Short Questions:
- Find the size of interior angle of each angle of the following regular polygons in degrees and grades:
- Pentagon
The interior angle of a regular pentagon is
\(\text{Interior angle} = \frac{(n-2) \cdot 180^\circ}{n}\)
Here, \( n = 5 \), therefore
\(\text{Interior angle} = \frac{(5-2) \cdot 180^\circ}{5} = \frac{3 \cdot 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ\)
- Hexagon
The interior angle of a regular hexagon is
\(\text{Interior angle} = \frac{(n-2) \cdot 180^\circ}{n}\)
Here, \( n = 6 \), therefore
\(\text{Interior angle} = \frac{(6-2) \cdot 180^\circ}{6} = \frac{4 \cdot 180^\circ}{6} = \frac{720^\circ}{6} = 120^\circ\)
- Heptagon
The interior angle of a regular heptagon is
\(\text{Interior angle} = \frac{(n-2) \cdot 180^\circ}{n}\)
Here, \( n = 7 \), therefore
\(\text{Interior angle} = \frac{(7-2) \cdot 180^\circ}{7} = \frac{5 \cdot 180^\circ}{7} = \frac{900^\circ}{7} \approx 128.57^\circ\)
- Octagon
The interior angle of a regular octagon is
\(\text{Interior angle} = \frac{(n-2) \cdot 180^\circ}{n}\)
Here, \( n = 8 \), therefore
\(\text{Interior angle} = \frac{(8-2) \cdot 180^\circ}{8} = \frac{6 \cdot 180^\circ}{8} = \frac{1080^\circ}{8} = 135^\circ\)
- Nonagon
The interior angle of a regular nonagon is
\(\text{Interior angle} = \frac{(n-2) \cdot 180^\circ}{n}\)
Here, \( n = 9 \), therefore
\(\text{Interior angle} = \frac{(9-2) \cdot 180^\circ}{9} = \frac{7 \cdot 180^\circ}{9} = \frac{1260^\circ}{9} = 140^\circ\)
- Decagon
The interior angle of a regular decagon is
\(\text{Interior angle} = \frac{(n-2) \cdot 180^\circ}{n}\)
Here, \( n = 10 \), therefore
\(\text{Interior angle} = \frac{(10-2) \cdot 180^\circ}{10} = \frac{8 \cdot 180^\circ}{10} = \frac{1440^\circ}{10} = 144^\circ\)
- Dodecagon
The interior angle of a regular dodecagon is
\(\text{Interior angle} = \frac{(n-2) \cdot 180^\circ}{n}\)
Here, \( n = 12 \), therefore
\(\text{Interior angle} = \frac{(12-2) \cdot 180^\circ}{12} = \frac{10 \cdot 180^\circ}{12} = \frac{1800^\circ}{12} = 150^\circ\)
- Pentagon
- Find the size of exterior angles of the following regular polygons in degrees and grades:
- Pentagon
The exterior angle of a regular pentagon is
\(\text{Exterior angle} = \frac{360^\circ}{n}\)
Here, \( n = 5 \), therefore
\(\text{Exterior angle} = \frac{360^\circ}{5} = 72^\circ\)
- Hexagon
The exterior angle of a regular hexagon is
\(\text{Exterior angle} = \frac{360^\circ}{n}\)
Here, \( n = 6 \), therefore
\(\text{Exterior angle} = \frac{360^\circ}{6} = 60^\circ\)
- Heptagon
The exterior angle of a regular heptagon is
\(\text{Exterior angle} = \frac{360^\circ}{n}\)
Here, \( n = 7 \), therefore
\(\text{Exterior angle} = \frac{360^\circ}{7} \approx 51.43^\circ\)
- Octagon
The exterior angle of a regular octagon is
\(\text{Exterior angle} = \frac{360^\circ}{n}\)
Here, \( n = 8 \), therefore
\(\text{Exterior angle} = \frac{360^\circ}{8} = 45^\circ\)
- Pentagon
- Find the sum of interior angles of the following polygons in degrees:
- Quadrilateral
The sum of the interior angles of a quadrilateral is
\(\text{Sum} = (n - 2) \cdot 180^\circ\)
Here, \( n = 4 \), therefore
\(\text{Sum} = (4 - 2) \cdot 180^\circ = 2 \cdot 180^\circ = 360^\circ\)
- Pentagon
The sum of the interior angles of a pentagon is
\(\text{Sum} = (n - 2) \cdot 180^\circ\)
