1. **Stating the problem:** We need to determine 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:** - $x \geq 2$ is a vertical line at $x=2$ and the region to the right. - $y \leq 8$ is a horizontal line at $y=8$ and the region below. - $x - y \leq 2$ can be rewritten as $y \geq x - 2$ (by adding $y$ and subtracting 2 on both sides). 3. **Finding the vertices of the solution region:** The solution region is the intersection of these three half-planes, which forms a polygon. - 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. **Plotting these points:** The vertices are $(2,0)$, $(2,8)$, and $(10,8)$. 5. **Determining the triangle type:** - Calculate side lengths: - Between $(2,0)$ and $(2,8)$: $$ \sqrt{(2-2)^2 + (8-0)^2} = 8 $$ - Between $(2,8)$ and $(10,8)$: $$ \sqrt{(10-2)^2 + (8-8)^2} = 8 $$ - Between $(2,0)$ and $(10,8)$: $$ \sqrt{(10-2)^2 + (8-0)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} $$ - Check for right angle using Pythagoras: $$ 8^2 + 8^2 = 64 + 64 = 128 $$ $$ (8\sqrt{2})^2 = 128 $$ Since $$8^2 + 8^2 = (8\sqrt{2})^2$$, the triangle is right-angled. - Check if isosceles: Two sides are equal (8 and 8), so it is an isosceles right triangle. 6. **Conclusion:** The solution region is a right-angled isosceles triangle. **Final answer:** $$\boxed{\text{segitiga siku-siku sama kaki}}$$ (option e)