1. **State the problem:** We are given four triangles with various side lengths and angles, and we need to find the value of $x$. 2. **Analyze each triangle:** - Triangle MLN is a right triangle with sides $ML=8$ and hypotenuse $MN=10$. - Triangle PST is a right triangle with legs $PS = x+2$ and $ST = x-1$. - Triangle FDE has angle $E=40^\circ$, sides $DE = x-3$ and $FD = x+1$. - Triangle ABC has angle $B=40^\circ$, sides $AB=6$ and $AC=12$. 3. **Use the Pythagorean theorem on triangle MLN:** $$ML^2 + LN^2 = MN^2$$ $$8^2 + LN^2 = 10^2$$ $$64 + LN^2 = 100$$ $$LN^2 = 36$$ $$LN = 6$$ 4. **Triangle PST is right angled at S, so apply Pythagorean theorem:** $$PS^2 + ST^2 = PT^2$$ $$(x+2)^2 + (x-1)^2 = PT^2$$ We don't know $PT$, so let's check if PST is similar to MLN by comparing ratios. 5. **Check similarity between MLN and PST:** Sides of MLN: 8, 6, 10 Sides of PST: $x+2$, $x-1$, $PT$ Ratio of legs in MLN: $\frac{6}{8} = 0.75$ Ratio of legs in PST: $\frac{x-1}{x+2}$ Set equal for similarity: $$\frac{x-1}{x+2} = 0.75$$ $$4(x-1) = 3(x+2)$$ $$4x - 4 = 3x + 6$$ $$4x - 3x = 6 + 4$$ $$x = 10$$ 6. **Check triangle FDE and ABC for similarity:** Both have angle $40^\circ$. If triangles are similar, ratios of sides opposite the angle should be equal. In ABC, side opposite angle $B$ is $AC=12$, adjacent is $AB=6$. Ratio $\frac{AC}{AB} = 2$ In FDE, sides are $DE = x-3$ and $FD = x+1$. Set ratio equal: $$\frac{x-3}{x+1} = 2$$ $$x - 3 = 2(x + 1)$$ $$x - 3 = 2x + 2$$ $$-3 - 2 = 2x - x$$ $$-5 = x$$ 7. **Since $x$ must be positive for side lengths, discard $x=-5$.** 8. **Final answer:** From step 5, $x=10$ satisfies the problem conditions. **Answer:** $\boxed{10}$