
How many diagonals does a convex quadrilateral have?
[a] 1
[b] 2
[c] 3
[d] 4
Answer
616.5k+ views
Hint: Recall that a diagonal is a line segment formed by two points of a polygon such that the points are not connected by the side of the polygon. Use the definition of a convex polygon and think how many diagonals can a convex quadrilateral have.
Complete step-by-step answer:
We know that a diagonal of a polygon is a line segment formed by joining two points of the polygon which are not connected by the side of the polygon.
A convex polygon is a polygon having the following property:
If any two points A and B are inside the polygon, then the whole line segment AB is inside the polygon. Hence a convex quadrilateral has all angles smaller than $180{}^\circ $.
ABCD is an example of a convex quadrilateral. It is clear from the picture that only the pairs of point (A, C) and (B, D) are not connected by a side.
Hence ABCD has two diagonals AC and BD.
Hence a convex quadrilateral has 2 diagonals.
Hence option [b] is correct.
Note: Alternatively, we know that the number of diagonals in a polygon of n sides is given by
$\dfrac{\left( n \right)\left( n-3 \right)}{2}$.
Put n= 4, we get
The number of diagonals in a quadrilateral $=\dfrac{4\left( 1 \right)}{2}=2$
Hence option [b] is correct.
Complete step-by-step answer:
We know that a diagonal of a polygon is a line segment formed by joining two points of the polygon which are not connected by the side of the polygon.
A convex polygon is a polygon having the following property:
If any two points A and B are inside the polygon, then the whole line segment AB is inside the polygon. Hence a convex quadrilateral has all angles smaller than $180{}^\circ $.
ABCD is an example of a convex quadrilateral. It is clear from the picture that only the pairs of point (A, C) and (B, D) are not connected by a side.
Hence ABCD has two diagonals AC and BD.
Hence a convex quadrilateral has 2 diagonals.
Hence option [b] is correct.
Note: Alternatively, we know that the number of diagonals in a polygon of n sides is given by
$\dfrac{\left( n \right)\left( n-3 \right)}{2}$.
Put n= 4, we get
The number of diagonals in a quadrilateral $=\dfrac{4\left( 1 \right)}{2}=2$
Hence option [b] is correct.
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