Chapter 7.3, Problem 3PSA

Solution

a.

The Number of diagonals can be drawn in pentagon.

The Number of diagonals can be drawn are $5$

Given: Pentagon

Concept used: Here, we are using the formula

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

Calculation: Here, the formula is Diagonals = $\frac{n(n-3)}{2}$ and $n=5$

Now, $\frac{5(5-3)}{2}$

➔ $\frac{10}{2}$

➔ $5$

Conclusion: Hence, the Number of diagonals can be drawn are $5$

b.

To find the number of diagonals can be drawn in hexagon.

The Number of diagonals can be drawn are $9$

Given: Hexagon

Concept used: Here, we are using the formula

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

Calculation: Here, the formula is Diagonals = $\frac{n(n-3)}{2}$ and $n=6$

Now, $\frac{6(6-3)}{2}$

➔ $\frac{18}{2}$

➔ $9$

Conclusion: Hence, the Number of diagonals can be drawn are $9$

c.

The number of diagonals can be drawn in a quadrilateral.

The Number of diagonals can be drawn are $2$

Given: Quadrilateral

Concept used: Here, we are using the formula

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

Calculation: Here, the formula is Diagonals = $\frac{n(n-3)}{2}$ and $n=4$

Now, $\frac{4(4-3)}{2}$

➔ $\frac{4}{2}$

➔ $2$

Conclusion: Hence, the Number of diagonals can be drawn are $2$

d.

To find the number of diagonals can be drawn in a triangle.

The Number of diagonals can be drawn are $0$

Given: Triangle

Concept used: Here, we are using the formula

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

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

Now, $\frac{3(3-3)}{2}$

➔ $\frac{0}{2}$

➔ $0$

Conclusion: Hence, the Number of diagonals can be drawn are $0$