First time I solved a preMath question. It is very interesting

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome, Devi. Keep it up 👍 Love and prayers from the USA! 😀

Let s be the side of the blue square，s+t be the side of the large square, then st=12, t(s+t)=st+t^2=21, so t^2=21-12=9, t=3 and s=4, therefore the answer is 4^2=16, done.😄

@PreMath

Жыл бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Here's another approach, no peeking: let x be the side of the unknown small square and x + y the side of the big square. Then we have x^2 = (x + y)^2 - (12 + 21) = (x + y)^2 - 33; Looking at the small upper right rectangle, xy = 12; so y = 12/x; and we now have two independent equations. Substituting 12/x for y in the first equation, x^2 = (x + 12/x)^2 - 33; expanding, x^2 = x^2 + 144/x^2 + (2x)(12/x) - 33; simplifying, x^2 = x^2 + 144/x^2 + 24 - 33; subtracting x^2 from both sides and rearranging 144/x^2 - 9 = 0; or, 144/x^2 = 9; multiplying both sides by x^2, 9(x^2) = 144; and finally, x^2 = 144/9 = 16. This is the area of the unknown square. Check: x = √16 = 4; y = 12/x = 12/4 = 3; lower rectangle : y(x + y) = 3(4 + 3) = 21; upper right rectangle: xy = (4)(3) = 12; entire large square: (x + y)^2 = (4 + 3)^2 = 7^2 = 49; 21 + 12 + 16 = 49. Cheers. 🤠

Cool solution when you break it down that way. Thanks!

Two squares mean that the area in the bottom left corner is the same as the area in the top right corner, which is 12 cm². Therefore, the area of the square in the bottom right corner is 21 cm² - 12 cm² = 9 cm². So, the side length of the square in the bottom right corner is 3 cm. This means that the blue square must have an area of 16 cm². The big square covers an area of (4 + 3)² = 49 cm² (= 16 + 2 * 12 + 9 cm²).

@davidtipton514

Жыл бұрын

I saw it this way as well, and it's only a few short steps!

Good Mornin Master Thanks Sir

area of Blue Square = x^2 area of Green Rectangle = xy = 12 area of Yellow Rectangle = (x+y)y = 21 xy+y^2=21 y^2=9 y=3 x=4 4^2=4＊4=16 16cm^2

I have watched many math channels, you sir are the best. You're a blessing to many math students and to those who love math.

Thank you PreMath for giving me many good moments and keeping my brain alive 🎉😊

Very enjoyable , Thanks PreMath . Thanks Sir .

It seems there is a much easier way. A rectangle Area 12 must have sides of either 12x1, 6x2 or 4x3. The rectangle Area 21 can only have sides of 7x3 OR 21x1. Logically, therefore, the sides of the large square are 7x7, or 49 units. 49-(21+12)=16.. this mental math took me about 30 seconds

@jxod4296

3 ай бұрын

Just because it worked for the given example doesn’t mean it will work all the time. One shouldn’t assume that the sides of the green and yellow rectangles are integers. If the yellow rectangle’s area is 18 then your method doesn’t work.

Nice easy one, thanks, my mind is I think getting sharper because of your training 👍🏻

I love these puzzles. Thanks for posting! Bye

I found it! It's right there!

You have a great voice...great work as well..... Really appreciate..... there were so many "times" ...and said sonorously....... was nice .....😀

Very well explained👍 Thanks for sharing😊😊

My approach was a bit different, but I got the same answer. I felt a much amused Thank u. Have a good day.

Thanks Professor, regardless of the difficulty of the problem your style of explanation is unchanged, and your voice is very relaxing.❤

@PreMath

Жыл бұрын

Many thanks! So nice of you, my dear friend You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@bigm383

Жыл бұрын

@@PreMath 🥂👍😀

let A A be the blue side square and B the remaining length of the larger square. you have AxB=12 and B*(A+B)=21 this get you to AxB + B*B = 21 B*B=21-12=9 so B=3 12/B=A (first formula) so A=4 and area is 16.

AB=AD=a, the side of the blie square is b. Green Area A1= b.(a-b) =12 Yellow area A2= a.(a-b)=21 A1/A2= b (a=b)/a(a-b) = 12/21= 4/7 a=7b/4 b (7.b/4 -b) = 12 b.3b=12.4 b^2=4.4=16 so blue Area A= b.b=16

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Thanks for video.Good luck sir!!!!!!!!

