The five rectangles are congruent to each other: this means that the ratio of their sides is 1 : 2. (Are congruent rectangles different? Maybe in position...) This means that the length of the rectangle ABCD is 5 units while its width is 2 units. side a = 10 units, side b = ? units 5 : a = 2 : b => 5 : 10 = 2 : b 5b = 10 * 2 = 20 b = 20 / 5 = 4 units A(triangle ABE) = 1/2 * ab = 1/2 * 10 * 4 = 20 units
Same thing happened to me. Came into the problem ready to use variables and got the answer 20 in like 2 minutes, then spent 15 more because I genuinely didn’t believe that I got the answer that quickly
Diagram is not drawn to scale, that can throw you for a loop. But it's an easy problem, I solved it as Presh did. You can tell by the length of the video, this is an easy one.
This is a fine example of what I call a minimal-information problem. Spoiler alert. Without loss of generality, we can safely assume an example case in which all five rectangles are 4-unit-by-2-unit rectangles, with the central one rotated 90 degrees with respect to the others. That makes the calculation trivial. The triangle must have half the area of a 10-unit-by-4-unit rectangle, or the area of five 2-unit-by-2-unit squares. A = 5 ∙ 4 = 20 square units. I'm not suggesting that this is a good method, but it is certainly an effective one.
Since all of the rectangles are congruent with each other, given their orientation, they must have side ratios of 2:1. Two rectangles side by side are as wide as one is long. Since the total width of the construction is given as 10 units, then the actual sizes of the rectangles will be, from left to right, 2k + 1k + 2k = 10, where k is the constant on which the 2:1 ratio acts to give the actual size values. Solving for k gives us 2, so the rectangles are each 2×4 units in size. That means that the total construction is 10×4, or 40 square units, and as a triangle is half the area of a rectangle with the same dimensions, the shaded area is 20 square units.
I did it using the areas, (2a+b)*2b=5ab, setting (2a+b) to 10 as stated, and then you have the area of the rectangle. The triangle is just half of it. Very easy one for Presh.
I presume the diagram is intentionally misleading and not to scale. The only way the rectangles can be congruent is for them to be identical with the long sides to be twice the short sides (2w = l). That means each rectangle is actually 2 squares with edge lengths of w. That makes the large rectangle 5w in length, and 2w in width. Since the actual length is 10, the width is 4, with an area of 40. Any triangle in the rectangle as shown must be half that, or 20
@@warlock415 Yes, though it is the _only_ reason this problem is difficult at first glance. Without it, the solution would be patently obvious. Most other no-to-scale diagrams are there to make a tricky problem harder, or lead you on a wild goose chase :)
Before watching the video: The ration of the sides of the rectangles can be chosen arbitrarily. The more the middle rectangle is elongated, the more elongates is the whole resulting structure. Thus the ratio of the sides of the whole structure can be arbitrarily chosen. Let k be the ratio of the rectangles inside and s be the bottom length of the whole structure (which in the task is set to 10). Then the area of the whole structure is s^2 * k / (1 + k^2). The area of the triangle is 1/2 that area. Thus the task at hand is underdetermined. ...Now i'm watching the video. My guess is that something in the initial picture is missing (would not be the first time - i already commented on another occasion on such missing detail)... Aha: I had a wrong memory of the meaning of "congruency" in mind. Good to have that rectified :D
I spent quite a while on this one because I forgot congruent meant they have the same sides, not just the same area, so I went back to the video and in 5 seconds I understood and got the answer
The rectangles being congruent mean that the ratio of their sides is the same. If you call the vertical sides of rectangles R1,R2,R4 and R5 a, and the horizontal side b, l is equal to 2a, and 2a / w would be equal to b / a. But from there, how do you go to l = b and w = a? Some explanation is needed where only a 1:2 rectangle would fit there.
