A speed line L4 passes through E and H. The centres are A and D, with radii 5cm and 4cm respectively. AB = 10cm.

Katika pembetatu ya kulia $\triangle AMC$, tunatumia theorem ya Pythagoras:

$$AC^2 = AM^2 + CM^2$$ $$5^2 = x^2 + CM^2$$ $$CM^2 = 25 - x^2 \quad (Eq. 1)$$

Katika pembetatu ya kulia $\triangle BMC$, $BM = AB - AM = 10 - x$. Tunatumia theorem ya Pythagoras:

$$BC^2 = BM^2 + CM^2$$ $$8^2 = (10-x)^2 + CM^2$$ $$CM^2 = 64 - (10-x)^2 \quad (Eq. 2)$$

1 na Eq. 2 ili kupata thamani ya $x$.

$$25 - x^2 = 64 - (10-x)^2$$ $$25 - x^2 = 64 - (100 - 20x + x^2)$$ $$25 - x^2 = 64 - 100 + 20x - x^2$$ $$25 = -36 + 20x$$ $$25 + 36 = 20x$$ $$61 = 20x$$ $$x = \frac{61}{20} = 3.05 \text{ cm}$$

Tumia $x = 3.05$ cm katika Eq. 1:

$$CM^2 = 25 - (3.05)^2$$ $$CM^2 = 25 - 9.3025$$ $$CM^2 = 15.6975$$ $$CM = \sqrt{15.6975} \approx 3.9620 \text{ cm}$$

Urefu wa kamba ya kawaida CD ni mara mbili ya $CM$:

$$CD = 2 \times CM = 2 \times 3.9620 = 7.924 \text{ cm}$$

Urefu wa kamba ya kawaida CD ni $\boxed{\text{7.924 cm}}$.

Sehemu ii): Kokotoa eneo la eneo lenye kivuli.
Eneo lenye kivuli ni jumla ya maeneo ya sehemu mbili za duara (segments) zinazoundwa na kamba CD. Tunatumia $\pi \approx 3.142$.

Kwa duara A (radius $r_A = 5$ cm):
Katika $\triangle AMC$:

$$\cos(\angle CAM) = \frac{AM}{AC} = \frac{3.05}{5} = 0.61$$ $$\angle CAM = \arccos(0.61) \approx 52.40^\circ$$

Pembe ya katikati $\theta_A = \angle CAD = 2 \times \angle CAM = 2 \times 52.40^\circ = 104.80^\circ$.

Kwa duara B (radius $r_B = 8$ cm):
$BM = 10 - x = 10 - 3.05 = 6.95$ cm.
Katika $\triangle BMC$:

$$\cos(\angle CBM) = \frac{BM}{BC} = \frac{6.95}{8} = 0.86875$$ $$\angle CBM = \arccos(0.86875) \approx 29.68^\circ$$

Pembe ya katikati $\theta_B = \angle CBD = 2 \times \angle CBM = 2 \times 29.68^\circ = 59.36^\circ$.

Eneo la sekta CAD:

$$\text{Eneo la sekta CAD} = \frac{\theta_A}{360^\circ} \times \pi r_A^2$$ $$= \frac{104.80}{360} \times 3.142 \times 5^2$$ $$= \frac{104.80}{360} \times 3.142 \times 25 \approx 22.847 \text{ cm}^2$$

Eneo la pembetatu CAD:

$$\text{Eneo la } \triangle CAD = \frac{1}{2} \times CD \times AM$$ $$= \frac{1}{2} \times (7.924) \times (3.05) \approx 12.084 \text{ cm}^2$$

Eneo la sehemu ya duara A = Eneo la sekta CAD - Eneo la $\triangle CAD$

$$= 22.847 - 12.084 = 10.763 \text{ cm}^2$$

Eneo la sekta CBD:

$$\text{Eneo la sekta CBD} = \frac{\theta_B}{360^\circ} \times \pi r_B^2$$ $$= \frac{59.36}{360} \times 3.142 \times 8^2$$ $$= \frac{59.36}{360} \times 3.142 \times 64 \approx 33.150 \text{ cm}^2$$

Eneo la pembetatu CBD:

$$\text{Eneo la } \triangle CBD = \frac{1}{2} \times CD \times BM$$ $$= \frac{1}{2} \times (7.924) \times (6.95) \approx 27.538 \text{ cm}^2$$

Eneo la sehemu ya duara B = Eneo la sekta CBD - Eneo la $\triangle CBD$

$$= 33.150 - 27.538 = 5.612 \text{ cm}^2$$

Jumla ya eneo lenye kivuli = Eneo la sehemu ya duara A + Eneo la sehemu ya duara B

$$= 10.763 + 5.612 = 16.375 \text{ cm}^2$$

Eneo lenye kivuli ni $\boxed{\text{16.38 cm}^2}$ (kwa sehemu mbili za desimali).