1. **Problem statement:** We have a regular hexagon ABCDEF with all sides 4 cm and center O. (a) Find angle $\angle AB\hat{C}$. (b) Given $OB=OC$, find the area of triangle $OBC$. (c) Find the area of the whole hexagon. --- 2. **Key formulas and facts:** - Sum of interior angles of an n-sided polygon: $$(n-2) \times 180^\circ$$ - Each interior angle of a regular polygon: $$\frac{(n-2) \times 180^\circ}{n}$$ - Area of triangle with base $b$ and height $h$: $$\frac{1}{2}bh$$ - Area of regular polygon with side length $s$ and number of sides $n$: $$\frac{1}{2} n s a$$ where $a$ is the apothem (distance from center to side). - In a regular hexagon, the apothem $a = s \times \frac{\sqrt{3}}{2}$. --- 3. **Step (a): Calculate $\angle AB\hat{C}$** - The hexagon has $n=6$ sides. - Each interior angle is $$\frac{(6-2) \times 180^\circ}{6} = \frac{4 \times 180^\circ}{6} = 120^\circ$$. - $\angle AB\hat{C}$ is an interior angle at vertex B, so its size is $$120^\circ$$. --- 4. **Step (b): Find area of triangle $OBC$ given $OB=OC$** - $O$ is center, $B$ and $C$ are adjacent vertices. - In a regular hexagon, the radius (distance from center to vertex) equals the side length, so $$OB = OC = 4$$ cm. - The central angle $\angle BOC$ between vertices B and C is $$\frac{360^\circ}{6} = 60^\circ$$. - Triangle $OBC$ is isosceles with sides $OB=OC=4$ and included angle $60^\circ$. - Area formula for triangle with two sides $a,b$ and included angle $\theta$: $$\frac{1}{2}ab\sin(\theta)$$. - Substitute: $$\frac{1}{2} \times 4 \times 4 \times \sin(60^\circ) = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}$$ cm$^2$. --- 5. **Step (c): Find area of whole hexagon** - Hexagon can be divided into 6 congruent triangles like $OBC$. - Area of hexagon = 6 times area of one triangle. - Area = $$6 \times 4\sqrt{3} = 24\sqrt{3}$$ cm$^2$. --- **Final answers:** - (a) $\angle AB\hat{C} = 120^\circ$ - (b) Area of triangle $OBC = 4\sqrt{3}$ cm$^2$ - (c) Area of hexagon = $24\sqrt{3}$ cm$^2$