A company makes and sells candles of different shapes and sizes - Leaving Cert Mathematics - Question 7 - 2022

The volume of a cone is given by:

$V = \frac{1}{3}πr^2h$

For the small candle:

$12 = \frac{1}{3}πr_s^2h_s$

Let h_s = h. Rearranging gives:

$r_s^2 = \frac{12}{\frac{1}{3}πh} = \frac{36}{πh}$

For the large candle with height 2h:

$150π = \frac{1}{3}πr_l^2(2h)$

This simplifies to:

$r_l^2 = \frac{150}{\frac{2}{3}h} = \frac{225}{h}$

We see that:

$\frac{r_l^2}{r_s^2} = k^2$

Substituting in the expressions for $r_s^2$ and $r_l^2$ gives:

$k^2 = \frac{\frac{225}{h}}{\frac{36}{πh}} = \frac{225π}{36} = \frac{25π}{4}$

Therefore, to find k:

$k = \sqrt{\frac{225}{36}} = \frac{15}{6} = \frac{5}{2}$

The length of arc AB can be found using the formula:

$\text{Arc length} = \frac{θ}{360} \times 2πr$

where θ is the central angle. Here:

$θ = 216°$

Setting up the equation:

$\text{Arc length} = \frac{216}{360} \times 2πr = \frac{3}{5} \times 2πr = \frac{6}{5}πr$

Given that |OA| = 8 cm is the slant height of the cone, we can relate this to the radius. Using the relationship of a cone:

$r = 8 \times \sin(108°)$

In conclusion, by applying trigonometric properties, the specific radius can be derived.

The volume of a sphere is given by:

$V = \frac{4}{3}πr^3$

Substituting in the value for the radius, we find:

$V = \frac{4}{3}π(2.7)^3$

Calculating gives:

$V \approx 82.4479 \, \text{cm}^3 \approx 82.448 \, \text{cm}^3 \, \text{(to 3 d.p.)}$

The area of the base of the candle is:

$A = πr^2$

Setting this equal to 5.4 cm²:

$\pi r^2 = 5.4$

Solving for r, we find:

$r = \sqrt{\frac{5.4}{π}}$

Next, we apply the Pythagorean theorem to determine l:

$l^2 + r^2 = (2.7)^2$

Thus:

$l = 2.7 - \sqrt{(2.7)^2 - r^2}$

Finally, calculate l to find its numerical value, rounding to the nearest mm.