myhelper: RS Aggarwal solution class 8 chapter 14 Polygons Exercise 14A

Exercise 14A

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Q1 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Question 1:

Find the measure of each exterior angle of a regular
(i) pentagon
(ii) hexagon
(iii) heptagon
(iv) decagon
(v) polygon of 15 sides.

Answer 1:

Exterior angle of an n-sided polygon = ${(\frac{360}{n})}^{o}$
(i) For a pentagon: $n = 5 ∴ (\frac{360}{n}) = (\frac{360}{5}) = 72^{o}$

(ii) For a hexagon: $n = 6 ∴ (\frac{360}{n}) = (\frac{360}{6}) = 60^{o}$

(iii) For a heptagon: $n = 7 ∴ (\frac{360}{n}) = (\frac{360}{7}) = 51 . 43^{o}$

(iv) For a decagon: $n = 10 ∴ (\frac{360}{n}) = (\frac{360}{10}) = 36^{o}$

(v) For a polygon of 15 sides: $n = 15 ∴ (\frac{360}{n}) = (\frac{360}{15}) = 24^{o}$

Q2 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Question 2:

Is it possible to have a regular polygon each of whose exterior angles is 50°?

Answer 2:

Each exterior angle of an n-sided polygon = ${(\frac{360}{n})}^{o}$
If the exterior angle is 50°, then:

$\frac{360}{n} = 50 \Rightarrow n = 7.2$

Since n is not an integer, we cannot have a polygon with each exterior angle equal to 50°.

Q3 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Question 3:

Find the measure of each interior angle of a regular polygon having
(i) 10 sides
(ii) 15 sides.

Answer 3:

For a regular polygon with n sides:
$Each interior angle = 180 - \{Each exterior angle\} = 180 - (\frac{360}{n})$

(i) For a polygon with 10 sides:
$Each exterior angle = \frac{360}{10} = 36^{o} \Rightarrow Each interior angle = 180 - 36 = 144^{o}$

(ii) For a polygon with 15 sides:
$Each exterior angle = \frac{360}{15} = 24^{o} \Rightarrow Each interior angle = 180 - 24 = 156^{o}$

Q4 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Question 4:

Is it possible to have a regular polygon each of whose interior angles is 100°?

Answer 4:

Each interior angle of a regular polygon having n sides = $180 - (\frac{360}{n}) = \frac{180 n - 360}{n}$

If each interior angle of the polygon is 100°, then:

$100 = \frac{180 n - 360}{n} \Rightarrow 100 n = 180 n - 360 \Rightarrow 180 n - 100 n = 360 \Rightarrow 80 n = 360 \Rightarrow n = \frac{360}{80} = 4.5$

Since n is not an integer, it is not possible to have a regular polygon with each interior angle equal to 100°.

Q5 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Question 5:

What is the sum of all interior angles of a regular
(i) pentagon
(ii) hexagon
(iii) nonagon
(iv) polygon of 12 sides?

Answer 5:

Sum of the interior angles of an n-sided polygon = $(n - 2) \times 180 °$

(i) For a pentagon:
$n = 5 ∴ (n - 2) \times 180 ° = (5 - 2) \times 180 ° = 3 \times 180 ° = 540 °$

(ii) For a hexagon:
$n = 6 ∴ (n - 2) \times 180 ° = (6 - 2) \times 180 ° = 4 \times 180 ° = 720 °$

(iii) For a nonagon:
$n = 9 ∴ (n - 2) \times 180 ° = (9 - 2) \times 180 ° = 7 \times 180 ° = 1260 °$

(iv) For a polygon of 12 sides:
$n = 12 ∴ (n - 2) \times 180 ° = (12 - 2) \times 180 ° = 10 \times 180 ° = 1800 °$

Q6 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Question 6:

What is the number of diagonals in a
(i) heptagon
(ii) octagon
(iii) polygon of 12 sides?

Answer 6:

Number of diagonal in an n-sided polygon = $\frac{n (n - 3)}{2}$
(i) For a heptagon:

$n = 7 \Rightarrow \frac{n (n - 3)}{2} = \frac{7 (7 - 3)}{2} = \frac{28}{2} = 14$

(ii) For an octagon:

$n = 8 \Rightarrow \frac{n (n - 3)}{2} = \frac{8 (8 - 3)}{2} = \frac{40}{2} = 20$

(iii) For a 12-sided polygon:

$n = 12 \Rightarrow \frac{n (n - 3)}{2} = \frac{12 (12 - 3)}{2} = \frac{108}{2} = 54$

Q7 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Question 7:

Find the number of sides of a regular polygon whose each exterior angle measures:
(i) 40°
(ii) 36°
(iii) 72°
(iv) 30°

Answer 7:

Sum of all the exterior angles of a regular polygon is $360^{o}$ .

(i)
$Each exterior angle = 40^{o} Number of sides of the regular polygon = \frac{360}{40} = 9$

(ii)
$Each exterior angle = 36^{o} Number of sides of the regular polygon = \frac{360}{36} = 10$

(iii)
$Each exterior angle = 72^{o} Number of sides of the regular polygon = \frac{360}{72} = 5$

(iv)
$Each exterior angle = 30^{o} Number of sides of the regular polygon = \frac{360}{30} = 12$

Q8 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Question 8:

In the given figure, find the angle measure x.

Answer 8:

Sum of all the interior angles of an n-sided polygon = $(n - 2) \times 180 °$

$m ∠ A D C = 180 - 50 = 130^{o} m ∠ D A B = 180 - 115 = 65^{o} m ∠ B C D = 180 - 90 = 90^{o} m ∠ A D C + m ∠ D A B + m ∠ B C D + m ∠ A B C = (n - 2) \times 180 ° = (4 - 2) \times 180 ° = 2 \times 180 ° = 360 ° \Rightarrow m ∠ A D C + m ∠ D A B + m ∠ B C D + m ∠ A B C = 360 ° \Rightarrow 130^{o} + 65^{o} + 90^{o} + m ∠ A B C = 360 ° \Rightarrow 285^{o} + m ∠ A B C = 360^{o} \Rightarrow m ∠ A B C = 75^{o} \Rightarrow m ∠ C B F = 180 - 75 = 105^{o}$
∴ x = 105

Q9 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Question 9:

Find the angle measure x in the given figure.

Answer 9:

For a regular n-sided polygon:
Each interior angle = $180 - (\frac{360}{n})$
In the given figure:
$n = 5 x ° = 180 - \frac{360}{5} = 180 - 72 = 108^{o}$
∴ x = 108

RS Aggarwal solution class 8 chapter 14 Polygons Exercise 14A

Exercise 14A

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Q1 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Q2 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Q3 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Q5 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Q6 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Q7 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Q8 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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Q9 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons

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