we are given that
two triangles are similar
so, the ratio of their sides must be same
we get
![(2)/(6) =(x)/(x+2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/sw30xpzco05vl0lu0yp446wwtt1ir3wwl4.png)
now, we can solve for x
step-1: Cross multiply both sides
![2(x+2) =6x](https://img.qammunity.org/2019/formulas/mathematics/high-school/d5pnwxbpm3gqyuvo23lfjedjj7bbqghr5c.png)
step-2: Simplify left side
![2x +4 =6x](https://img.qammunity.org/2019/formulas/mathematics/high-school/irlz6hpb4avrd10diwzavh70tndl8o23x4.png)
step-3: Subtract both sides by 2x
![2x +4-2x =6x-2x](https://img.qammunity.org/2019/formulas/mathematics/high-school/7e5voe2u9xhkj9qd6azkeaxl7duw82mpub.png)
![4 =4x](https://img.qammunity.org/2019/formulas/mathematics/high-school/7wl5iycl0adhbozlehf1x61r2p2qlnre0c.png)
step-4: Divide both sides by 4
![x=1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/etd77jui6mm2z2ngebg2jbgevvrgge83wx.png)
so,
............Answer