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What are the interior angles of a 40 sided polygon?
The interior angles of a polygon can be calculated using the formula ((n - 2) \times 180) degrees, where (n) is the number of sides. For a 40-sided polygon, this would be ((40 - 2) \times 180 = 38 \times 180 = 6840) degrees. Therefore, the sum of the interior angles of a 40-sided polygon is 6840 degrees. Each interior angle in a regular 40-sided polygon would then be (6840 \div 40 = 171) degrees.
What is the sum of interior angles of a 40 sided polygon?
180*(40-2) = 180*38 = 6840 degrees.
What is the total measure of the angles of a forty sided polygon?
Exterior angles add up to 360 degrees Interior angles add up to 6840 degrees
Does a 40 sided polygon have 6840 diagonals?
The sum of the angles of a regular n-sided polygon is equal to (n - 2) x 180 degrees.Therefore, the sum of the angles of a regular 40-sided polygon (tetradecagon) is equal to (40 - 2) x 180 = 6840 degrees. Therefore, 6840 refers to the sum of the degrees, not the number of diagonals.The number of diagonals of an n-sided polygon is given by n(n-3)/2. So a 40-sided polygon has 40*37/2=740 diagonals.
What is the sum of the interior of a 40 gon?
6840 degrees
How many sides on a polygon with the sum of the interior angles is 6840?
To find the number of sides on a polygon with a sum of interior angles of 6840 degrees, we can use the formula: (n-2) * 180 = 6840, where n represents the number of sides. Solving for n, we get n = (6840 / 180) + 2 = 38. Therefore, the polygon has 38 sides.
If the sum of the interior angles of a polygon is 6840 how many sides does it have?
sum interior angles = (number of side - 2) × 180° → number of sides = sum interior angles ÷ 180° + 2 = 6840° ÷ 180° + 2 = 40 It has 40 sides.
What are the sum of the interior angles of a polygon with 40 sides?
The sum of the interior angles of a polygon can be calculated using the formula: [ \text{Sum of interior angles} = (n - 2) \times 180^\circ ] where (n) is the number of sides of the polygon. For a polygon with 40 sides ((n = 40)): [ \text{Sum of interior angles} = (40 - 2) \times 180^\circ = 38 \times 180^\circ = 6840^\circ ] Thus, the sum of the interior angles of a polygon with 40 sides is **6840 degrees Read more....tinyurl com/22enuvst
What is the sum of the interior angles of a polygon with 40 sides?
The sum of the interior angles of any regular polygon of n sides is equal to 180(n - 2) degrees. 6840 degrees
What are the interior angles of a 40 sided polygon?
The interior angles of a polygon can be calculated using the formula ((n - 2) \times 180) degrees, where (n) is the number of sides. For a 40-sided polygon, this would be ((40 - 2) \times 180 = 38 \times 180 = 6840) degrees. Therefore, the sum of the interior angles of a 40-sided polygon is 6840 degrees. Each interior angle in a regular 40-sided polygon would then be (6840 \div 40 = 171) degrees.
What is the interior of a 40 side polygon?
The interior angles of a 40 sided polygon add up to 6840 degrees
What is the sum of interior angles of a 40 sided polygon?
180*(40-2) = 180*38 = 6840 degrees.
What is the total measure of the angles of a forty sided polygon?
Exterior angles add up to 360 degrees Interior angles add up to 6840 degrees
What is the number of sides of a regular polygon if one angle measures 171 degrees?
The formula to find the sum of the angles of a polygon is: 180(n - 2) Since we know that one angle of this regular polygon is 171 degrees, we can find what kind of polygon it will be, using the formula: 180(n - 2) = 171n 180n - 360 = 171n 180n - 171n = 360 9n = 360 n = 360/9 n = 40 Thus the polygon is a 40-sided polygon. Check: 180(40 - 2) = 180(38) = 6840, which is the sum of the angles of the polygon. 6840/40 = 171 degrees, which is the measure of one of the angles of that polygon .
Interior angles of a 40-gon?
The sum of the interior angles of an n-gon is (n-2)*180 degrees. So, for a 40-gon, the sum of the interior angles would be 38*180 = 6840 degrees. If the 40-gon was not regular that is as far as you could go. But if it was a regular n-gon, then all its interior angles are equal, and each would be of 6840/40 = 171 degrees.
Does a 40 sided polygon have 6840 diagonals?
The sum of the angles of a regular n-sided polygon is equal to (n - 2) x 180 degrees.Therefore, the sum of the angles of a regular 40-sided polygon (tetradecagon) is equal to (40 - 2) x 180 = 6840 degrees. Therefore, 6840 refers to the sum of the degrees, not the number of diagonals.The number of diagonals of an n-sided polygon is given by n(n-3)/2. So a 40-sided polygon has 40*37/2=740 diagonals.
What is the sum of the interior of a 40 gon?
6840 degrees
