I did it slightly differently. I used variables: x = length of green circle y = height of green circle Therefore, the rectangle that hangs over 8×7 rectangle has dimensions (x-2)(y-2) and the rectangle that hangs over red 6×7 rectangle has dimensions (x-3)(y-2) From these we get: (x-2)(y-2) + 36 = 8×7 → xy - 2x - 2y = 16 (x-3)(y-2) + 26 = 6×7 → xy - 2x - 3y = 10 Subtracting we get: *y = 6* then substituting into either equation we get: *x=7* Area of green rectangle = 7 × 6 = *42*
I solved it a little bit differently than you but I'm excited to get one of these right without peaking for a strategy tip. I calculated the missing segments of the green squares by calling their components a, b, c, and d and then solving by setting opposite sides equal
The horizontal line where the red & blue rectangles overlap = y The vertical line where the red & blue rectangles overlap = x Area of known value for blue rectangle = (8*7) - [(y + 3) (z + 2)] = 36 14 = yz + 2y + 3z (equation 1) Area of known value for red rectangle = (7*6) - [(2 + y) (z + 2)] = 26 12 = yz + 2y + 2z (equation 2). Subtract equation 2 from equation 1. z = 2. Substitute z = 2 into equation, to find y = 2. Area green rectangle = (2 + y + 3)(2 + z + 2) = 7*6= 42 units²
7 * 6 = 42. 42 - 26 = 16. There are only 2 possibilities for 16 area. 4x4 & 8x2. Ruling out 8x2. That area must be 4x4. So Green rectangle dimensions will be 4+3 & 4+2 So area = 42
I solved this problem differently. If you connect the corners of the green and red rectangles, you get a 9 x 9 square. Draw a diagonal line from the upper left-hand corner of this square to its lower right-hand corner. This divides the square in half, and the line will be at 45°. The new quadrilateral in the upper left-hand corner has a height of 2, but since the 45° line passes through the intersection of the green and red rectangles, that quadrilateral is shown to be a 2 x 2 square. The same is also true of the new quadrilateral in the lower right-hand corner -- it is a 3 x 3 square. From here you can calculate the rest of the line lengths with simple arithmetic, get the area of the green rectangle, and double-check the areas of the blue-green overlap and the green-red overlap.
We suppose that the green has an area G=Base*Height=b*h → Area of the green and blue overlap = (7*8)-36=56-36=20 → (b-2)(h-2)=20 → (b*h)-2b-2h=16 → G=2b+2h+16 → Green and red overlap area = (6*7)-26=16 → (b-3)(h-2)=16 → (b*h)-2b-3h=10 → G=2b+3h+10 → 2b+2h+16=2b+3h+10 → h=6 → (b-3)(h-2)=16 → (b-3)(6-2)=16 → b=28/4=7 → Green area=G=b*h=7*6=42 Cheers
42 = Answer let x = the horizontal line of the green rectangle, and y = vertical line; hence xy = the area of the green rectangle. Since if turned blue in a complete rectangle, its area would be 56 (8x7), but the blue area= 36 (given), then (x-2)(y-2) = 20 -2x-2y+ 4 + xy =20 xy =16+ 2x + 2y equation P Since if turned red in a complete rectangle, its area would be 42 (6x7), but the red area =26 (given) then (x-3)(y-2) = 16 -2x-3y + 6 + xy =16 xy =10+ 2x + 3y equation R 16 + 2x + 2y = 10 + 2x + 3y (equation P and R are equal) 6 = y hence the vertical side of the green rectangle= 6 units 6x =16 + 2x + 12 ( substituting 6 in equation P) 4x =28 x =7 hence the horizontal side of the green rectangle = 7 units Area = 7 x 6 =42 Answer.
Fabulous question sir
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I stared at this for SO LONG before continuing with the video -- what an elegant and clever solution! Thank you for posting this!
Thanks for video.Good luck sir!!!!!!!!!!!!!!
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Your sums are very helpful . Thanks..
Great explanation! Thanks PreMath sir!