Here, \( n = 5 \), therefore
\(\text{Sum} = (5 - 2) \cdot 180^\circ = 3 \cdot 180^\circ = 540^\circ\)
- Hexagon
The sum of the interior angles of a hexagon is
\(\text{Sum} = (n - 2) \cdot 180^\circ\)
Here, \( n = 6 \), therefore
\(\text{Sum} = (6 - 2) \cdot 180^\circ = 4 \cdot 180^\circ = 720^\circ\)
- Octagon
The sum of the interior angles of an octagon is
\(\text{Sum} = (n - 2) \cdot 180^\circ\)
Here, \( n = 8 \), therefore
\(\text{Sum} = (8 - 2) \cdot 180^\circ = 6 \cdot 180^\circ = 1080^\circ\)
- Quadrilateral
- From the given exterior angles of regular polygons find the number of sides:
- 60°
Given the exterior angle is \(60^\circ\), the number of sides \(n\) of the regular polygon is
Exterior angle\(= \frac{360^\circ}{n} \)
or\(\displaystyle n = \frac{360^\circ}{\text{Exterior angle}} = \frac{360^\circ}{60^\circ} = 6\)
Therefore, the polygon is a hexagon with 6 sides. - 120°
Given the exterior angle is \(120^\circ\), the number of sides \(n\) of the regular polygon is
Exterior angle\(= \frac{360^\circ}{n} \)
or\(\displaystyle n = \frac{360^\circ}{\text{Exterior angle}} = \frac{360^\circ}{120^\circ} = 3\)
Therefore, the polygon is a triangle with 3 sides. - 72°
Given the exterior angle is \(72^\circ\), the number of sides \(n\) of the regular polygon is
Exterior angle\(= \frac{360^\circ}{n} \)
or\(\displaystyle n = \frac{360^\circ}{\text{Exterior angle}} = \frac{360^\circ}{72^\circ} = 5\)
Therefore, the polygon is a pentagon with 5 sides. - 36°
Given the exterior angle is \(36^\circ\), the number of sides \(n\) of the regular polygon is
Exterior angle\(= \frac{360^\circ}{n} \)
or\(\displaystyle n = \frac{360^\circ}{\text{Exterior angle}} = \frac{360^\circ}{36^\circ} = 10\)
Therefore, the polygon is a decagon with 10 sides.
- 60°
- Find the (i) interior and (ii) exterior angle of the given regular polygons in degrees and radian:
- Octagon
In a regular octagon, we have (\(n = 8\))
(i) Interior angle:
In degrees: \(\displaystyle \frac{(8-2) \cdot 180^\circ}{8} = \frac{1080^\circ}{8} = 135^\circ\)
In radians: \(135^\circ \times \frac{\pi}{180^\circ} = \frac{3\pi}{4} \text{ rad}\)
(ii) Exterior angle:
In degrees: \(\displaystyle \frac{360^\circ}{8} = 45^\circ\)
In radians: \(45^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{4} \text{ rad}\)
- Hexagon
In a regular hexagon, we have (\(n = 6\))
(i) Interior angle:
In degrees: \(\displaystyle \frac{(6-2) \cdot 180^\circ}{6} = \frac{720^\circ}{6} = 120^\circ\)
In radians: \(120^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{3} \text{ rad}\)
(ii) Exterior angle:
In degrees: \(\displaystyle \frac{360^\circ}{6} = 60^\circ\)
In radians: \(60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3} \text{ rad}\)
- Pentagon
For a regular pentagon, we have (\(n = 5\))
(i) Interior angle:
In degrees: \(\displaystyle \frac{(5-2) \cdot 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ\)
In radians: \(108^\circ \times \frac{\pi}{180^\circ} = \frac{3\pi}{5} \text{ rad}\)
(ii) Exterior angle:
In degrees: \(\displaystyle \frac{360^\circ}{5} = 72^\circ\)
In radians: \(72^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{5} \text{ rad}\)
- Octagon
Long Questions:
- The angles of a pentagon are in the ratio 2 : 3 : 4 : 5 : 6. Find the angles in
- degrees
- grades.
