1. **State the problem:** We have a circle with center O and a triangle ABC inscribed in it. The segment AB is the diameter of the circle. (a) Given the area of the circle is $256\pi$ cm², find the length of the diameter $|AB|$. (b) Given angle $y = 68^\circ$ at vertex A, find the angles $x$ at vertex C and $z$ at vertex B. 2. **Formula for the area of a circle:** $$\text{Area} = \pi r^2$$ where $r$ is the radius. 3. **Find the radius and diameter:** Given area $= 256\pi$, so $$\pi r^2 = 256\pi$$ Divide both sides by $\pi$: $$\cancel{\pi} r^2 = 256 \cancel{\pi}$$ $$r^2 = 256$$ Take the square root: $$r = \sqrt{256} = 16$$ Diameter $|AB| = 2r = 2 \times 16 = 32$ cm. 4. **Use the property of a triangle inscribed in a circle with diameter as one side:** The angle opposite the diameter is a right angle. So, angle $x$ at vertex C is $90^\circ$. 5. **Use the triangle angle sum rule:** Sum of angles in triangle ABC is $180^\circ$. Given $y = 68^\circ$, $x = 90^\circ$, find $z$: $$x + y + z = 180^\circ$$ $$90 + 68 + z = 180$$ $$158 + z = 180$$ $$z = 180 - 158 = 22^\circ$$ **Final answers:** - Diameter $|AB| = 32$ cm - Angle $x = 90^\circ$ - Angle $y = 68^\circ$ - Angle $z = 22^\circ$