Question

In a square $ABCD\text{if}\text{AB = 5x +}\frac{1}{3}\text{and}BC=x+\frac{19}{2}\text{find}x$

Answer 1

Given $AB=5x+\frac{1}{3}$ and $BC=x+\frac{19}{2}$

But $AB=BC,\therefore ABCD$ is a square

$\therefore 5x+\frac{1}{3}=x+\frac{19}{2}$

$\Rightarrow 4x=\frac{19}{2}-\frac{1}{3}=\frac{57-2}{6}=\frac{55}{6}$

$\Rightarrow x=\frac{55}{24}$