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Area is 132

Good morning from PATRAS, GREECE. I saw your clever and more geometrical solution. My solution, i think, is more "learning" with using more Algebra. Let's name ABCD the given square and KLMN the green quadrilateral in which: AK=a, BL=b, CM=c and DN=d. Then AL=16-b, BM=16-c, CN=16-d and DK=16-a. Let's (X) the aria of KLMN. Then, (X)=(ABCD)-S (1) where S=the sum of the areas all the 4 right triangles. We know that (ABCD)= 16^2=256CM^2 (2) (AKL)=1/2[a(16-b)]=(16a-ab)/2. In the same way, (BLM)=(16b-bc)/2, (CMN)=(16c-cd)/2 and (DKN)=(16d-ad)/2. Adding S=(AKL)+(BLM)+(CMN)+(DKN)=...={16(a+b+c+d)+[(a+c)+(b+d)]}/2=> S=112cm^2 (3) So, replacing on (1) by (2), (3) we easily have (X)=144cm^2.

a = 2 + b (b+2)² + b² = 10² b² + 4b + 4 + b² = 100 2b² + 4b - 96 = 0 ∆ = 16 + 768 = 784 √784 = 28 -4 + 28 /4 = 6 b = 6 > a = 8 A = (6+8)² = 196 cm²

To make it more simple, it is twice the area of the middle square minus the area of the small square, ie 2×10^2-2^2=196

Thank you 😊

A = 10 cm x 2 = 20 cm B = 5 cm x 2 = 10 cm Therefore, A + B = 20 cm + 10 cm = 30 cm

Use Paithagorus theorem and common sense. 8+6=14 14x14=196 easy

100 (square inside is also blue)

Thanks