The figure below is a cone ABC from which BCXY remained after the small cone AXY was cut off. Given that EX = 4cm, DB = 12cm and DE = 15cm, calculate the height AE, of the smaller cone AXY and the volume of XBCY, the shape that remained.

Koni AXY na koni ABC zinafanana. Hivyo, uwiano wa rediasi zao ni sawa na uwiano wa kimo chao.
Tunajua rediasi ya koni ndogo $r_1 = EX = 4$ cm na rediasi ya koni kubwa $r_2 = DB = 12$ cm.
Kimo cha koni ndogo ni $AE = h_1$. Kimo cha koni kubwa ni $AD = h_2$.
Pia tunajua $DE = h_2 - h_1 = 15$ cm.

$$\frac{r_1}{r_2} = \frac{h_1}{h_2}$$ $$\frac{4}{12} = \frac{h_1}{h_2}$$ $$\frac{1}{3} = \frac{h_1}{h_2}$$

Hii inamaanisha $h_2 = 3h_1$.

Tumia uhusiano $h_2 - h_1 = 15$ na badilisha $h_2$ na $3h_1$.

$$3h_1 - h_1 = 15$$ $$2h_1 = 15$$ $$h_1 = \frac{15}{2}$$ $$h_1 = 7.5 \text{ cm}$$

Kimo cha koni ndogo AE ni $\boxed{\text{7.5 cm}}$.

$$h_2 = 3h_1 = 3 \times 7.5$$ $$h_2 = 22.5 \text{ cm}$$

Rediasi ya koni kubwa $r_2 = 12$ cm.
Rediasi ya koni ndogo $r_1 = 4$ cm.
Kimo cha koni ndogo $h_1 = 7.5$ cm.

Fomula ya ujazo wa koni ni $V = \frac{1}{3}\pi r^2 h$. Tumia $\pi = 3.142$.

$$V_2 = \frac{1}{3}\pi r_2^2 h_2$$ $$V_2 = \frac{1}{3} \times 3.142 \times (12)^2 \times 22.5$$ $$V_2 = \frac{1}{3} \times 3.142 \times 144 \times 22.5$$ $$V_2 = 3.142 \times 48 \times 22.5$$ $$V_2 = 3.142 \times 1080$$ $$V_2 = 3393.36 \text{ cm}^3$$

$$V_1 = \frac{1}{3}\pi r_1^2 h_1$$ $$V_1 = \frac{1}{3} \times 3.142 \times (4)^2 \times 7.5$$ $$V_1 = \frac{1}{3} \times 3.142 \times 16 \times 7.5$$ $$V_1 = 3.142 \times 16 \times 2.5$$ $$V_1 = 3.142 \times 40$$ $$V_1 = 125.68 \text{ cm}^3$$

Ujazo wa frustum ni tofauti kati ya ujazo wa koni kubwa na ujazo wa koni ndogo.

$$V_{\text{frustum}} = V_2 - V_1$$ $$V_{\text{frustum}} = 3393.36 - 125.68$$ $$V_{\text{frustum}} = 3267.68 \text{ cm}^3$$

Ujazo wa XBCY ni $\boxed{\text{3267.68 cm}^3}$.