A convex polygon has the property that the number of its diagonals equals three times its number of sides. Find the measure of each exterior angle.

Acha $n$ iwe idadi ya pande za poligoni.
Idadi ya diagonals, $D$, inatolewa na fomula:

$$D = \frac{n(n-3)}{2}$$

Kutokana na swali, idadi ya diagonals ni mara tatu ya idadi ya pande:

$$D = 3n$$

Sasa, weka fomula hizi mbili sawa:

$$\frac{n(n-3)}{2} = 3n$$

Zidisha pande zote mbili kwa 2:

$$n(n-3) = 6n$$

Panua upande wa kushoto:

$$n^2 - 3n = 6n$$

Peleka maneno yote upande mmoja:

$$n^2 - 3n - 6n = 0$$ $$n^2 - 9n = 0$$

Toa $n$ kama kigezo cha kawaida:

$$n(n-9) = 0$$

Hii inatoa suluhisho mbili: $n=0$ au $n=9$. Kwa kuwa poligoni haiwezi kuwa na pande 0, tunachukua $n=9$.
Poligoni ina pande 9.

Kwa poligoni ya kawaida yenye pande $n$, kipimo cha kila pembe ya nje hutolewa na fomula:

$$\text{Pembe ya nje} = \frac{360^\circ}{n}$$

Badilisha $n=9$ kwenye fomula:

$$\text{Pembe ya nje} = \frac{360^\circ}{9}$$ $$\text{Pembe ya nje} = 40^\circ$$

Kipimo cha kila pembe ya nje ni:

$$40^\circ$$