How many diagonals does each of the following have? $\left( a \right)$ A convex quadrilateral $\left( b \right)$ A regular hexagon $\left( c \right)$ A triangle.
Answer
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Hint: In this question we have to find the number of diagonals for the specific given options, a diagonal is a line segment joining any two opposite corners of a polygon whether it is a square, rhombus, regular hexagon etc. The general formula for the number of diagonals of n-sided polygon is $\dfrac{{n\left( {n - 3} \right)}}{2}$. Use this concept to get the answer.
Complete step-by-step answer: The general formula for number of diagonals (d) in any figure is (n-3) multiply by the number of vertices and divide by 2. $ \Rightarrow d = \dfrac{{n\left( {n - 3} \right)}}{2}$ $\left( a \right)$ A convex quadrilateral As we know in a convex quadrilateral there are four sides (see figure) $ \Rightarrow n = 2$ Therefore number of diagonals in a convex quadrilateral is $ \Rightarrow d = \dfrac{{n\left( {n - 3} \right)}}{2} = \dfrac{{4\left( {4 - 3} \right)}}{2} = \dfrac{4}{2} = 2$ $\left( b \right)$ A regular hexagon As we know in a convex quadrilateral there are six sides (see figure) $ \Rightarrow n = 6$ Therefore number of diagonals in a convex quadrilateral is $ \Rightarrow d = \dfrac{{n\left( {n - 3} \right)}}{2} = \dfrac{{6\left( {6 - 3} \right)}}{2} = \dfrac{{6 \times 3}}{2} = 9$ $\left( c \right)$ A triangle. As we know in a convex quadrilateral there are three sides (see figure) $ \Rightarrow n = 3$ Therefore number of diagonals in a convex quadrilateral is $ \Rightarrow d = \dfrac{{n\left( {n - 3} \right)}}{2} = \dfrac{{3\left( {3 - 3} \right)}}{2} = \dfrac{0}{2} = 0$ So, these are the required diagonals in the following figures.
Note: Whenever we face such types of problems the key concept is the basic understanding of the definition of diagonal of a polygon. The gist of direct formula for diagonal of a n-sided polygon will eventually help you in getting the right track to reach the answer accurately.
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