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Q: What is the perimeter of a regular polygon when each side is 4 cm and has 90 diagonals showing work?

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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.

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

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

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

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

Area of the rhombus: 0.5*7.5*10 = 37.5 square cm Perimeter using Pythagoras: 4*square root of (3.75^2 plus 5^2) = 25 cm

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

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

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.

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

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.

Perimeter = 29 cm so each side is 7.25 cm. The triangle formed by the diagonal and two sides has sides of 7.25, 7.25 and 11.8 cm so, using Heron's formula, its area is 24.9 square cm. Therefore, the area of the rhombus is twice that = 49.7 square cm.

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