Surface area and volume - WJECPyramids and cones - Higher tier only

Volume = \(\frac{1}{3}\) × 20 × 15 = \(\frac{1}{3}\) × 300 = 100 cm³

Volume = \(\frac{1}{3}\) × area of base × height

Volume = \(\frac{1}{3}\) × 30 mm² × 7.5 mm = 75 mm³

\(\text{Area of a square} ~=~ \text{length} \times \text{width}\)

\(\text{Area of a triangle} ~=~ \frac {1} {2} \times \text{base} \times \text{height}\)

2. \(\text{Volume of a cone} ~=~ \frac {1}{3} ~ \pi \times \text{r}^{2} \times \text{h}\)

Volume = \(\frac {1}{3} ~ \pi \) × 6² × 15 = 565.4866776 cm³

3. \(\text{Curved surface area of a cone} ~=~ \pi \times \text{r}\times \text{l}\)

Curved area = \(\pi \times \text{6} \times \text{l}\)

\(\text{l}^{2} ~=~ \text{r}^{2} ~+~ \text{h}^{2}\)

\(\text{l}^{2}\) = 6 ² + 15 ² = 261 cm

\(\text{l}~=~\sqrt{261} ~=~ \text{16.15549442 cm}\) (don’t round this value until the end of the calculation)

Curved area = \(\pi \times \text{6} \times \text{16.15549442}\) = 304.5238955 cm²

4. Area of the base = \(\pi \times \text{r}^{2}\) = \(\pi\) x 6² = 113.0973355 cm²