Part a) By similiar triangle we note that: $BQ = AK=AR=BP=m$ in consequence: $AL=BL=BN=AN=n$, such that: $m+n=5$ and by the law of cossines: $LN^2=AN^2 + AL^2 -2.AN.AL.cos\angle LAN$ $\Rightarrow 26=2n^2 - 2n^2.cos(180^0 - \angle LAK)$, but: $cos(180^0 - \angle LAK)=-cos\angle LAK=-\dfrac{m}{n}$ substituing: $n^2 + n.m - 13=0$ choosing the positive quadratic solution:$n=\dfrac{-m + \sqrt{m^2 + 52}}{2}$: $\Rightarrow \dfrac{-m + \sqrt{m^2 + 52}}{2} + m=5$ $\Rightarrow m^2 + 52=m^2 -20m + 100$ $\Rightarrow 20m=48 \implies m=2,4$ If I understand rightly the overlap area is the sum of the area of the triangles $QLB$ and $RAN$,that's defined by: $\dfrac{BQ}{2} + \dfrac{AR}{2}=m=2.4$ Part b) Knowing that $n=\sqrt{m^2 + 1}=2,6$ and $\angle ALB=\angle LAK$, by using again the law of cossines: $AB^2=AL^2 + BQ^2 -2.AL.BQ.cos\angle ALB$ $\Rightarrow AB^2 =2n^2 -2n^2.cos\angle LAK$ $\Rightarrow AB^2=1,04 \implies AB=\dfrac{\sqrt{26}}{5}$