Define the interior angle theorem.\[\]
Answer
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Hint: We recall the definition of interior angle which the angle subtended by two sides by a polygon and the statement of interior angle theorem “The sum of interior angles of a polygon with $n$ number of sides is $\left( n-2 \right){{180}^{\circ }}$.”\[\]
Complete step-by-step answer:
Note: We note that the interior angle theorem is different from exterior angle theorem which is only defined for only triangles and state that “The exterior angle of any triangle is greater than its angles and is equal to sum of remote interior angles ” . The exterior angle sum theorem states that “ The sum of exterior angles of any polygon is ${{360}^{\circ }}$”.
Complete step-by-step answer:
We know that a polygon is a closed curve made from only line segments called sides and the point of intersection of sides are called vertices. The region enclosed by all the sides of a polygon is called interior and the angles subtended by any two sides in the interior is called an interior angle. If we take any two points in the interior and join them to obtain none of the points lying outside the interior then the polygon is called convex polygon otherwise called concave polygon. \[\]
The interior angle theorem states that “The sum of measures of interior angles of a polygon with $n$ number of sides is $\left( n-2 \right){{180}^{\circ }}$ ”. Let us check the statement is true for a triangle or not.\[\]
We know the triangle has three sides $\left( n=3 \right)$ and the sum of the interior angles is ${{180}^{\circ }}$. The triangle is always a convex polygon. We apply the theorem and get the sum of angles as $\left( n-2 \right){{180}^{\circ }}=\left( 3-2 \right){{180}^{\circ }}={{180}^{\circ }}$. So the triangle satisfies the theorem. Now let us check for quadrilateral.\[\]
The quadrilateral above ABCD is convex quadrilateral and the quadrilateral below PQRS is a concave quadrilateral. We know that for both the quadrilaterals the sum of angles is ${{360}^{\circ }}$. We apply the theorem for quadrilateral with number of sides $n=4$ and have $\left( n-2 \right){{180}^{\circ }}=\left( 4-2 \right){{180}^{\circ }}={{360}^{\circ }}$. So the quadrilateral satisfies the theorem. We can similarly check for pentagon$\left( n=5 \right)$, hexagon$\left( n=6 \right)$ and so on. \[\]
Note: We note that the interior angle theorem is different from exterior angle theorem which is only defined for only triangles and state that “The exterior angle of any triangle is greater than its angles and is equal to sum of remote interior angles ” . The exterior angle sum theorem states that “ The sum of exterior angles of any polygon is ${{360}^{\circ }}$”.
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