Find the area of the middle rectangle

Can you solve this geometry brain teaser?

Note: I was emailed a similar problem but I am unable to locate the original source. It seems to be in the style of Bright Side quizzes, but often those are based on problems circulating. Please let me know if can identify the original source and I will give credit. Thanks!

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[agnisain]

That's one of the easiest question you have ever posted. Thanks buddy to boost our confidence up.

[DavidChristmas]

Honestly, this was easy enough to solve just by looking at the thumbnail of the video!

[cshanek]

This has probably been said in other comments but the difference between rectangle on left and right is 1m along x, which equates to a difference in area of 5 square meters, which means Y is 5m. This info alone is enough to solve for everything and this approach *seems* even easier than the video approach and other approaches mentioned in comments. This would also work for non-integers.

[jcb]

I actually did this one in my head. Thanks for the confidence boost.

[danielgteaching]

If the area of a rectangle increases by 5 when the width increases by 1, then the height must be 5/1.
Once you know the height, everything else is solved pretty well instantly.

[forstudypurpose]

Every one is only thinking about how difficullt he can solve math
No ome is talking about how many effort he is taking to animate these animations

[gondikhushalreddy]

I have solved problem in another way
let 'X' be area of rectangle
Then we have
(20+X)/7=(25+X)/8
8X+160=175+7X
X=15

[HisMajesty]

This one was easy enough to do in my head
He’s going easy on us now 😆

[StopColdYou]

Although I can understand your explanation, I find it personally overcomplicated . Here is a much simpler decision of only few rows:
Lets name the vertical side of that rectangle "Z". Then 25 + x = 8 x Z and 20 + x = 7 x Z. If we substract the two equations, then Z=5m. If we replace in the first equation, then 25 + x = 40. Then the middle rectangle have an area of 15m2. also the smaller side have a length of 15 7 5 = 3m. There is another one (a bit longer) short solution.

[Korpionix]

That’s the easiest MYD question I’ve ever seen.

[rhamzellibay]

Solved it in 10 seconds. The fact that an increase of 1 meter with the length of the second and third rectangles gives it 5 more square meters than the first rectangle already means that the width is 5. Then 7*5 = 35 but the first rectangle is 20 square meters then the middle is 15.

[applealvin]

I felt so proud of myself until I saw everyone thinks it’s easy :p

[chinareds]

I'm sure there's an algebraic way of solving it but, I immediately saw that the numbers work if you make the height of the rectangles 5, which would give the left rectangle a width of 4, the middle rectangle a width of 3, and the right rectangle a width of 5. Therefore the middle rectangle must have an area of 15.

The algebriac method works for any values of areas for the left and right rectangles, so it would be a better problem if the areas weren't so easily factorable into 4x5 and 5x5.

[Brookesey]

Yay, I finally got one... I don't care if it was the easiest thing ever

[XionAzura]

I did it a little differently. 25 is 5 squared, so that tells you that the height is 5 and the length of the right rectangle is also 5. 8-5=3 so the length of the blue rectangle is 3. 5*3= 15, so the blue rectangle is 15 square meters.

[hkyeung]

I solved it this way:
Let s = vertical length of the rectangles, A = Area of blue rectangle.
Left rectangle + blue rectangle: 7s = 20+A
Right rectangle + blue rectangle: 8s = 25+A
Solving them we get s = 5 and A = 15

[shangri-la-la-la]

You know the times when you just look at a problem, don't even know how you are getting the answer but know it is 15?

[davittabatadze]

*"Don't forget the m²"* 😂

[zat]

Oh good. I can solve this one.

I solved it differently. My equations were:

The area of the left and middle rectangle is equal to 7 times the side y or
20 + x = 7y

and the area of the middle and right rectangle equals 8 times y or
25 + x = 8y

Subtracted the first from the second to get y = 5 and substituted y to get x = 15.

[eugeneclemente]

I have another solution,
combine 1st & 2nd rectangles and we'll have equation 1 which is
7y = A1 + A2
7y = 20 + A2
then, combine 2nd and 3rd rectangles to have equation 2,
8y = A2 + A3
8y = A2 + 25
then subtract eq 2 to 1
8y = A2 + 25
- 7y = 20 + A2
y = 5
Then A2 is solved; 7(5) = 20 + A2; A2 = 15sqm.