1. **Stating the problem:** We need to find the shape of the solution region for the system of inequalities: $$x \geq 2, \quad y \leq 8, \quad x - y \leq 2$$ 2. **Understanding the inequalities:** - $x \geq 2$ means the region is to the right of the vertical line $x=2$. - $y \leq 8$ means the region is below the horizontal line $y=8$. - $x - y \leq 2$ can be rewritten as $y \geq x - 2$, meaning the region is above the line $y = x - 2$. 3. **Finding the vertices of the region:** The feasible region is bounded by these three lines. Let's find their intersection points. - Intersection of $x=2$ and $y=8$: $$ (2,8) $$ - Intersection of $x=2$ and $y = x - 2$: Substitute $x=2$ into $y = x - 2$: $$ y = 2 - 2 = 0 $$ So point is: $$ (2,0) $$ - Intersection of $y=8$ and $y = x - 2$: Set $8 = x - 2$: $$ x = 10 $$ So point is: $$ (10,8) $$ 4. **Shape formed:** The three points $(2,0)$, $(2,8)$, and $(10,8)$ form a triangle. 5. **Determining the type of triangle:** Calculate the lengths of the sides: - Between $(2,0)$ and $(2,8)$: $$ \sqrt{(2-2)^2 + (8-0)^2} = \sqrt{0 + 64} = 8 $$ - Between $(2,8)$ and $(10,8)$: $$ \sqrt{(10-2)^2 + (8-8)^2} = \sqrt{64 + 0} = 8 $$ - Between $(2,0)$ and $(10,8)$: $$ \sqrt{(10-2)^2 + (8-0)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} $$ 6. **Check angles:** Using the Pythagorean theorem: $$ 8^2 + 8^2 = 64 + 64 = 128 $$ $$ (8\sqrt{2})^2 = 128 $$ Since the sum of the squares of the two shorter sides equals the square of the longest side, the triangle is right-angled. 7. **Check if isosceles:** Two sides are equal (8 and 8), so it is an isosceles right triangle. **Final answer:** The solution region forms a right isosceles triangle. **Answer choice:** e. segitiga siku-siku sama kaki