Find the area of the square

[@Bayerwaldler]

Cool solution. I didn't use the fact that the inner triangle is a right triangle. I came up with the equation : (x-sqrt(5-x^2))^2 + (x-sqrt(4-x^2))^2 = 1. After some straight forward but messy calculations I ended up with the quadratic equation: 65u^2 - 16*18u + 16^2 = 0 where u =x^2, which can be factored as follews: (5u-16)(13u-16) = 0. That leads to u=16/5 and u= 16/13. Only the first solution is admissable. Not very elegant but lots of fun 😉

[@xualain3129]

My solution with trigonometry is as follows: AB=x=2*cos (alpha) ….(1) AD=c=2*sin(alpha)+1*cos(alpha) ….(2) Equating (1) and (2) 2*cos(alpha)=2*sin(alpha)+cos(alpha) --> tan(alpha)=1/2 ->sec(alpha)=sqrt(1+tan(alpha ^2) sec(alpha)=sqrt(5)/2 -->cos(alpha)=2/sqrt(5) from (1) x=2*2/sqrt(5)=4/sqrt(5) Area of the square =x*x=16/5

[@Samwa351]

correct

[@jacquespictet5363]

Your final calculations are not independent. Maybe you should use one of them and confirm it with BCF: (3x/4)(3x/4)+x*x=25x*x/16=5, thus x*x=16/5. Or Calculate the sum of the 4 triangles.

[@MrBrain4]

Interestingly, it turns out that the shaded triangle is also similar to the other two similar triangles.

[@himo3485]

x²+(2x)²=2² 5x²=4 x²=4/5 side of the square : 2x area of the square : 2x * 2x = 4x² = 4*4/5 = 16/5

[@michaelstockley7897]

As the triangles BAE and BEF are similar, AB/2 = 2/sqrt(5) this solved easily to the answer of 16/5. The solution proposed seems too complex

[@ajitandyokothakur7191]

Your written characters are not legible. Please make them larger. Thanks. Dr. Ajit Thakur (USA).

[@mathu6514]

thank you for the comment, I'll remember to enlarge my handwriting next time.