The figure below shows two intersecting circles with radius 9 cm with centre O1 and O2. Find the length of the common code AB.

Acha $M$ iwe sehemu ya makutano ya kodi ya kawaida AB na mstari unaounganisha vituo $O_1O_2$.
Kodi ya kawaida AB imegawanywa katika sehemu mbili sawa na $M$, hivyo $AM = MB$. Acha $AM = h$.
Acha $O_1M = x$. Kisha $O_2M = O_1O_2 - x = 9 - x$.

Katika pembetatu ya kulia $\triangle O_1MA$:

$$r_1^2 = O_1M^2 + AM^2$$ $$4^2 = x^2 + h^2$$ $$16 = x^2 + h^2 \quad (1)$$

Katika pembetatu ya kulia $\triangle O_2MA$:

$$r_2^2 = O_2M^2 + AM^2$$ $$9^2 = (9-x)^2 + h^2$$ $$81 = (9-x)^2 + h^2 \quad (2)$$

Tenga $h^2$ kutoka mlinganyo (1): $h^2 = 16 - x^2$.
Badilisha $h^2$ katika mlinganyo (2):

$$81 = (9-x)^2 + (16 - x^2)$$ $$81 = (81 - 18x + x^2) + 16 - x^2$$ $$81 = 81 - 18x + 16$$ $$0 = -18x + 16$$ $$18x = 16$$ $$x = \frac{16}{18} = \frac{8}{9} \text{ cm}$$

Badilisha thamani ya $x$ katika mlinganyo (1):

$$h^2 = 16 - \left(\frac{8}{9}\right)^2$$ $$h^2 = 16 - \frac{64}{81}$$ $$h^2 = \frac{16 \times 81 - 64}{81}$$ $$h^2 = \frac{1296 - 64}{81}$$ $$h^2 = \frac{1232}{81}$$

Pata mzizi wa mraba wa $h^2$:

$$h = \sqrt{\frac{1232}{81}} = \frac{\sqrt{1232}}{\sqrt{81}}$$

Ili kurahisisha $\sqrt{1232}$: $1232 = 16 \times 77$, hivyo $\sqrt{1232} = \sqrt{16 \times 77} = 4\sqrt{77}$.

$$h = \frac{4\sqrt{77}}{9} \text{ cm}$$

Urefu wa kodi ya kawaida AB ni $2h$:

$$AB = 2 \times \frac{4\sqrt{77}}{9} = \frac{8\sqrt{77}}{9} \text{ cm}$$

Urefu wa kodi ya kawaida AB ni $\frac{8\sqrt{77}}{9} \text{ cm}$.