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How many diagonals does a hexagon have?
A hexagon has 9 diagonals. Each vertex of a n-sided polygon can be connected to n - 3 others with diagonals. Thus n(n - 3) possible diagonals. However, when Vertex A is connected to vertex C, vertex C is also connected to vertex A, thus each diagonal is counted twice. Thus: number_of_diagonals = n(n - 3)/2 = 6(6-3)/2 = 6x3/2 = 9
In a regular polygon the ratio of the measure of the exterior angle to the measure of the adjacent interior angle is 4 to 1 how many sides does the polygon have Please show your work?
This is not possible for a normal regular polygon. (A regular polygon has all equal angles and all equal sides. A normal polygon has no intersecting edges.)The smallest regular polygon is an equilateral triangle (a three sided polygon), whose exterior angle measure is twice the measure of its interior angle. A four-sided polygon (a square) has equal interior and exterior angle measures of 90⁰. Starting from a five-sided polygon, the exterior angle measure is smaller than the interior angle measure.Let's assume that the given information is true. So we need to verify it.Let's say that the interior angle of the regular polygon has a measure of x degrees, and the measure of the exterior angle of that polygon is 4x degrees.Since the sum of the interior and the exterior angles of the polygon is 180 degrees (a straight line), the interior angle is 36 degrees.4x + x = 1805x = 1805x/5 = 180/5x = 36The sum of the angles of a polygon = 180⁰(n - 2), where n is the number of the sides of the polygon.The measure of one of the angles of a polygon = 180⁰(n - 2)/n. Substituting the angle measure of 36⁰ into this formula, we have:36⁰ = 180⁰(n - 2)/n (multiply by n to both sides)36⁰n = 180⁰(n - 2)36⁰n = 180⁰n - 360⁰ (add 360⁰ and subtract 36⁰n to both sides)360⁰ = 144⁰n (divide by 144⁰ to both sides)2.5 = n !!That means that a such normal polygon does not exist.
Eight less than twice a number is fourteen what is the number?
let the number be (x) Therefore:-2x-8=14=>x-4=7(Dividing both sides by 2)=>x=7+4(Adding 4 to both sides)=>x=11Ans:-The required number is 11.
Difference of eight times a number and twice the number?
twice the difference of three times a number and eight
What is twice a number plus 27 in an algebraic expression?
Twice a number can be represented as 2x, where x is the unknown number. Adding 27 to this expression gives 2x + 27. Therefore, "twice a number plus 27" can be written as the algebraic expression 2x + 27.
What polygon has twice as many diagonals as sides?
An heptagon has 7 sides and 14 diagonals
How many diagonals does polygon have?
In a polygon with n sides, we have n(n-3)/2 diagonals. In a convex polygon with n sides, you can draw n-3 diagonals from each vertex, but you are counting each one twice you so you need to divide by do. That is why we have n(n-3) divided by 2
What is the number of sides of polygons which has 44diagonals?
11 sides The diagonal of a polygon is a line segment that joins two non adjacent vertices. So for a n-sided polygon, each vertex can be joined to n-3 other sides. So there ought to be n x (n-3) diagonals for a n-sided polygon; however, each diagonal is counted twice, once by the vertex at each end. So the number of diagonals is: diagonals = n (n-3) / 2 So for 44 diagonals: 44 = n (n-3) / 2 88 = n2 - 3n n2 - 3n - 88 = 0 (n - 11)(n + 8) = 0 n = 11 or -8 Since the number of sides can't be negative, the number of sides is 11, a hendecagon (or undecagon)
How many diagonals can be drawn from one vertex of a polygon of 50 sides?
47 sides. Take a vertex of an n-sided polygon. There are n-1 other vertices. It is already joined to its 2 neighbours, leaving n-3 other vertices not connected to it. Thus n-3 diagonals can be drawn in from each vertex. For n=50, n-3 = 50-3 = 47 diagonals can be drawn from each vertex. The total number of diagonals in an n-sided polygon would imply n-3 diagonals from each of the n vertices giving n(n-3). However, the diagonal from vertex A to C would be counted twice, once for vertex A and again for vertex C, thus there are half this number of diagonals, namely: number of diagonals in an n-sided polygon = n(n-3)/2.
A polygon has 20 diagonals how many sides does it have?
It is an octagon with 8 sides. Let n be the number of sides = number of vertices. Then each vertex can be connected to n-3 other vertices, giving n(n -3) connections. However, this counts each vertex pair connection twice, so there are: n(n - 3) / 2 diagonals. Thus if there are 20 diagonals: n(n - 3) / 2 = 20 n2 - 3n = 40 n2 - 3n - 40 = 0 (n + 5)(n - 8) = 0 so n = -5 or 8. As a polygon cannot have a negative number of sides, n = 8.
How do you find diagonals in a polygon?
Suppose a polygon has n vertices (and sides). From each vertex, a diagonal can be drawn to all vertices, excluding itself and the two adjacent vertices. So n-3 diagonals can be drawn from each vertex. Multiplying by the full complement of n vertices gives n(n-3). However, as things stand we have counted each diagonal twice: once at both ends. Dividing by two gives the actual number of diagonals. number of diagonals = n(n-3)/2
What is a polygon that has twice as many sides as another polygon you have fewer than 8 sides what are you?
hexagon
What polygon has twice as many sides as another polygon but have fewer than 8 sides?
A hexagon and a triangle.
How many diagonals will an -sided polygon have?
An n-sided polygon wil have n*(n-3)/2 diagonals.Consider joining each vertex to every other vertex. That gives potentially n-1 vertices. However, two of these will be sides of the polygon and so not diagonals. So each vertex gives rise to (n-3) diagonals. There are n vertices in the polygon and so that gives n*(n-3) diagonals. But, this method counts each diagonal twice: once from each end and so the correct answer is n*(n-3)/2.
What is the formula for the number of diagonals in a polygon?
Suppose a polygon has n vertices (and sides).From each vertex, a diagonal can be drawn to all vertices, excluding itself and the two adjacent vertices. So n-3 diagonals can be drawn from each vertex.Multiplying by the full complement of n vertices gives n(n-3). However, as things stand we have counted each diagonal twice: once at both ends. Dividing by two gives the actual number of diagonals.number of diagonals = n(n-3)/2
What is the name of a polygon that has twice as many sides as a quadrilateral?
octagon-8 sides
What is a polygon that has twice as many sides as another polygon that has fewer than 8 sides?
a hexagon ( triangle has 3 and a hexagon has 6)
