Geometry Question by Mithil Shah # 1

Medians divide sides into halves and are themselves split 2:1. So from the blue triangles we get:

$b^2+4a^2=\frac{1}{2}$

$4b^2+a^2=\frac{3}{4}$

Solving these leads to: $a^2=\frac{1}{6}, b^2=\frac{1}{12}$

Remaining side is from the pink triangle $\sqrt{4a^2+4b^2}=\sqrt{4\times(\frac{1}{6}+\frac{1}{12})}=1$

A slightly different method from @Marta Reece :

$a^{2}+4b^{2}=\frac{3}{4}$

$b^{2}+4a^{2}=\frac{1}{2}$

Upon adding the two equations,

$5(a^{2}+b^{2})=\frac{5}{4}$

The desired result is:

$4(a^{2}+b^{2})=\frac{4}{5}\times\frac{5}{4}=1=\sqrt{\boxed{1}}$