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.😄
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. 🤠
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).
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.
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
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.
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 √ )
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
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
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).
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
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
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
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
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.
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
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.
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
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
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!
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!
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
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)
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.
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
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 :)
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
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First time I solved a preMath question. It is very interesting
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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.😄
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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. 🤠
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.
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).
I saw it this way as well, and it's only a few short steps!
Cool solution when you break it down that way. Thanks!
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.
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
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.
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
Thanks Professor, regardless of the difficulty of the problem your style of explanation is unchanged, and your voice is very relaxing.❤
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Thank you very much for this nice geometric puzzle and your excellent explanation!
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 √ )
Nice easy one, thanks, my mind is I think getting sharper because of your training 👍🏻
Good Mornin Master
Thanks Sir
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
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Thank you PreMath for giving me many good moments and keeping my brain alive 🎉😊
My approach was a bit different, but I got the same answer. I felt a much amused
Thank u. Have a good day.
You have a great voice...great work as well..... Really appreciate.....
there were so many "times" ...and said sonorously....... was nice .....😀
Very enjoyable ,
Thanks PreMath .
Thanks Sir .
Nice and awesome, many thanks, Sir!
ab = 12
12 + b(b) = 21 → b = 3 → a = 4 → a(a) = 16
I love these puzzles. Thanks for posting! Bye
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
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
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Very well explained👍
Thanks for sharing😊😊
I found it! It's right there!
Very neat. Never learned that criss-cross method. Knowing all the areas, is it possible to then determine the side lengths?
Yes, absolutely!
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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).
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
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
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
You could assume the values are integers and see if it works since the product of 21 is two prime numbers.
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
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
Assuming integer numbers:
21+12=33
Next square numbers are 36; 49; 64; etc.
36-33=3
49-33=16=4² 😊
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.
شكر
نضعDC=X ونحدد بدلالة X وبطريقتين ضلع المربع الأزرق نجد X=7 وبالتالي فإن المساحة المطلوبة هي 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
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
(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
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.
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
Good question
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
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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!
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!
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
3 x 7 = 21. 3 x 4 = 12. 7 - 3 = 4. 4 x 4 = 16 cm^2
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I did this in my head.
3*7=21. 7*7=49 7-3=4 4*4=16
9m^
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
using whole numbers the bottom is 3x7 therefore the uppermright is 3x4 and the remaining is 4x4
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.
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.
The similarity (proportion) technique did the trick
16 Cm² ( = 4 Cm × 4 Cm )
By observation, the area is 7×7 = 49
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
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.
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16 cm2
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.
3×4=12
3×7=21
=>4×4=16
16
Ans : 16 sq units
Area of blue shade=4x4=16cm²
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)
What could be solved in 30 secs, you solved in 8 mins. great.
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.
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
Green is 3x4. Yellow is 7x3. Blue is (7-3)x4 = 16 😅 Is this ok?
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 :)
@@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...
Area of? =16
Must put in video begining than abcd is also a square. :(
16cmsq
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
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!!
Troppo cervellotico😨
Very Poor and Boaring solution , You have given .it just takes 4 steps to solve.
16
16