I did it slightly differently. I used variables:
x = length of green circle
y = height of green circle
Therefore, the rectangle that hangs over 8×7 rectangle has dimensions (x-2)(y-2)
and the rectangle that hangs over red 6×7 rectangle has dimensions (x-3)(y-2)
From these we get:
(x-2)(y-2) + 36 = 8×7 → xy - 2x - 2y = 16
(x-3)(y-2) + 26 = 6×7 → xy - 2x - 3y = 10
Subtracting we get: *y = 6*
then substituting into either equation we get: *x=7*
Area of green rectangle = 7 × 6 = *42*
Another great question. Thank you
Area of blue (8x7=56) 56- 36=20 factors of 20 = 1,2,4,5,10,20. 4 and 5 only possible factors to satisfy the problem 2 +5=7 and 2+ 4=6 6x7=42
Nice approach, though it requires to assume that the sides of the 20 sq. Units rectangle are integer numbers.
Good video, it is a fine solution and interpretation!
I solved it a little bit differently than you but I'm excited to get one of these right without peaking for a strategy tip. I calculated the missing segments of the green squares by calling their components a, b, c, and d and then solving by setting opposite sides equal
Excellent question sir
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Another great video!
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Very interesting I like to solve this problem
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The horizontal line where the red & blue rectangles overlap = y
The vertical line where the red & blue rectangles overlap = x
Area of known value for blue rectangle =
(8*7) - [(y + 3) (z + 2)] = 36
14 = yz + 2y + 3z (equation 1)
Area of known value for red rectangle =
(7*6) - [(2 + y) (z + 2)] = 26
12 = yz + 2y + 2z (equation 2).
Subtract equation 2 from equation 1.
z = 2. Substitute z = 2 into equation, to find y = 2.
Area green rectangle =
(2 + y + 3)(2 + z + 2) = 7*6=
42 units²
thnku good question
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Wow that was amazing
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7 * 6 = 42. 42 - 26 = 16. There are only 2 possibilities for 16 area. 4x4 & 8x2. Ruling out 8x2. That area must be 4x4. So Green rectangle dimensions will be 4+3 & 4+2 So area = 42
I solved this problem differently. If you connect the corners of the green and red rectangles, you get a 9 x 9 square. Draw a diagonal line from the upper left-hand corner of this square to its lower right-hand corner. This divides the square in half, and the line will be at 45°. The new quadrilateral in the upper left-hand corner has a height of 2, but since the 45° line passes through the intersection of the green and red rectangles, that quadrilateral is shown to be a 2 x 2 square. The same is also true of the new quadrilateral in the lower right-hand corner -- it is a 3 x 3 square. From here you can calculate the rest of the line lengths with simple arithmetic, get the area of the green rectangle, and double-check the areas of the blue-green overlap and the green-red overlap.
We suppose that the green has an area G=Base*Height=b*h → Area of the green and blue overlap = (7*8)-36=56-36=20 → (b-2)(h-2)=20 → (b*h)-2b-2h=16 → G=2b+2h+16 → Green and red overlap area = (6*7)-26=16 → (b-3)(h-2)=16 → (b*h)-2b-3h=10 → G=2b+3h+10 → 2b+2h+16=2b+3h+10 → h=6 → (b-3)(h-2)=16 → (b-3)(6-2)=16 → b=28/4=7 → Green area=G=b*h=7*6=42
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answer 42
Area=98-62+6=36+6=42 sq unit
42 = Answer
let x = the horizontal line of the green rectangle, and y = vertical line; hence xy = the area of the green
rectangle.
Since if turned blue in a complete rectangle, its area would be 56 (8x7), but the blue area= 36 (given), then
(x-2)(y-2) = 20
-2x-2y+ 4 + xy =20
xy =16+ 2x + 2y equation P
Since if turned red in a complete rectangle, its area would be 42 (6x7), but the red area =26
(given) then
(x-3)(y-2) = 16
-2x-3y + 6 + xy =16
xy =10+ 2x + 3y equation R
16 + 2x + 2y = 10 + 2x + 3y (equation P and R are equal)
6 = y
hence the vertical side of the green rectangle= 6 units
6x =16 + 2x + 12 ( substituting 6 in equation P)
4x =28
x =7
hence the horizontal side of the green rectangle = 7 units
Area = 7 x 6 =42 Answer.
I don't see how it can be built
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