•length of rect
=sqrt3
Hence area ot shaded region=
2sqroot3
Here's one simple way, using your labelling.
Spoiler alert.
We can imagine extending AP and CQ to meet at point D, such that ABCD is a rectangle with side-lengths of a and a + b, whose area is a(a + b) = a² + ab. Then we need to find the area of triangle ADC, and subtract from it the area of triangle PDQ.
Area of triangle ABC (visible) = a² / 2 + ab / 2 = area of triangle ADC.
Area of triangle PDQ (invisible) = a² / 2 − ab / 2.
Yellow area = a² / 2 + ab / 2 − (a² / 2 − ab / 2)
= ab / 2 + ab / 2
= ab.
But ab is also the area of each of our two initial rectangles, so the yellow area is 2√3 square units.
Ah, it turns out that this is pretty much exactly what you did. Something of your teaching must have rubbed off on me. Thanks, as always.
Good morning sir
Though the area of both rectangles is same,how can we assume that the sides are same ?
How I solved it:
Let M be the point along BC where the two rectangles meet.
Let QC = x and AB = y
Area ACQP = Area ABCQP - Area triangle ABC
= (Area ABMP + Area Trapezoid PMCQ) - Area Triangle ABC
= 2√3 sqcm
( since area of PMCQ = area triangle ABC = 0.5(x+y)y )
Let's assume that this puzzle should have only one valid solution.
Since we can't find the value of a and b, let's assume they are equal.
Having two squares forming a domino, point D is equal to point Q.
Since AC is a diagonal of the domino, ABC is half the area of ABCD.
I did the same. When a question lacks information (and we assume from the question that there is a correct solution) simply adjust the lengths within the constraints of the information given, to give the simplest possible example, which is then easy to solve. 😀
@@ajbonmg According to the diagram, the two rectangles were not defined as being identical. With identical rectangles, we already have a selection of special cases. Of these, the case where both rectangles are squares is unique and easy to visualize. This is good enough for a quick assessment of the problem, but the real proof for the general case is even simpler...😉
Did it like this, the area of the large triangle minus the area of the small triangle.
I solved it from the preview image before watching the video, so I didn't know that the two rectangles had the same dimensions. It doesn't matter, the two could be different and the area is still 2 root 3.
Also an odd puzzle, challenging😅. Let a,b be the dimension of the rectangle, then the area of ABCQ is (a+b)^2/2, the area of ABC is (a+b)b/2, thus the answer is (a+b)a/2=(a^2+ab)/2=(a^2+2root 3)/2, so undetermined, for unknown a.🤔
so what?
Oh, I see my mistake. 🤣😂🤣
The diagram makes it look as though 2*sqrt3 is the area of each of the blue *trapezoids*, not rectangles. Not until I started watching the video did I realize that the number is supposed to be the area of each entire rectangle -- even though the corner of each rectangle is yellow, not blue.
Güzel çözüm
Interesting, but theres no need to calculate the area of the rectangle, the answer is simply the difference in areas of the 2 right angle triangles
The youtube has sent a popup message conveying that I had posted a comment on PreMath channel, which they didnot like. As, I am not having any memory of the said comment, I can not submit any reason etc on that account. I have, therefore unsubscribed PreMath to avoid any such future recurrence. Thanks for the association for so long with the channel of my like.
I know you're going to solve it, but I can't help thinking you bit off a lot to chew here. That's mind bending stuff right there. I'm still trying to figure out how it works. Lol. A little bit advanced for me, I have to admit.
funny math