You are substituting the meaning of "similar" where it says "congruent". Congruent figures can be mapped on each other using only rigid transformations, so reflection across a line, translation and rotation. Once you introduce scaling transformations, the objects are no longer congruent, they are only similar. @@brunogrieco5146
Once the fact that the dimensions on all of them were the same clicked with me, it just fell right into place. And then I spent a minute going "DUUUUUUUH!" because it was so painfully obvious at that point.
Ooh, I worked this one out in my head without using a formula to solve it. When I apply my logic to how I worked it out in my head, I can conclude that the math is correct. Now I want to learn how to apply the formulas that traditional math uses so I can translate my logic and thought process onto paper when seeing more of these problems 😄
Normally I can't do these in my head, but this one I could. I think it's because I already knew the tricks involved. There are a couple of visual tricks going on that try to make it more confusing than it needs to be, but if you catch them then it's simple.
Other than being on CD, the location of E is not specified. So we are free to slide it up to point C making the blue triangle a right triangle and the problem becomes even easier than it already is.
The area of the triangle is Base (AB) x height/2 The length of the height is the same as BC. So the area of the triangle is just half the area of the big rectangle. Proof: Imagine the height // to BC passing through E and intersecting AB in F. Now you have 2 right triangles and AF+FB=AB
I figured out the answer mentally although my calculations were more and I used the fact that area of triangle ABE = area of triangle ABD=half the area of rectangle ABCD
This is not presented to scale. The rectangles don't look congruent (same shape and size). The middle one looks longer than the other 4. So, ignore the picture to some degree - it's deceptive.
Hight of of R3 is twice that of all the others. Ratio of all the rectangles are 2:1 AD =(2 +1 +2)k =10 5k =10 k=2 All rectangles are 4×2 AB =(1 +1)k =(2)(2) =4 Area =(AB ×AD)/2 =(4×10)/2 =20
I got it in a different way using Area instead of side lengths Area of the entire shape = 10*2w Area of the entire shape = 5* Area of each rectangle = 5*l*w 5*l*w = 10*2*w => l*w = 4*w =>l = 4 10 = 2*4 + w => w = 2 2*2*10 / 2 = 20
I did the same math to find w and l, and then I logiced that since the location of point E was not specified, it must not matter, so if I put it up in the corner, ABE is therefore half the rectangle, thus, 20 square whatevers.
Is it wrong to call these triangles congruent when they don't have the exact same measurements? Is it more correct to simply say their areas are equal?
@@FalseNoob777visually, R3 is very clearly not congruent to the other four. Mathematically, it can be, but the point is that it looks visually different, which could throw off the reader if they don't pay attention.
I think vertical w not equal to horizontal w Same with l, vertical l not equal with horizontal l But I agree horizontal l is 4 and horizontal w is 2 The vertical l or vertical w may vary, but adopt 2w=l Just my opinion
This is a simple problem (only mental math needed) once you take the statement about congruence seriously, even though as drawn, the rectangles clearly don't have identical dimensions. All it takes is pretending that they do. :)
@@Richie_P I know that scale doesn't match, so even not look at it. But still, how get from point A to B, that is what matters. There is ABCD rectangle divided on 5 equal by area rectangles. ABCD one dimmension is 10. That is all. Still same question is not answered.
Congruent means the same, but moved by a combination of translation, rotation, and reflection across a line. In this problem the 5th rectangle is not only the same area as the others, it also has the same ratio between its sides. That's one of the properties of congruence.@@hvnterblack
I spotted the 2l + w = 10 long before l = 2w even though I knew I needed another equation lol. Dunce cap time! Or perhaps bedtime instead of watching math videos at 2 AM.
I forgot what congruent means, though I guess at least I knew from the get go the area would be half of the rectangle as a whole? Once he explained what congruent meant it kinda just clicked. Kinda embarrassing that I had already forgotten congruent though XD
Answered from thumbnail alone before watching the video. The area of the big rectangle is 40. And the area of the shaded triangle is 20. I'll reveal my methods if correct.
