Find the area of the square and length x #geometryskills #mathpuzzles #thinkoutsidethebox

Find the area of the blue region and the length x

This will be so much appreciated.

#mathpuzzles #geometryskills #thinkoutsidethebox

[fried_ady]

Let the other leg of the side of the triangle (base 4)=f. So, the triangle directly above the base 4 triangle will have similar sides, since 180 degrees will equal 90 plus two complementary angles of the triangles, meaning they share the same complementary angles themselves (sum of angles in a triangle). Since the big unshaded is a square itself, the length will be 7. Thus, in respect to 4 and f, the side of the blue square will equal the square root of 4^2 + f^2. Thus, by similarity, this side, say k, over the base 4, is equal to the hypotenuse of the triangle directly above (7-f) over k. So, we get that (7-f)/k = k/4. Rearranging, we get 4(7-f)=k^2. So, 28-4f=(sqr 4^2 + f^2). So, 28-4f=16+f^2. This is a quadratic, giving f^2+4f-12=0. By product and sum, f^2+4f-12=(f+6)(f-2), giving f=2. That means the area of the square, k^2, equals the square of the square root of 4^2 + f^2, or simply 16 + f^2. Substituting f, we get 16+4, = 20 units squared. As for x, by similarity (top right corner, complementary angles) x over 7 equals 2/4, or x=7(1/2), = 7/2.

[lisepenot]

My name is lise . I'm 16 years old and I want to become a mathematicuan but I am a average student. Can you give advice to become good as at mathematics and how do i can achieve your mathematics skills please ?

[murdock]

Nice!
a = AD = DE
BC = 7; AB = 4 → BD = k → CD = 7 - k → cos⁡(φ) = k/a = a/(7 - k) →
k = √(a^2 - 16) → 7 - k = 7 - √(a^2 - 16) →
cos⁡(φ) = 4/a = a/(7 - √(a^2 - 16)) → a^2 - 16 = 49 + (1/16)a^4 - (7/2)a^2
m ∶= a^2 → m^2 - 72 = -1040 → (m - 36)^2 = 256 →
m1 = 36 + 16 = 52 > 49 ≠ solution;
m2 = 36 - 16 = 20 → a = √20 = 2√5 → a^2 = m = 20
btw: k = 2 → sin⁡(φ) = k/a = √5/5 → φ ≈ 26, 565°