The sum of interior angles of a pentagon is:
\((5 - 2) \times 180^\circ = 540^\circ\)
Let the angles be \(2x, 3x, 4x, 5x, 6x\).
Then:
\(2x + 3x + 4x + 5x + 6x = 20x = 540^\circ\)
\(\Rightarrow x = \dfrac{540^\circ}{20} = 27^\circ\)
(a) Angles in degrees:
\(2x = 54^\circ\)
\(3x = 81^\circ\)
\(4x = 108^\circ\)
\(5x = 135^\circ\)
\(6x = 162^\circ\)
(b) Angles in grades (1° = \(\frac{10}{9}\) grades):
\(54^\circ = 54 \times \dfrac{10}{9} = 60^g\)
\(81^\circ = 81 \times \dfrac{10}{9} = 90^g\)
\(108^\circ = 108 \times \dfrac{10}{9} = 120^g\)
\(135^\circ = 135 \times \dfrac{10}{9} = 150^g\)
\(162^\circ = 162 \times \dfrac{10}{9} = 180^g\)
- The angles of a hexagon are in the ratio of 1 : 2 : 3 : 4 : 5 : 5. Find the angles in
- degrees
- grade
The sum of interior angles of a hexagon is:
\((6 - 2) \times 180^\circ = 720^\circ\)
Let the angles be \(x, 2x, 3x, 4x, 5x, 5x\).
Sum of ratios: \(1x + 2x + 3x + 4x + 5x + 5x = 20x\)
So:
\(20x = 720^\circ \Rightarrow x = \dfrac{720^\circ}{20} = 36^\circ\)
(a) Angles in degrees:
\(1x = 36^\circ\)
\(2x = 72^\circ\)
\(3x = 108^\circ\)
\(4x = 144^\circ\)
\(5x = 180^\circ\)
\(5x = 180^\circ\)
(b) Angles in grades (since \(1^\circ = \dfrac{10}{9}^g\)):
\(36^\circ = 40^g\)
\(72^\circ = 80^g\)
\(108^\circ = 120^g\)
\(144^\circ = 160^g\)
\(180^\circ = 200^g\)
\(180^\circ = 200^g\)
- The angles of an octagon are in the ratio of 1 : 2 : 3 : 2 : 2 : 5 : 6 : 6. Find the angles in
- degrees
- radian.
The sum of interior angles of an octagon is:
\((8 - 2) \times 180^\circ = 1080^\circ\)
Let the angles be \(1x , 2x , 3x , 2x , 2x , 5x, 6x , 6 x\).