Yeah, 40 and 20. I immediately saw that the rectangles we're twice as high as wide. If the long side of the big rectangle is 10, this means the small rectangles are 4, 2, 4 on that edge. Which means the small rectangles are 4×2, which means the large rectangle is 4×10. This makes the area of the large rectangle 40. And, the easy part most of the others seem to have missed A triangle with the same base and height as a rectangle will be precisely half it's area, no matter the angles. So, the area of the triangle is 20. Now to watch the video.
For this video the thumbnail and the problem presented matched. There have been numerous videos where information essential to finding a solution did not appear in the thumbnail. I've been bitten by that mismatch a number of times.@@cooltaylor1015
SOLUTION HAS A FLAW they are congruent, doesn't mean they are equal if the dimensions of the central rectangle are L and W that mean dimensions of other rectangles maybe 2L and 2W else you need to give proof that those rectangles are equal
In the thumbnail it says “equal”, not congruent. Solve that problem instead. 😏 I suppose “equal” is to be interpreted as having equal areas so the triangle has an area that is 2.5 rectangles.
All we need is height, perpendicular or not so the answer would be the same wherever along CD the apex lies. Even if we extend CD either way in a straight line and put the apex anywhere along it the answer would be the same. That's triangles for you.
The height IS perpendicular. The height is a line segment from E to the base AB parallel to AD. That line segment has length 10 and is perpendicular to the base AB.
The height is perpendicular to the base because the base is one side of the rectangle and the height is measured as the length of a perpindicular side of the rectangle. The height is not the same thing as a side of the triangle, unless the triangle happens to be a right triangle.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
The five rectangles are congruent to each other: this means that the ratio of their sides is 1 : 2. (Are congruent rectangles different? Maybe in position...)
This means that the length of the rectangle ABCD is 5 units while its width is 2 units.
side a = 10 units, side b = ? units
5 : a = 2 : b => 5 : 10 = 2 : b
5b = 10 * 2 = 20
b = 20 / 5 = 4 units
A(triangle ABE) = 1/2 * ab = 1/2 * 10 * 4 = 20 units
I solved in the same way
Solved in mind in a few minutes by the exactly same way
Quite surprised this one is easy. So much that I had to double check my calculation because it finished too fast
Made me feel smart
Fr
Same thing happened to me. Came into the problem ready to use variables and got the answer 20 in like 2 minutes, then spent 15 more because I genuinely didn’t believe that I got the answer that quickly
Diagram is not drawn to scale, that can throw you for a loop. But it's an easy problem, I solved it as Presh did.
You can tell by the length of the video, this is an easy one.
Finally an easy one. Thanks.
This is a fine example of what I call a minimal-information problem.
Spoiler alert.
Without loss of generality, we can safely assume an example case in which all five rectangles are 4-unit-by-2-unit rectangles, with the central one rotated 90 degrees with respect to the others. That makes the calculation trivial. The triangle must have half the area of a 10-unit-by-4-unit rectangle, or the area of five 2-unit-by-2-unit squares. A = 5 ∙ 4 = 20 square units. I'm not suggesting that this is a good method, but it is certainly an effective one.
Since all of the rectangles are congruent with each other, given their orientation, they must have side ratios of 2:1. Two rectangles side by side are as wide as one is long.
Since the total width of the construction is given as 10 units, then the actual sizes of the rectangles will be, from left to right, 2k + 1k + 2k = 10, where k is the constant on which the 2:1 ratio acts to give the actual size values.
Solving for k gives us 2, so the rectangles are each 2×4 units in size. That means that the total construction is 10×4, or 40 square units, and as a triangle is half the area of a rectangle with the same dimensions, the shaded area is 20 square units.
I did it using the areas, (2a+b)*2b=5ab, setting (2a+b) to 10 as stated, and then you have the area of the rectangle. The triangle is just half of it. Very easy one for Presh.
