The interior angles of a polygon are x°, 1/2x, (x+40)°, 110°, 135°, 160°. Find the value of x. 12. The sum of interior angles of a regular polygon is 1800°. Find the size of each exterior angle.

Oktagoni ina pande 8 ($n=8$). Jumla ya pembe za ndani za poligoni hutolewa na fomula $(n-2) \times 180^\circ$.

$$\text{Jumla} = (8-2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ$$

$$80^\circ + \frac{1}{2}x + (3x+40^\circ) + 110^\circ + 135^\circ + 160^\circ + (200^\circ+10^\circ) + 185^\circ = 1080^\circ$$

Kwanza, jumlisha namba zote zisizo na $x$:

$$80 + 40 + 110 + 135 + 160 + 210 + 185 = 920^\circ$$

Kisha, jumlisha vigezo vyenye $x$:

$$\frac{1}{2}x + 3x = \frac{1}{2}x + \frac{6}{2}x = \frac{7}{2}x$$

Sasa, weka jumla ya pembe sawa na $1080^\circ$:

$$\frac{7}{2}x + 920^\circ = 1080^\circ$$

Toa $920^\circ$ kutoka pande zote mbili:

$$\frac{7}{2}x = 1080^\circ - 920^\circ$$ $$\frac{7}{2}x = 160^\circ$$

Zidisha pande zote mbili kwa 2:

$$7x = 160^\circ \times 2$$ $$7x = 320^\circ$$

Gawanya pande zote mbili kwa 7:

$$x = \frac{320}{7}$$ $$x \approx 45.714^\circ$$

Thamani ya $x$ ni:

$$\boxed{x = \frac{320}{7}^\circ}$$

Jumla ya pembe za ndani za poligoni hutolewa na fomula $(n-2) \times 180^\circ$.
Tunajua jumla ni $1800^\circ$:

$$(n-2) \times 180^\circ = 1800^\circ$$

Gawanya pande zote mbili kwa $180^\circ$:

$$n-2 = \frac{1800}{180}$$ $$n-2 = 10$$

Ongeza 2 kwa pande zote mbili:

$$n = 10 + 2$$ $$n = 12$$

Poligoni ina pande 12.

Jumla ya pembe za nje za poligoni yoyote ya mbonyeo ni $360^\circ$. Kwa poligoni ya kawaida, pembe zote za nje ni sawa.
Ukubwa wa kila pembe ya nje = $\frac{\text{Jumla ya pembe za nje}}{\text{Idadi ya pande}}$

$$\text{Kila pembe ya nje} = \frac{360^\circ}{n}$$ $$\text{Kila pembe ya nje} = \frac{360^\circ}{12}$$ $$\text{Kila pembe ya nje} = 30^\circ$$

Ukubwa wa kila pembe ya nje ni:

$$\boxed{30^\circ}$$