@PreMath

Жыл бұрын

You are very welcome! Thanks for your continued love and support! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Very neat. Never learned that criss-cross method. Knowing all the areas, is it possible to then determine the side lengths?

@PreMath

Жыл бұрын

Yes, absolutely! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

At a glance, if all side lengths are integers then 21=7*3 and 12 =4*3 hence large square side = 7 and Area of blue shaded area is 7 * 7 - 21 -12 = 16 cm^2, a square of side 4 cm^2

I assigned the variable X to represent the length of the sides of the square in centimeters. Then the remaining value on the top of the larger square would be 12/X. This would yield two equal expressions for the area of the larger square; x^2 + 12 + 21 = (x + 12/x)^2 . This would be reduced to x^2 + 33 = x^2 + 24 + 144/x^2. This would simplify to 33=24+144/x^2. Finally 9=144/x^2 ; 9x^2 = 144 ; x^2 = 16 cm. sq. (The area of the blue square).

By extending the parallel to AD, the diagram becomes a visual proof that (x + y)² = x² + 2xy + y² So here we have y² = 9 and xy = 12. Therefore, the area of the blue square is (12 / √9)² = 16 cm²

Nice and awesome, many thanks, Sir! ab = 12 12 + b(b) = 21 → b = 3 → a = 4 → a(a) = 16

I did this in my head.

I think sit that rectangle 21 area is formed by 7x3 , the other little triangle 12 is formed by 4x3 , so the length of square is (7-3) = 4 , so area of square is 16

Good question

@Vedant-Goyal

Жыл бұрын

Another way could be : Yellow rectangle area = 21 , only 7cm x 3cm possible. Green rectangle area = 12 = 2x6 = 3x4 . Acc. to situation only 3x4 is possible to make blue region a square . So, area of blue region = 4² = 16cm²

@PreMath

Жыл бұрын

Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Prolongamos el lateral izquierdo verde hasta DC y ABCD queda dividido en cuatro celdas 》Si AB=a+b 》Áreas celdas: (a^2), (a×b=12), (a×b=12), (b^2=21-12=9) 》a^2/12 = 12/9 》Área cuadrado azul =a^2 =12×12/9 =16 》》Otra solución: b^2=9》b=3》a=12/3=4》a^2=16 Gracias y un saludo.

Designate: the side of the blue square is 'a'; the width of the green rectangle is "b". Squares: ab=12; b(a+b)=21 ab+b^2= 21; 12+b^2= 21 b^2= 9; b=3; a=4; С=16

شكر نضعDC=X ونحدد بدلالة X وبطريقتين ضلع المربع الأزرق نجد X=7 وبالتالي فإن المساحة المطلوبة هي 16

The similarity (proportion) technique did the trick

a² is área of blue square b² is area if big square Area of yellow rectangle: Ay = b. (b-a) = 21 Area of green rectangle: Ag = a. (b-a) = 12 Dividing: b/a = 7/4 ; b = 7/4 a Ag = a. (7/4 a - a) = 3/4 a² =12 a² = 4. 12 / 3 Area of blue square: A = a² = 16 cm² ( Solved √ )

using whole numbers the bottom is 3x7 therefore the uppermright is 3x4 and the remaining is 4x4

Let length of square be a. Let the breadth of 12 cm^2 rectangle be b. ab=12 (a+b)b = 21 b•b =21 -12 b=3 a=4 area of blue square = 16

I just assumed the answer would be an integer solution. We have the smaller square plus 2 rectangles = a larger square. So we find a² = b² + 12 + 21 where a & b are integers. There are only 2 solutions where a² -b² = 33. a =17, b = 16 and a = 7, b = 4. 16 and 17 obviously won't work due to the layout so the solution must be 7 and 4. 16 is the answer.

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@notovny

Жыл бұрын

Yeah, I think this puzzle's difficulty is downgraded a bit by using a semiprime for the yellow triangle's area. 12 and 21 are basically solve-on-sight numbers where the first intuitive guesses are the answer.

I made s the length of the yellow rectangle. I made a the length of the yellow rectangle. a * s =21. Length of green rectangle is s-a. Width of green rectangle =b. (s-a)*b =12. The blue square has dimensions of (s-a) * (s-b). In my system of equations s-b = s-a, so a =b. (s-a)*a =12 and s= 21/a. (21/a - a) *a becomes a^2 - 9 = 0 a=3 or a = -3 (rejected), so a = 3. b=3. s-a =4 and s-b =4. The area of the small blue square = 4 *4 = 16.

