Chapter 7.3, Problem 13PSB

Solution

a.

The name of polygon which contains $14$ diagonals.

The name of polygon which contains $14$ diagonals is heptagon.

Given: Number of diagonals are 14.

Concept used: Here, we are using the formula,

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

Calculation: Here, the formula is $\frac{n(n-3)}{2}$

➔14 = $\frac{n(n-3)}{2}$

➔ $28=n(n-3)$

➔ $28=n^2-3n$

➔ $-n^2+3n+28=0$

➔ $-n^2+7n-4n+28=0$

➔ $-n(n-7)-4(n-7)=0$

➔ $(-n-4)(n-7)=0$

So, n = $7$ or $-4$

Since, a polygon cannot have a negative number of sides, we know that this polygon must have $7$ sides.

Conclusion: Hence, the name of polygon which contains $14$ diagonals is heptagon.

b.

The name of polygon which contains $35$ diagonals.

The name of polygon which contains $35$ diagonals is decagon.

Given: Number of diagonals are $35$ .

Concept used: Here, we are using the formula,

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

Calculation: Here, the formula is $\frac{n(n-3)}{2}$

➔ $35=\frac{n(n-3)}{2}$

➔ $70=n(n-3)$

➔ $-n^2+3n+70=0$

➔ $-n^2-7n+10n+70=0$

➔ $-n(n+7)+10(n+7)=0$

➔ $(-n+10)(n+7)=0$

So, n = $10$ or $-7$

Since, a polygon cannot have a negative number of sides, we know that this polygon must have $10$ sides.

Conclusion: Hence, the name of polygon which contains $35$ sides is decagon.

c.

The name of polygon which contains $209$ diagonals.

The name of polygon which contains $209$ diagonals is doicosagon.

Given: Number of diagonals are $209$ .

Concept used: Here, we are using the formula,

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

Calculation: Here, the formula is $\frac{n(n-3)}{2}$

➔ $209=\frac{n(n-3)}{2}$

➔ $418=n(n-3)$

➔ $-n^2+3n+418=0$

➔ $-n^2+22n-19n+418=0$

➔ $-n(n-22)-19(n-22)=0$

➔ $(-n-19)(n-22)=0$

So, n = $-19$ or $22$

Since, a polygon cannot have a negative number of sides, we know that this polygon must have $22$

sides.

Conclusion: Hence, the name of polygon which contains $22$ sides is doicosagon.