Chapter 5.4, Problem 21PSC

Solution

a.

To find: The number of diagonals a triangle can have

A triangle has $0$ diagonals.

Given:

A triangle.

Concept used:

Here, we are using the formula:

Diagonal of a polygon = $\frac{n\left(n-3\right)}{2}$

(Where, “n” is side)

Calculation:

We know that,

A triangle has $3$ sides.

So, diagonals of a triangle = $\frac{n\left(n-3\right)}{2}$

(Where, “n” is side)

➜ $\frac{3\left(3-3\right)}{2}$

➜ $0$

Conclusion: Hence, a triangle has $0$ diagonals.

b.

To find: The number of diagonals a quadrilateral can have

A quadrilateral has $2$ diagonals.

Given:

A quadrilateral.

Concept used:

Here, we are using the formula:

Diagonal of a polygon = $\frac{n\left(n-3\right)}{2}$

(Where, “n” is side)

Calculation:

We know that,

A quadrilateral has $4$ sides.

So, diagonals of a quadrilateral = $\frac{n\left(n-3\right)}{2}$

(Where, “n” is side)

➜ $\frac{4\left(4-3\right)}{2}$

➜ $2$

Conclusion: Hence, a quadrilateral has $2$ diagonals.

c.

To find: The number of diagonals a five-sided polygon can have

A five-sided polygon have $5$ diagonals.

Given:

Afive-sided polygon

Concept used:

Here, we are using the formula:

Diagonal of a polygon = $\frac{n\left(n-3\right)}{2}$

(Where, “n” is side)

Calculation:

So, diagonals of a five-sided polygon = $\frac{n\left(n-3\right)}{2}$

(Where, “n” is side)

➜ $\frac{5\left(5-3\right)}{2}$

➜ $5$

Conclusion: Hence, a five-sided polygon have $5$ diagonals.

d.

To find: The number of diagonals a six-sided polygon can have

A six-sided polygon have $9$ diagonals.

Given:

A five-sided polygon

Concept used:

Here, we are using the formula:

Diagonal of a polygon = $\frac{n\left(n-3\right)}{2}$

(Where, “n” is side)

Calculation:

So, diagonals of a six-sided polygon = $\frac{n\left(n-3\right)}{2}$

(Where, “n” is side)

➜ $\frac{6\left(6-3\right)}{2}$

➜ $9$

Conclusion: Hence, a six-sided polygon have $9$ diagonals.

e.

To find: The number of diagonals meet at one vertex of a polygon with n sides.

The number of diagonals is $n\left(n-3\right)$

Given:

A polygon have n sides.

Concept used:

Here, we are using the formula:

Diagonal of a polygon = $\frac{n\left(n-3\right)}{2}$

(Where, “n” is side)

Explanation:

We know that,

From each vertex we can draw $n-3$ diagonals.

So, the total number of diagonals is $n\left(n-3\right)$

Conclusion: Hence, the number of diagonals is $n\left(n-3\right)$

f.

To find: The number of vertices an n-sided polygon can have

The number of vertices is n

Given:

An n-sided polygon.

Explanation:

An n-sided polygon can have n vertices and from each vertex we can draw $n-3$ diagonals.

Conclusion: Hence, the number of vertices is n

g.

To find: The number of diagonals an n-sided polygon can have

The number of diagonals is $n\left(n-3\right)$

Given:

An n-sided polygon.

Explanation:

An n-sided polygon can have n vertices and from each vertex we can draw $n-3$ diagonals.

So, the total number of diagonals = $n\left(n-3\right)$

Conclusion: Hence, the number of diagonals is $n\left(n-3\right)$