Mathematics — Standard (041) — CBSE Class 10 board question

Given: Cube side = 14 cm; largest cone carved from one face has radius r = 7 cm, height h = 14 cm.

Volume of remaining solid:

Volume of cube = $14^3 = 2744 \text{ cm}^3$

Volume of cone = $\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 14 = \dfrac{1}{3} \times \dfrac{22}{7} \times 49 \times 14 = \dfrac{1}{3} \times 2156 = 718.67 \text{ cm}^3$

Volume of remaining solid = 2744 − 718.67 = 2025.33 cm³

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Surface area of remaining solid:

Slant height of cone: $l = \sqrt{r^2 + h^2} = \sqrt{49 + 196} = \sqrt{245} = 7\sqrt{5} = 7 \times 2.2 = 15.4 \text{ cm}$

Surface area = 5 faces of cube + base of cube (with circular hole) + CSA of cone

$$= 5 \times 14^2 + (14^2 - \pi r^2) + \pi r l$$

$$= 5 \times 196 + (196 - \tfrac{22}{7} \times 49) + \tfrac{22}{7} \times 7 \times 15.4$$

$$= 980 + (196 - 154) + 338.8$$

$$= 980 + 42 + 338.8$$

Surface area of remaining solid = 1360.8 cm²

Source: Chapter 12, Surface Areas and Volumes (Sections 12.2 & 12.3)

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• The largest cone carved from a face of a cube of side 14 cm has r = 7 cm (half the face) and h = 14 cm (full side).
• Volume = Volume of cube − Volume of cone. Straightforward subtraction.
• Surface area is trickier: the 5 untouched faces contribute fully; the 6th face loses the circular base of the cone, so add $(14^2 - \pi r^2)$; then add the curved surface area of the cone ($\pi r l$). The circular base of the cone is not added separately because it is open (carved out).
• Examiners award marks at each step: identifying r & h, slant height, each area component, and final addition — show all steps clearly.