I presume the diagram is intentionally misleading and not to scale. The only way the rectangles can be congruent is for them to be identical with the long sides to be twice the short sides (2w = l). That means each rectangle is actually 2 squares with edge lengths of w. That makes the large rectangle 5w in length, and 2w in width. Since the actual length is 10, the width is 4, with an area of 40. Any triangle in the rectangle as shown must be half that, or 20
All diagrams cannot be assumed to be to scale unless so stated.
@@warlock415 Yes, though it is the _only_ reason this problem is difficult at first glance. Without it, the solution would be patently obvious. Most other no-to-scale diagrams are there to make a tricky problem harder, or lead you on a wild goose chase :)
Before watching the video: The ration of the sides of the rectangles can be chosen arbitrarily. The more the middle rectangle is elongated, the more elongates is the whole resulting structure. Thus the ratio of the sides of the whole structure can be arbitrarily chosen.
Let k be the ratio of the rectangles inside and s be the bottom length of the whole structure (which in the task is set to 10). Then the area of the whole structure is s^2 * k / (1 + k^2). The area of the triangle is 1/2 that area.
Thus the task at hand is underdetermined.
...Now i'm watching the video. My guess is that something in the initial picture is missing (would not be the first time - i already commented on another occasion on such missing detail)...
Aha: I had a wrong memory of the meaning of "congruency" in mind. Good to have that rectified :D
The rectangles are not only congruent, but identical.
I spent quite a while on this one because I forgot congruent meant they have the same sides, not just the same area, so I went back to the video and in 5 seconds I understood and got the answer
i was sort of in the same boat. i guess being out of school almost 60 years has a price. 😀
same 🤣
I thought so too, especially when the rectangles shown were NOT the same size!
I'm glad when he tosses in an easy question every so often.
The rectangles being congruent mean that the ratio of their sides is the same. If you call the vertical sides of rectangles R1,R2,R4 and R5 a, and the horizontal side b, l is equal to 2a, and 2a / w would be equal to b / a. But from there, how do you go to l = b and w = a?
Some explanation is needed where only a 1:2 rectangle would fit there.
Congruent does not mean the ratio of the sides is the same; that is "similar". Congruent means all rectangles are the same size.
@@MichaelOnines That would mean that R1 = R2 = R3... not R1 ~ R2 ~ R3... (sorry, no congruent character)
You are substituting the meaning of "similar" where it says "congruent". Congruent figures can be mapped on each other using only rigid transformations, so reflection across a line, translation and rotation.
Once you introduce scaling transformations, the objects are no longer congruent, they are only similar. @@brunogrieco5146
A harder way of asking the area of the triangle is "the area of the blue shaded area of the rectangle".
Very true. Presh made this already simple puzzle even more simple by needlessly using the word "triangle" to describe the shape of the shaded area.
Once the fact that the dimensions on all of them were the same clicked with me, it just fell right into place.
And then I spent a minute going "DUUUUUUUH!" because it was so painfully obvious at that point.
Ooh, I worked this one out in my head without using a formula to solve it. When I apply my logic to how I worked it out in my head, I can conclude that the math is correct. Now I want to learn how to apply the formulas that traditional math uses so I can translate my logic and thought process onto paper when seeing more of these problems 😄
Normally I can't do these in my head, but this one I could. I think it's because I already knew the tricks involved. There are a couple of visual tricks going on that try to make it more confusing than it needs to be, but if you catch them then it's simple.
Other than being on CD, the location of E is not specified. So we are free to slide it up to point C making the blue triangle a right triangle and the problem becomes even easier than it already is.
Now we know how the area formula was figured out :)
The area of the triangle is Base (AB) x height/2
The length of the height is the same as BC.
So the area of the triangle is just half the area of the big rectangle.
Proof: Imagine the height // to BC passing through E and intersecting AB in F.
Now you have 2 right triangles and AF+FB=AB
Nice Content for all ages and ability!
