1. **State the problem:** We have a rectangular block with dimensions 15 cm by 8 cm by 6 cm. A circular hole of diameter 4 cm is cut through the block along the 15 cm length. We need to find the surface area of the resulting object to the nearest square centimetre. 2. **Calculate the original surface area of the block:** The surface area of a rectangular block is given by: $$2(lw + lh + wh)$$ where $l=15$, $w=8$, and $h=6$. Calculate each term: $$lw = 15 \times 8 = 120$$ $$lh = 15 \times 6 = 90$$ $$wh = 8 \times 6 = 48$$ Sum: $$120 + 90 + 48 = 258$$ Original surface area: $$2 \times 258 = 516 \text{ cm}^2$$ 3. **Calculate the surface area removed and added by the hole:** The hole is a cylinder with diameter 4 cm, so radius $r=2$ cm, and height $h=15$ cm. - The two circular faces of the hole remove area from the block's surface: Each circular face area: $$\pi r^2 = \pi \times 2^2 = 4\pi$$ Two faces area removed: $$2 \times 4\pi = 8\pi \approx 25.13 \text{ cm}^2$$ - The curved surface area of the cylinder (hole) is added to the surface area: $$2\pi r h = 2 \times \pi \times 2 \times 15 = 60\pi \approx 188.50 \text{ cm}^2$$ 4. **Calculate the new surface area:** New surface area = Original surface area - area of two circular faces + curved surface area of hole $$516 - 25.13 + 188.50 = 679.37 \text{ cm}^2$$ 5. **Round to the nearest square centimetre:** $$\boxed{679 \text{ cm}^2}$$