Answer

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**Hint**: To solve this type of problem first we have to draw pentagons which have 5 sides. Now applying the formula for the number of diagonals that can be formed. The formula is

Number of diagonals \[=\dfrac{n\left( n-3 \right)}{2}\].

**:**

__Complete step-by-step answer__Step 1: Take 5 points like A, B, C, D, E.

Step 2: Join AB, BC, CD, DE, EA.

Step 3: Join AC, AD, BE, BD, CE.

Now applying the formula for number of diagonals,

Number of diagonals\[=\dfrac{n\left( n-3 \right)}{2}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)

The number of sides of a pentagon is 5.

Now substituting the value of 5 in (a).

We get,

\[=\dfrac{5\left( 5-3 \right)}{2}\]

\[=\dfrac{5\left( 2 \right)}{2}\]

\[=5\].

Therefore the number of diagonals that can be drawn is 5.

**So, the correct answer is “Option C”.**

**Note**: By applying the above formula we can find the number of diagonals for any polygon.Triangle is an exception to this rule, due to the shape of the triangle , it does not have any diagonals. Take care while drawing figures.

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