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What is an interior angle of a regular polygon having 90 diagonals showing key stages of work?
Wiki User
∙ 11y ago
It must have 15 sides to comply with the formula:-
0.5*(152-(3*15)) = 90 diagonals
So: 360/15 = 24 degrees and 180-24 = 156 degrees
Therefore: each interior angle measures 156 degrees
Wiki User
∙ 11y ago
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What is the size of each individual interior angle of a polygon having 464 diagonals showing key stages of work?
Providing that it is a regular polygon then let its sides be x: So: 0.5*(x2-3x) = 464 diagonals Then: x2-3x-928 = 0 Solving the equation: x = 32 sides Total sum of interior angles: 30*180 = 5400 degrees Each interior angle: (5400+360)/180 = 168.75 degrees
What are the total diagonals in an irregular polygon whose interior angles add up to 1980 degrees showing all work?
Suppose the polygon has V vertices.Then sum of interior angles is (V - 2)*180 degrees = 1980 degrees => V - 2 = 1980/180 = 11 => V = 13 A polygon with V vertices has V*(V-3)/2 = 13*10/2 = 65 diagonals.
What are the sizes of each interior angle of a regular polygon that has 170 diagonals showing work?
1 Let the sides be n and use the formula: 0.5*(n2-3n) = diagonals 2 So: 0.5*(n2-3n) = 170 => which transposes to: n2-3n-340 = 0 3 Solving the above quadratic equation gives n a positive value of 20 4 So the polygon has 20 sides and (20-2)*180 = 3240 interior angles 5 Each interior angle measures: 3240/20 = 162 degrees
What is the measure of each interior angle of a regular polygon that has a total sum of 275 diagonals showing work with answer?
Let the number of sides be n and so:- If: 0.5*(n^2 -3n) = 275 Then: n^2 -3n -550 = 0 Solving the above quadratic equation: n has positive value of 25 Each interior angle: (25-2)*180/25 = 165.6 degrees
How many sides does an irregular polygon have when its interior angles add up to 5040 degrees showing work?
It is: (5040+360)/180 = 30 sides
What is the size of each individual interior angle of a polygon having 464 diagonals showing key stages of work?
Providing that it is a regular polygon then let its sides be x: So: 0.5*(x2-3x) = 464 diagonals Then: x2-3x-928 = 0 Solving the equation: x = 32 sides Total sum of interior angles: 30*180 = 5400 degrees Each interior angle: (5400+360)/180 = 168.75 degrees
What are the total diagonals in an irregular polygon whose interior angles add up to 1980 degrees showing all work?
Suppose the polygon has V vertices.Then sum of interior angles is (V - 2)*180 degrees = 1980 degrees => V - 2 = 1980/180 = 11 => V = 13 A polygon with V vertices has V*(V-3)/2 = 13*10/2 = 65 diagonals.
What are the sizes of each interior angle of a regular polygon that has 170 diagonals showing work?
1 Let the sides be n and use the formula: 0.5*(n2-3n) = diagonals 2 So: 0.5*(n2-3n) = 170 => which transposes to: n2-3n-340 = 0 3 Solving the above quadratic equation gives n a positive value of 20 4 So the polygon has 20 sides and (20-2)*180 = 3240 interior angles 5 Each interior angle measures: 3240/20 = 162 degrees
What is the measure of each interior angle of a regular polygon that has a total sum of 275 diagonals showing work with answer?
Let the number of sides be n and so:- If: 0.5*(n^2 -3n) = 275 Then: n^2 -3n -550 = 0 Solving the above quadratic equation: n has positive value of 25 Each interior angle: (25-2)*180/25 = 165.6 degrees
How many sides does an irregular polygon have when it has 230 diagonals showing work?
Let its sides be x and use the formula: 0.5*(x squared-3x) = 230 So: x squared-3x-460 = 0 Solving the quadratic equation gives x positive value of 23 Therefore the polygon has 23 sides irrespective of it being a regular or an irregular polygon. Check: 0.5*(23^2-(3*23)) = 230 diagonals
How many sides does an irregular polygon have if it has 252 diagonals showing work?
Let its sides be x and rearrange the diagonal formula into a quadratic equation:- So: 0.5(x^2-3x) = 252 Then: x^2-3n-504 = 0 Solving the quadratic equation: gives x a positive value of 24 Therefore the polygon has 24 sides irrespective of it being irregular or regular
How many sides does an irregular polygon have when its interior angles add up to 5040 degrees showing work?
It is: (5040+360)/180 = 30 sides
What is the total sum of its interior angles of a polygon that has 189 diagonals showing work?
Using the diagonal formula when n is number of sides :- If: 0.5*(n^2-3n) = 189 Then multiplying both sides by 2 and subtracting both sides by 2*189 So: n^2-3n-378 = 0 Solving the above quadratic equation gives n a positive value of 21 Sum of interior angles: (21-2)*180 = 3420 degrees
What is the perimeter of a regular polygon when each side is 4 cm and has 90 diagonals showing work?
Let the number of sides be x and use the diagonal formula:- If: 0.5*(x^2 -3x) = 90 then x^2 -3x = 90*2 If: x^2 -3x = 180 then x^2 -3x -180 = 0 Using the quadratic equation formula gives a positive value of 15 for x Therefore perimeter of the polygon: 15*4 = 60 cm
How many sides does a regular polygon have when it has 4752 diagonals showing work?
Consider a regular polygon with n sides (and n vertices). Select any vertex. This can be done in n ways. There is no line from that vertex to itself. The lines from the vertex to the immediate neighbour on either side is a side of the polygon and so a diagonal. The lines from that vertex to any one of the remaining n-3 vertices is a diagonal. So, the nuber of ways of selecting the two vertices that deefine a diaginal seem to be n*(n-3). However, this process counts each diagonal twice - once from each end. Therefore a regular polygon with n sides has n*(n-3)/2 diagonals. Now n*(n-3)/2 = 4752 So n*(n-3) = 9504 that is n2 - 3n - 9504 = 0 using the quadratic equation, n = [3 + sqrt(9 + 4*9504)]/2 = 99 sides/vertices. The negative square root in the quadratic formula gives a negative number of sides and that answer can be ignored.
What is the total number of degrees in a 16 sided polygon?
2520 to get this. You must get the number of sides(16) subtract it by 2, then multiply it by 180. Subtracting it by two is actually the number of triangles inside the polygon showing the possible number of all available non-intersecting diagonals. so in general. The number of non-intersecting triangles multiplied by 180 degrees.( which is the number of degrees in one triangle.
What is the perimeter of a rhombus when one of its diagonals is greater than the other diagonal by 5 cm with an area of 150 square cm showing key aspects of work?
Let the diagonals be x+5 and x:- If: 0.5*(x+5)*x = 150 sq cm Then: x2+5x-300 = 0 Solving the above by means of the quadratic equation formula: x = +15 Therefore: diagonals are 15 cm and 20 cm The rhombus has 4 interior right angle triangles each having an hypotenuse Dimensions of their sides: 7.5 and 10 cm Using Pythagoras' theorem: 7.52+102 = 156.25 Its square root: 12.5 cm Thus: 4*12.5 = 50 cm which is the perimeter of the rhombus Note: area of any quadrilateral whose diagonals are perpendicular is 0.5*product of their diagonals