It’s very simple if 5 rectangles are congruent instead of equal area only.
How do we know 5 rectangles are congruent? If they are not congruent, there are many solutions.
@@byungchunkim2702the problem tells you they are congruent.
Mathtastic as always!🎉
As simple as it looks, just 2 equations n done!
1) 2l+b = 10
2) 10b² = 5lb
So, l = 4, b = 2 (b value is enough)
Area = 0.5*2(2)*10 = 20
I figured out the answer mentally although my calculations were more and I used the fact that area of triangle ABE = area of triangle ABD=half the area of rectangle ABCD
Nice job.
I followed this one completely
This is not presented to scale. The rectangles don't look congruent (same shape and size). The middle one looks longer than the other 4. So, ignore the picture to some degree - it's deceptive.
Hight of of R3 is twice that of all the others.
Ratio of all the rectangles are 2:1
AD =(2 +1 +2)k =10
5k =10
k=2
All rectangles are 4×2
AB =(1 +1)k =(2)(2) =4
Area =(AB ×AD)/2
=(4×10)/2
=20
I got it in a different way using Area instead of side lengths
Area of the entire shape = 10*2w
Area of the entire shape = 5* Area of each rectangle = 5*l*w
5*l*w = 10*2*w => l*w = 4*w =>l = 4
10 = 2*4 + w => w = 2
2*2*10 / 2 = 20
Missed the 2w=l trick. The rest is straightforward
I solved this in my head in a minute :D
Same here. And that is pretty rare for me on this channel. So I would consider this as one of the easier problems on this channel.
Small rectangles are X*Y
With 2X+Y=10 and 2Y=X
The Surface S=2Y*10/2=10Y
Let's find Y:
4Y+Y=10 ie Y=2 (and X=4)
Surface is then S=20
I finished this faster than you could say jackknife
Wow, that was a good one!
@@English_shahriar1 ΒΟΤ
length of rectangles must be 2x of the width, so 2L+(1/2)L=10 so L=4, area = (Lx10)/2=20
I did the same math to find w and l, and then I logiced that since the location of point E was not specified, it must not matter, so if I put it up in the corner, ABE is therefore half the rectangle, thus, 20 square whatevers.
It's so confusing that the congruent rectangles are not even remotely congruent in the diagram, for no apparent reason.
probably to emphasize that "diagram not to scale" is extremely common
Is it wrong to call these triangles congruent when they don't have the exact same measurements? Is it more correct to simply say their areas are equal?
I see now that its not to scale.
These *rectangles* are congruent.
Trivial once you see that l=2w. But that's the trick.
Once again, picture is not to scale, since no way R3 is congruent to the other rectangles.
It doesn't need to be up to scale until and unless you have maths on your side
Why can't it?
R3 can be absolutely congruent to others
Think of it as 5 rectangles of same dimensions whose length is double the widht arranged in the given way
You should never assume something is to scale unless explicitly stated
@@FalseNoob777visually, R3 is very clearly not congruent to the other four. Mathematically, it can be, but the point is that it looks visually different, which could throw off the reader if they don't pay attention.
Easy one today 🎉
This was nice
I think vertical w not equal to horizontal w
Same with l, vertical l not equal with horizontal l
But I agree horizontal l is 4 and horizontal w is 2
The vertical l or vertical w may vary, but adopt 2w=l
Just my opinion
I really need to start pausing the video when you say so. I figure out where to start, and then you're on how to solve it.
Sorry, but the Rs look like they are side lengths, not areas (or the shape itself).
I got it in like 30 sec so i had to double checked if I was wrong on anything ☠️
I did it exactly the same way
The answer is 0.5 * (2 S) * (2 L + S) where 2 L + S == 10 and L == 2 S
This is a simple problem (only mental math needed) once you take the statement about congruence seriously, even though as drawn, the rectangles clearly don't have identical dimensions. All it takes is pretending that they do. :)
20
Done this one without using a paper :)
Wait. How you concluded that l=2w? 4 rectangles are same, one in middle is different. Thats all, so?