Assuming integer numbers: 21+12=33 Next square numbers are 36; 49; 64; etc. 36-33=3 49-33=16=4² 😊

Other easier method: Let's green square area be: ab=12 (so a is also the lenght of the blue square and b is the shorter lenght of the green square). The blue square area is: a^2 Square ABCD area is: a^2+12+21=(a+b)^2 a^2+12+21=a^2+b^2+2ab a^2-a^2+12+21-2ab=b^2 12+21-24=b^2 9=b^2 b=3 From ab=12, a=4 Blue square area is a^2=4^2=16

(a+12/a)'2 = a'2 + 12 + 21 , a'2 + 24 + 144/a'2 = a'2 + 33 , 9 = 144/a'2 , a'2 = 16

What could be solved in 30 secs, you solved in 8 mins. great.

Answer 4^2 = 16 Could solve by letting the blue square side = x; hence the green sides are x, and 12/x; and hence the large square sides are x + 12/x; hence the area of square ABCD = (x+12/x)^2, or x^2 + 144/x^2 + 24. But this area = x^2 + 12 + 21; hence x^2 +144/x^2 + 24 =x^2 +12 + 21 144/x^2 = 33-24 (12/x)^2 = 9 (12/x)^2 =3^2 12/x = 3, the square root of both sides 12 = 3x cross multiply x=4 , the side of the blue Hene its area = 16

@devondevon4366

Жыл бұрын

You could assume the values are integers and see if it works since the product of 21 is two prime numbers.

9m^

By observation, the area is 7×7 = 49

16 Cm² ( = 4 Cm × 4 Cm )

If blue square is length X and large square is length X+Y, then Ysquared = 21-12 IE 9 so y=3, and XY =12, so X=4. Area of blue square =16 is much simpler!

The problem is over when a(b-a) = 12. so (b-a)sq = 9 so b-a = 3. (b-a)*a = 12 so a = 4. so b = 7 & Blue area = 16.

(b-a)*( b-a)=9 Or b-a =3 Now b*(b-a)= 21 Thus b = 21/3= 7 And a= b-3=4 😮 Thus given square = 4*4 = 16

3 x 7 = 21. 3 x 4 = 12. 7 - 3 = 4. 4 x 4 = 16 cm^2

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Your solution was very complicated!! With an area of 21 sq cm, the only possible dimensions for the yellow rectangle would be 7x3. Therefore, DC=7cm and the total area would be 49 sq cm. 49sq cm - 21sq cm - 12 sq cm = 16 sq cm. Simple solution!

@WernHerr

Жыл бұрын

Where did you find in the specification that the solutions must be integers? Also e.g. 5 and 4.2 result in 21 as area

x^2 - y^2 = 33 x^2 área great cuadrado y^2 área small cuadrado (x - y)(x + y) = 33 x(x - y) = 21 (a) x - y = 21/x (x - y)y = 12 (b). x - y = 12/y 21/x = 12/y 12x = 21y x = 21y/12 (c) (x - y)y = 12 (21y/12 - y)y = 12 ((21y - 12y)/12)y = 12 (9y)y = 144 y^2 = 144/9 = 16 De un chichombiano from BOGOTÁ D.C.

3*7=21. 7*7=49 7-3=4 4*4=16

16 cm2

16

Again, without drawing lines: a * (b - a) = 12 b - a = 12/a b * (b - a) = 21 b - a = 21/b 12/a = 21/b b * 12 = a * 21 b = 21a/12 b = 7a/4 a * (b - a) = 12 a * (7a/4 - a) = 12 a * (7a/4 - 4a/4) = 12 a * 3a/4 = 12 3a^2/4 = 12 3a^2 = 48 a^2 = 16

Area of blue shade=4x4=16cm²

LoL I just look at the shape and give random value (4;3) to the rectangle of 12cm square. So the blue square is 16cm2. Sides of the large square are 7cm

x^2 = ? x*y = 12 z *(x+y) = 21 x+y = x+z y =z x*y= 12 xy + y^2 = 21 x = 12/y y^2 + 12 = 21 y^2 = 9 y =3 x = 4 x^2 = 16