Sum of ratios: \(1x + 2x + 3x + 2x + 2x + 5x + 6x + 6 x= 27x\)
Then
\(27x = 1080^\circ \Rightarrow x = \dfrac{1080^\circ}{27} = 40^\circ\)
(a) Angles in degrees:
\(1x =1 \times 40= 40^\circ\)
\(2x =2 \times 40= 80^\circ\)
\(3x =3 \times 40= 120^\circ\)
\(2x =2 \times 40= 80^\circ\)
\(2x = 2 \times 40= 80^\circ\)
\(5x = 5 \times 40= 200^\circ\)
\(6x = 6 \times 40= 240^\circ\)
\(6x = 6 \times 40= 240^\circ\)
(b) Angles in radians (using \(\theta_{\text{rad}} = \theta_{\text{deg}} \cdot \frac{\pi}{180}\)):
\(40^\circ = 40 \times \dfrac{\pi}{180} = \dfrac{2\pi}{9}\)
\(80^\circ = 80 \times \dfrac{\pi}{180} = \dfrac{4\pi}{9}\)
\(120^\circ = 120 \times \dfrac{\pi}{180} = \dfrac{2\pi}{3}\)
\(80^\circ = 80 \times \dfrac{\pi}{180} = \dfrac{4\pi}{9}\)
\(80^\circ = 80 \times \dfrac{\pi}{180} = \dfrac{4\pi}{9}\)
\(200^\circ = 200 \times \dfrac{\pi}{180} = \dfrac{10\pi}{9}\)
\(240^\circ = 240 \times \dfrac{\pi}{180} = \dfrac{4\pi}{3}\)
\(240^\circ = 240 \times \dfrac{\pi}{180} = \dfrac{4\pi}{3}\)
- The ratio of interior and exterior angles of a regular polygon is 2 : 1. Find the number of sides of the polygon. Also express the angles in degrees and radian.
Given: Interior : Exterior = 2 : 1.
Let interior angle = \(2x\), exterior angle = \(x\).
Since they are supplementary:
\(2x + x = 180^\circ \Rightarrow 3x = 180^\circ \Rightarrow x = 60^\circ\)
Therefore:
- Exterior angle = \(60^\circ = 60 \times \dfrac{\pi}{180} = \dfrac{\pi}{3}\) rad
- Interior angle = \(120^\circ = 120 \times \dfrac{\pi}{180} = \dfrac{2\pi}{3}\) rad
Number of sides:
\(n = \dfrac{360^\circ}{\text{Exterior angle}} = \dfrac{360^\circ}{60^\circ} = 6\)
Hence, the polygon is a regular hexagon with 6 sides. - The ratio of interior and exterior angles of a regular polygon is 7 : 2. Find the number of sides. Also express the angles in degrees and radian.
Given: Interior : Exterior = 7 : 2.
Let interior angle = \(7x\), exterior angle = \(2x\).
Since they are supplementary:
\(7x + 2x = 180^\circ \Rightarrow 9x = 180^\circ \Rightarrow x = 20^\circ\)
Therefore:
- Interior angle = \(7x = 140^\circ = 140 \times \dfrac{\pi}{180} = \dfrac{7\pi}{9}\) rad
- Exterior angle = \(2x = 40^\circ = 40 \times \dfrac{\pi}{180} = \dfrac{2\pi}{9}\) rad
Number of sides:
\(n = \dfrac{360^\circ}{\text{Exterior angle}} = \dfrac{360^\circ}{40^\circ} = 9\)
Hence, the polygon is a regular nonagon with 9 sides. - Draw a regular octagon of side 3 cm. Find its interior and exterior angles in degrees, grades and radian.
In a regular octagon, we have \(n=8\)
Interior angle is
In degrees: \(\dfrac{(n-2)\times 180}{n}=\dfrac{(8-2)\times 180}{6}=\dfrac{(6)\times 180}{8}=135^\circ\)
In grades: \(135^\circ \times \dfrac{10}{9} = 150^g\)
In radians: \(135^\circ \times \dfrac{\pi}{180} = \dfrac{3\pi}{4}\)
Exterior angle is
In degrees: \(\dfrac{360^0}{n}=\dfrac{360^0}{8}=45^\circ\)
In grades: \(45^\circ \times \dfrac{10}{9} = 50^g\)
In radians: \(45^\circ \times \dfrac{\pi}{180} = \dfrac{\pi}{4}\)
The construction process is below.- Draw a line segment \(AB=3cm\)
- Draw \(135^0\) at B such that \(\measuredangle ABC=135^0\)
- Draw \(135^0\) at C such that \(\measuredangle BCD=135^0\)
- Similarly, draw a regular octagon \(ABCDEFGH\).
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