The problem states that all the rectangles are congruent. The picture is deceptive because it is not drawn to scale.
@@Richie_P I know that scale doesn't match, so even not look at it. But still, how get from point A to B, that is what matters. There is ABCD rectangle divided on 5 equal by area rectangles. ABCD one dimmension is 10. That is all. Still same question is not answered.
Congruent means the same, but moved by a combination of translation, rotation, and reflection across a line. In this problem the 5th rectangle is not only the same area as the others, it also has the same ratio between its sides. That's one of the properties of congruence.@@hvnterblack
@@MichaelOnines so 'conguence' means all are same shape, AxB? English is not my native, maybe missing word made me wrong. Btw, thanks for clearing it.
Yes. Congruence means they are the same shape, AxB. They are just in different positions.@@hvnterblack
Too easy 😮
What abad drawing!!!
No, it's abcd.
no 20 is incorrect
the width of the middle rectangle is not the same as the width of the other rectangles
I spotted the 2l + w = 10 long before l = 2w even though I knew I needed another equation lol. Dunce cap time! Or perhaps bedtime instead of watching math videos at 2 AM.
I forgot what congruent means, though I guess at least I knew from the get go the area would be half of the rectangle as a whole? Once he explained what congruent meant it kinda just clicked. Kinda embarrassing that I had already forgotten congruent though XD
You just have to throw yourself at it and put in some variables for this one 😃
00:08 "..as shown." What lie!
20?
Answered from thumbnail alone before watching the video.
The area of the big rectangle is 40.
And the area of the shaded triangle is 20.
I'll reveal my methods if correct.
Always a risk - the thumbnail often leaves out critical information you get from the full problem description.
Yeah, 40 and 20.
I immediately saw that the rectangles we're twice as high as wide.
If the long side of the big rectangle is 10, this means the small rectangles are 4, 2, 4 on that edge. Which means the small rectangles are 4×2, which means the large rectangle is 4×10.
This makes the area of the large rectangle 40.
And, the easy part most of the others seem to have missed
A triangle with the same base and height as a rectangle will be precisely half it's area, no matter the angles.
So, the area of the triangle is 20.
Now to watch the video.
Eh, the video presented the information as matter of fact, you should know this stuff.
So, probably, no one else thought of mentioning it.
For this video the thumbnail and the problem presented matched. There have been numerous videos where information essential to finding a solution did not appear in the thumbnail. I've been bitten by that mismatch a number of times.@@cooltaylor1015
SOLUTION HAS A FLAW
they are congruent, doesn't mean they are equal
if the dimensions of the central rectangle are L and W that mean dimensions of other rectangles maybe 2L and 2W
else you need to give proof that those rectangles are equal
What definition of "congruent" are you using?
picture is confusing.....
In the thumbnail it says “equal”, not congruent. Solve that problem instead. 😏 I suppose “equal” is to be interpreted as having equal areas so the triangle has an area that is 2.5 rectangles.
Nice
Let the longer edge of the small rectangle be a and the other b.
2a+b = 10
5ab = 10 * (ab/2) = 10 * 2b
a = 4, b = 2
Area = 10 * b = 20.
30 seconds
But perpendicular height we take, here height is not perpendicular
All we need is height, perpendicular or not so the answer would be the same wherever along CD the apex lies.
Even if we extend CD either way in a straight line and put the apex anywhere along it the answer would be the same.
That's triangles for you.
The height IS perpendicular. The height is a line segment from E to the base AB parallel to AD. That line segment has length 10 and is perpendicular to the base AB.
The height is perpendicular to the base because the base is one side of the rectangle and the height is measured as the length of a perpindicular side of the rectangle. The height is not the same thing as a side of the triangle, unless the triangle happens to be a right triangle.
❤
Waaaaaaa
@@English_shahriar1 alright,noted
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