Let the side length of the blue square be x and the side length of the entire square be y. As the blue square has height x, so does the green rectangle. By observation the width of the green rectangle is y-x, so the area is x(y-x). By observation, the height of the yellow rectangle is y-x, so the area is y(y-x). By observation, the section of the yellow rectangle directly below the blue square is the same area as the green rectangle, x(y-x), so its area is also 12. This means the remainder of the yellow triangle, with area (y-x)², has an area of 9. Therefore (y-x) = √9 = 3. As (y-x) = 3, then as the green rectangle has area x(y-x) = 12, then x = 12/3 = 4. Area of blue square = 4² = 16cm²

Let a= side length of the blue square and b= side length of square ABCD. (b-a)a= 12 ..................................(1) (b-a)b= 21...................................(2) If we divide (1) by (2), a/b=12/21=4/7 This gives, a=4b/7........................................(3) Substitute a into (1), (b-(4b/7))(4b/7)=12 (4b^2)/7-(16b^2)/49=12 7(4b^2)-16b^2=12*49 28b^2-16b^2=12*49 12b^2=12*49 b^2=49 >>>>>>>>>>>b=7cm..........................(4) IABCDI=b^2=49 sq.cm. Substitute b=7 into (2), (7-a)7=21 49-7a=21, 7a=28 a=4 cm...............................................................(5) Iblue squareI=a^2 =4^2 =16 sq.cm. and this is our answer. Check; Iblue squareI+Igreen squareI+Iyellow squareI=IABCDI 16+12+21= 7*7=49 sq.cm 49=49. Thanks for the puzzle professor.

another way...... yellow is 21 sq cm=>7x3 or 3x7....no other possibility...... means larger side is 7.....so whole rectangle is 49 sq cm 49 - 21-12 = 16sq cm....... so 4x4 16 sq cm is the area we are seeking to find

@WernHerr

Жыл бұрын

Where did you find in the specification that the solutions must be integers? Also e.g. 5 and 4.2 result in 21 as area

Blue=4*4=16 unit.

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome, Amitava. Keep it up 👍 Love and prayers from the USA! 😀

Ans : 16 sq units

16cmsq

3×4=12 3×7=21 =>4×4=16

Here is my incredibly lazy shortcut. Assume that 12=3x4 and 21=7x3. If the green rectangle is 3cm wide its height must be 4cm, which will also be the height of the blue square. The yellow rectangle must then have a width of 7cm and a height of 3cm, confirming the width of the blue square as 7-3=4cm. Thus the blue square has an area of 4x4=16cm^2. This is confirmed by the outer square having a side length of 7 and an area of 7x7=49cm^2, which is equal to the total area of the coloured shapes: 16+12+21.

Quick Solution: 7X3 are the only numbers to equate to 21. If the big rectangle is square, then length of blue square is 4 and its area is 16! (1 & 21 don't work)

Area of? =16

Troppo cervellotico😨

Green is 3x4. Yellow is 7x3. Blue is (7-3)x4 = 16 😅 Is this ok?

@fred_2021

Жыл бұрын

Yes, the numbers are such that they can be guessed immediately, without putting pen to paper, but if they weren't convenient integers that would be impossible for most of us :)

@neofolk3051

Жыл бұрын

@@fred_2021 Ah, That's right. Thanks!

easy. 12 cm2 is a 3x4 square, 21cm2 is a 7x3 square, making the last a 4 unit square. 16cm2...

b(a+b)=21 ab + b^2 =21 but ab=12 so b^2=9 B=3 so b=12/3 = 4 b^2=16 Don't make it so complicated. Also there is nothing in the diagram that indicated that ABCD is a square, there should be.

Must put in video begining than abcd is also a square. :(

You made solution too complicated, if areas of square is 9 then side is 3, you can calculate everything from there

You assumed that the big rectangle is square. But there is no information to lead to this assumption!!

This method is too complicated. Let the side of big square is s, the width of green rect is a . As blue one is a square, the height of yellow rect should be equal to a as well. the area of yellow rect is equal to sa= 21 (1) the area of green rect is a(s-a)=12 ==> sa-a^2=12 (2) (1)-(2) ==> sa-(sa-a^2)= 21-12 ==> a^2 =9 ==> a=3. Put a=3 into (1) ==> s=7. Then the area of blue square is (7-3)^ =16. very straightforward

Very Poor and Boaring solution , You have given .it just takes 4 steps to solve.

16

16